PART-A
1
Analytical Geometry in Three Dimensions
1.1
INTRODUCTION
In a plane, the position of a point is determined by two numbers x and y, obtained with
reference to two straight lines in the plane intersecting at right angles. The position of a
point (location) in space can be determined in terms of its perpendicular distances (known as
rectangular cartesian coordinates) from three mutually perpendicular planes (known as
coordinate planes). The lines of intersection of these three coordinate planes are known as
coordinate axes and their point of intersection, ‘the origin’.
The three dimensional analytic geometry is the study of the geometrical objects (regular
shapes or irregular shapes) in the three-dimensional Euclidean space by three cartesian
coordinates: The x-coordinate, the y-coordinate and the z-coordinate. Thus, the idea gathered
will be useful in the various branches of Engineering.
Definition 1 (Analytic Geometry). The geometry in which the position is represented analytically (by coordinates) and algebraic methods of reasoning are used for the
most part is known as analytic geometry.
Definition 2 (Solid Geometry). The branch of geometry which studies figures in
space (three dimensional space) whose plane sections are the figures studied in plane
elementary geometry, such as, cubes, spheres, polyhedrons and angles between planes is
called solid geometry.
1.2
CARTESIAN COORDINATES IN SPACE
Z
In the figure 1.1, we have
C
L
x = PL, y = PM, z = PN
M
(x, y, z)
P
The point P whose X-co-ordinate is = x, Y-co-ordinate
is = y, and Z-co-ordinate = z, is written P ≡ (x, y, z).
Y
O
The co-ordinate planes are the YOZ plane (or the X
B
plane), the ZOX plane (or the Y plane) and the XOY plane
N
(or the Z plane) the planes ZOX and XOY intersect along X A
OX or the x-axis; and so on.
Fig. 1.1
Thus, the co-ordinate axes are:
OX ≡ the x-axis, OY ≡ y-axis, OZ ≡ z-axis.
A plane divides space into two parts, and so, the three co-ordinate planes divide space
into 2.2.2 = 23 = 8 parts, called octants.
1
2
Engineering Mathematics – I
If P ≡ (x, y, z), then we have
x = LP = CM = OA = BN 


 y = MP = CL = OB = AN 
z = NP = AM = OC = BL 
Equations of co-ordinate planes and axes.
For any point on the YOZ plane, the perpendicular distance from the YOZ plane is
evidently equal to zero; that is, the x-co-ordinate is zero. Thus, x = 0 is the characteristic
property of all points on the YOZ plane. Hence, x = 0 is the equation of the YOZ plane.
Similarly y = 0 is the equation of ZOX; and z = 0 is the equation of the XOY plane.
The x-axis, or OX is the interesection of the planes ZOX and XOY ; i.e., of y = 0, and
z = 0. The co-ordinates of all points on the x-axis, satisfy the relations y = 0, z = 0. Hence, we
get,
Equations of the x -axis is : y = 0 = z 


Equations of the y -axis is : z = 0 = x 
Equations of the z -axis is : x = 0 = y 
For any point on the x-axis. only the x co-ordinate is present, since y = 0 and the z
co-ordinate is also zero. Hence we can write
any point on the x -axis ≡ ( x1, 0 , 0 ) 


any point on the y-axis ≡ ( 0 , y1 , 0 ) 
any point on the z -axis ≡ ( 0 , 0 , z1 ) 
All points on a plane parallel to the YOZ plane are at the same distance from that
plane. Hence, their x co-ordinates must be the same. So,
x = k , is a plane parallel to YOZ 


 y = k , is a plane parallel to ZOX 
z = k , is a plane parallel to XOY 
Similarly y = m, z = n ; represents all the points whose y and z co-ordinate remain
unchanged ; and only the x-co-ordinate changes. So, it gives a straight line parallel to the x-axis.
Thus y = m , z = n ; is a straight line|| x axis,


z = n , x = l ; is a straight line|| y axis ,


x = l , y = m ; is straight line|| z axis. 
For example, if P = (a, b, c), we can write down the coordinates of the gereral points in
the Fig. 1.1.
Evidently for points on the x-plane, x = 0 ; and for points on the x-axis, y = 0 = z.
Hence, we get,
O ≡ (0, 0; 0); A ≡ (a, 0, 0); B ≡ (0, b, 0); C ≡ (0, 0, c)
P ≡ (a, b, c); L ≡ (0, b, c); M ≡ (a, 0, c); N ≡ (a, b, 0)
(1) Write down the equations of the planes parallel to the co-ordinate planes, and
passing through the point P ≡ (2, 3, – 5).
Any plane parallel to the YOZ plane, is x = k; here P ≡ (2, 3, – 5) lies on the plane.
∴ 2 = k. Hence, the plane through P, parallel to YOZ is, x = 2. Similarly, plane through P
parallel to ZOX is, y = 3; and parallel to XOY is, z = – 5.
Dharm\C\N-ENGM\EMATH1.PM5
We can write down the equations of straight lines through P ≡ (2, 3, – 5), parallel to the
co-ordinate axes respectively are :
 ( i ) parallel to x axis : y = 3, z = − 5 


 ( ii ) parallel to y axis : z = − 5, x = 2 
( iiii ) parallel to z axis ; x = 2, y = 3 
1. Distance Formula
If P = (x1, y1, z1), Q = (x2, y2, z2) be any two points in space then the distance between
and B is given by
PQ =
( x2 − x1 )2 + ( y2 − y1 )2 + ( z2 − z1 )2
2. Division Formula
The coordinates of a point dividing the line joining
P(x1, y1, z1), Q(x2, y2, z2) in the ratio l : m is given by
 lx2 + mx1 ly2 + my1 lz2 + mz1 
,
,
R= 

l+m
l+m 
 l+m
 lx2 − mx1 ly2 − my1 lz2 − mz1 
,
,
R1 = 

l−m
l−m 
 l−m
(a) Coordinates of any point on the line segment PQ.
Let l : m = k : l or l/m = k in the internal division formula.
putting l = k m we have
(For internal division)
(For external division)
 kmx2 + mx1 kmy2 + my1 kmz2 + mz1 
,
,
R= 

km + m
km + m 
 km + m
 kx2 + x1 ky2 + y1 kz2 + z1 
,
,
R= 

k +1
k +1 
 k +1
i.e.,
WORKED EXAMPLES
1. Using the distance formula show that the points P(1, 2, 3), Q (3, 7, 7) and R (5, 12, 11) are
collinear.
Solution:
PQ2 = (x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2
= (3 – 1)2 + (7 – 2)2 + (7 – 3)2
= 45
\
PQ = 3 5
QR2 = (5 – 3)2 + (12 – 7)2 + (11 – 7)2 = 45
\
QR = 3 5
RP2 = (5 – 1)2 + (12 – 2)2 + (11 – 3)2 = 180.
RP = 6 5 .
Clearly,
PR = PQ + QR
So, P, Q, R are collinear.
Dharm\C\N-ENGM\EMATH1.PM5
PART-A
3
Analytical Geometry in Three Dimensions
4
Engineering Mathematics – I
2. Find the coordinates of the point which divides the line joining (2, –3, 1) and (3, 4, –5) in
the ratio 1 : 3.
Solution: Using the section formula,
(3, 4, – 5)
1
(2, – 3, 1)
3
A
1:3
B
(2, – 3, 1)
P
(3, 4, – 5)
Let P(x, y, z) be the required point,
then
x=
1⋅ 3 + 3 ⋅ 2 9
=
1+3
4
y=
1 ⋅ 4 + 3( − 3) − 5
=
1+3
4
and
z=
1( −5) + 3(1)
1
=−
1+3
2
9 − 5 −1
,
P=  ,
.
2 
4 4
3. Find the points of trisection of PQ where P = (3, – 6, 12), Q = (9, 3, 6).
Solution: Clearly, there are two points of intersection, R and S
Hence
where
and
PR : RQ = 1 : 2
PS : SQ = 2 : 1
∴
 (1) (9) + (2) (3) (1) (3) + 2( − 6) 1(6) + 2(12) 
,
,
R= 
 = (5, – 3, 10)
1+2
1+2
1+2 

 2 (9) + 1 (3) 2 (3) + 1 ( − 6) 2 (6) + 1 (12) 
,
,
S= 
 = (7, 0, 8).
2 +1
2 +1
 2 +1

Similar Problems
1. Find the coordinates of the point which divide internally and externally, the line
joining the point (a + b, a – b) to the point (a – b, a + b) in the ratio a : b.
2. Find the ratio in which the yz-plane divides the line joining the points A (–3, 4, 5) and
B(2, –1, –2) Also find its coordinates
3. Show that the points A(3, 4, 7), B = (5, 6, 2), C = (1, 2, 1), D = (–1, 0, 6) are the vertices
of the parallelogram.
1.3
DIRECTION COSINES AND DIRECTION RATIOS
Definition 1. If a , b , g are the direction angles made by a line with the positive direction of
x, y, z axes respectively, then l = cos a , m = cos b , and n = cos g are called direction cosines of
the line.
Direction cosines are not independent, when two of them are given, the third can be
found, except for sign, by use of the pythogorean relation,
cos2 a + cos2 b + cos2 g = 1.
Note: Direction cosines, (D.C’s) here referred to as actual D.C’s, denoted by (l : m : n)
Definition 2. Direction ratios (direction numbers), of a line in space: (D.R’s)
Any three numbers, not all zero, proportional to the direction cosines of the line, are
called direction ratios or direction numbers of the line.
Dharm\C\N-ENGM\EMATH1.PM5
If a line passes through the points (x1, y1, z1) and (x2, y2, z2), its direction numbers are
proportional to (or, D.r’s) are: (x2 – x1) : (y2 – y1) : (z2 – z1), and its direction cosines are
x2 − x1 y2 − y1 z2 − z1
,
,
D
D
D
when D =
( x 2 − x1 )2 + ( y2 − y1 )2 + ( z2 − z1 )2 , the distance between the points.
A Simple and Useful Result
Z
Let OP = r, if (l : m : n) are the actual D.C’s of OP, then,
from the figure:
x = r cos a = lr
y = r cos b = mr
z = r cos g = nr
Then the coordinates of P are (lr, mr, nr)
i.e.
P = (lr, mr, nr)
[VTU, Aug., 1999] [1]
If
P = (x, y, z)
Then the actual D.C’s of OP are:
x y z
 , ,  = (l ; m; n)
r r r
P(x, y, z)
r
O
a
A
X
Fig. 1.2
An important identity (relation) between the actual direction cosines (D.C′s)
(l : m : n).
Let OP = r = 1 unit
Z
Let AB be the given line whose D.C’s are (l : m : n).
Through the origin O = (0, 0, 0), draw a line OP parallel
B
P(l, m, n)
to AB. Hence, D.C’.s of OP are also (l : m : n)
: n)
1
(If the lines are parallel, then their D.C’s are equal).
(l : m
A
Now cut off OP = 1 unit. Then coordinates of P =
O
X
(l, m, n), (using [1], r = 1).
(0, 0, 0)
By distance formula:
Y
OP =
or
(l − o)2 + (m − o)2 + (n − o)2
Fig. 1.3
l2 + m2 + n2
Squaring both sides, we get
1=
l 2 + m2 + n2 = 1
[2]
Note: If a, b, c are D.R’s of a line, then a2 + b2 + c2 π 1
Other forms:
(a) We have
Then
⇒
Dharm\C\N-ENGM\EMATH1.PM5
x = r cos α
y = r cos β
z = r cos γ
r2 = (x2 + y2 + z2) = (r2 cos2α + r2 cos2β + r2 cos2γ )
= r2 (cos2α + cos2β + cos2γ )
cos2 α + cos2 β + cos2 γ = 1
[3]
PART-A
5
Analytical Geometry in Three Dimensions
6
Engineering Mathematics – I
(b) The identity [3] can also be written as
(1 – sin2 α ) + (1 – sin2 β ) + (1 – sin2 γ ) = 1
or
3 – (sin2 α + sin2 β + sin2 γ) = 1
⇒
sin2 α + sin2 β + sin2 γ = 2
[VTU, Mar., 2000] [4]
Rule to find the actual direction cosines of a line whose proportional D.C’.s (i.e. direction
ratios) are given:
If (a, b, c) be the D.R’s of a line, and if (l : m : n) be actual D.C’s of the line, then by an
algebraic principle, we have,
l2 + n2 + m2
m n
l
= =
=
=
b c
a
a2 + b2 + c2
l=
a
∑a
2
,m =
b
∑a
2
,n=
1
∑a
(Q l 2 + m2 + n2 = 1)
2
c
∑ a2
i.e. dividing the direction ratios a, b, c each by their Square root of sum of Squares viz.
∑ a 2 , to get the actual D.C’s
Q
(x2, y2, z2)
Projection of a Line
The join of two points p(x1, y1, z1) and
Q (x2, y2, z2) on a line with D.C’s are l : m : n is
equal to P¢Q¢ = l (x2 – x1) + m (y2 – y1) + n (z2 – z1).
\ Projection of PQ on a line is P¢Q¢ whose
D.C’s are (l : m : n) is l (x2 – x1) + m ( y2 – y1) + n
(z2 – z1) = P¢Q¢
P
(x1, y1, z1)
A
P¢
[VTU, Mar., 1999]
Z
1
L2
A (l1, m1, n1)
q
1
Y
O
X
= (l12 + m12 + n12 ) + (l 22 + m 22 + n 22 ) – 2(l1l2 + m1m2 + n1n2)
= 1 + 1 – 2 (l1l2 + m1m2 + n1n2)
Now from the triangle OAB, by cosine rule
OA 2 + OB 2 − AB 2
= 1 + 1 – [2 – 2 (l1l2 + m1m2 + n1n2)]
2 OA ⋅ OB
Dharm\C\N-ENGM\EMATH1.PM5
L1
(l2, m2, n2)
B
A O$ B = q .
cosq =
B
Fig. 1.4
Angle between Two Lines when Their D.Cs are
given
(i) If L 1, L 2 be the given lines and (l1 : m1 : n 1 ),
(l2 : m2 : n2) be their D.C’s. Let q be the angle between them,
through O, draw OA, OB ||el to L1 and L2 respectively,
then
To find an expression for the angle between the lines
whose direction cosines are (l1 : m1 : n1) and (l2 : m2 : n2)
Let OA = OB = 1 unit.
∴ Coordinates of A and B are (l1, m1, n1) and
(l2, m2, n2).
AB2 = (l2 – l1)2 + (m2 – m1)2 + (n2 – n1)2.
Q¢
(l : m : n)
Fig. 1.5
7
Analytical Geometry in Three Dimensions
[1]
(l1m2 − l2m1 )2 + (m2n1 − m1n2 )2 + (n2l1 − n1l2 )2
sin q =
[VTU, Mar., 1999]
Proof:
sin2 q = 1 – cos2 q
= 1 – [l1l2 + m1m2 + n1n2]2
(using [1])
= (l12 + m12 + n12 ) (l22 + m22 + n22 ) – (l1l2 + m1m2 + n1n2)2
= (l1m2 – m1l2)2 + (m1n2 – n1m2)2 + (n1l2 – l1n2)2
(Q l12 + m12 + n12 = 1 = l22 + m22 + n22 )
\
∑ ( l1 m2 − m1 l2 )2
sinq =
[2]
∑ ( l1 m2 − m1 l2 )2
sin θ
tanq =
=
l1 l2 + m1 m2 + n1 n2
cos θ
and
(iii) Condition of perpendicularity
If the two lines are ^r, then, q = 90°
\
cosq = cos 90° = 0
fi
l1l2 + m1m2 + n1n2 = 0
(iv) Condition of parallelism.
If the two lines are parallel, then, q = 0, so that sin q = 0
fi
(l1m2 – m1l2)2 + (m1n2 – n1m2)2 + (n1l2 – n2l1)2 = 0
l1
m1
U|
||
m
n
m n – n m = 0,
or
[3]
=
V|
m
n
n
l
||
n l – l n = 0,
or
=
n
l
W
Note: (i) If (a , b , c ) and (a , b , c ) are sets of number, then we have an identity
l1m2 – m1l2 = 0,
fi
1 2
1
1 2
1
or
l2
2
1 2
1
1
2
known as Lagrange’s identity given by:
2
=
m2
1
1
2
2
1
1
2
2
2
(a 12 + b 12 + c 12 ) (a 22 + b 22 + c 22 ) – (a1a2 + b1b2 + c1c2)2
= (a1b2 – a2b1)2 + (b1c2 – c1b2)2 + (c1a2 – a1c2)2
which is same as
Σ ( a1 b2 − a2 b1 )2
which is similar to the expression for sine in equation
[2] above.
(ii) The result for sine given in [2] above can be remembered as: write down the D.C’s
in two rows as:
m1 n1 l1 m1
R|
S| m
T
2
n2 l2
Dharm\C\N-ENGM\EMATH1.PM5
U|
V to get m n
m |W
1 2
2
– n1m2 etc. Square and add.
PART-A
fi
cos q = (l1l2 + m1m2 + n1n2)
Hence
q = cos–1 (l1l2 + m1m2 + n1n2)
(ii) Expression for sin q and tan q :
8
Engineering Mathematics – I
[1]
l1
fi
=
l2
m1
n1
=
m2
l12 + m12 + n12
=
n2
l22 + m22 + n22
=
1
1
fi
l1 = l2, m1 = m2, n1 = n2
(a) Find the angle between the two lines whose D.R’s are (a1 : b1 : c1) and (a2 : b2 : c2).
[VTU, Aug., 1999]
Let (l1 : m1 : n1) be the actual D.C’s of line whose D.R’s are (a1 : b1 : c1)
c1
b1
a1
\
l1 =
, m1 =
, n1 =
2
2
∑ a12
∑ a1
∑ a1
Let (l2 : m2 : n2) be the actual D.C’s of the line whose D.R’s are (a2 : b2 : c2),
b2
a2
c2
\
l2 =
, m2 =
, n2 =
2
∑ a2
∑ a22
∑ a22
Thus we have, from results [1], that
a1a2 + b1b2 + c1c2
cos q =
If the lines are ^r,
\
a12
+ b12 + c12
a22 + b22 + c22
q = 90°
cos 90° = 0 =
a1a2 + b1b2 + c1c2
∑ a12 ∑ a22
\
a1a2 + b1b2 + c1c2 = 0
If the lines are ||el,
\
l1 = l2, m1 = m2, n1 = n2
a1
or
∑ a12
\
a1
a2
c1
and
c2
\
Note:
a1
a2
a1
=
∑ a22
∑ a12
=
∑ a22
∑ a12
=
=
∑ a22
b1
b2
=
ly,
; |||
b1
b2
=
;
c1
c2
D.C’s of the x-axis = (1:0:0)
D.C’s of the y-axis = (0:1:0)
D.C’s of the z-axis = (0:0:1)
Dharm\C\N-ENGM\EMATH1.PM5
∑ a12
∑ a22
9
Analytical Geometry in Three Dimensions
PART-A
WORKED EXAMPLES
1. Find the coordinates of the foot of the ^r from
A = (1, 8, 4) to the line joining
B (0, – 11, 4) and C (2, – 3, 1)
(V.T.U.J/A, 2004)
Solution: Let D be the foot of the perpendicular from A on the line joining BC.
Let the point D divide BC in the ratio l : 1
 2λ − 3λ − 1 λ + 4 
,
,
P= 

λ +1 
 λ +1 λ +1
We find that the lines AD and BC are ^r.
Now, D.R’s of BC are:
\
(0 – 2) : (– 11 + 3) : (4 – 1)
– 2, – 8, 3
2, 8, – 3
i.e.
or,
and D.R’s of AD are
[1]
 2λ
  − 3 λ − 11
 λ +4

− 1 : 
− 8 : 
− 4

λ +1
  λ +1
  λ +1

λ − 1 − 11 λ − 19 − 3λ
,
,
λ +1
λ +1
λ +1
fi
fi
l –1, –11l –19, –3l
Since the lines are ^r, from [1] and [2], we get
[2]
2(l – 1) + 8(–11l – 19) – 3(–3l ) = 0
\
fi
l=–2
Hence the coordinates of D are (4, + 5, –2).
2. Find the actual direction cosines of the line joining the two points
P = (2, 3, –5) and Q = (5, 7, 7)
Solution: The D.R’s are equal to:
(5 – 2) : (7 – 3) : (7 – (– 5)) = (3 : 4 : 12)
The actual D.C’s are equal to
3
2
2
3 + 4 + 12
2
,
4
2
2
3 + 4 + 12
2
12
,
2
2
3 + 4 + 12
2
 3 , 4 , 12 
= 
.
 13 13 13 
3. Find the D.C’s of a st. line equally inclined to the three coordinate axes.
Solution: It is given that
a =b =g
\
cosa = cosb = cosg
We know that cos2a + cos2b + cos2g = 1
i.e.
Dharm\C\N-ENGM\EMATH1.PM5
3 cos2a = 1;
10
Engineering Mathematics – I
1
\
cosa = ±
\
cosa = cosb = cosg = ±
3
1
3
Hence, the D.C’s are
 1
1
1 
±
±
±
 = (1 : 1 : 1)
3
3
3

4. Find the D.C’s of the line perpendicular to the lines whose D.C’s are proportional to
(1, –1, 2) and (2, 1, –1)
Solution: If (l; m; n) be the actual D.C’s of the line, then since the line is ^r to the other two
lines, we have
and
l◊1 – 1◊m + 2◊n = 0
2l + 1 ◊ m – 1 ◊ n = 0
Solving, by Cross multiplications,
we get
fi
−m
l
n
,
,
1 − 2 −4 − 1 1 + 2
m n
l
= =
=
5
3
−1

fi D.C’s are  −

1
35
,
5
35
1
35
,
=
1
2
( −1 ) + 52 + 32
3 .

35 
5. Find the angle P of the triangle P Q$ R, where P = (2, 3, 5) Q = (3, 5, 2) and R = (5, 2, 3).
Solution: PQR forms a triangle. The angle P of the triangle is the angle between the lines
PQ and PR.
Now, D.R’s of PQ are:
(3 – 2) : (5 – 3) : (2 – 5) = (1 : 2 : – 3).
D.R’s of PR are:
(5 – 2) : (2 – 3) : (3 – 5) = (3 : – 1 : – 2)
\
cos P = ±
=
a1 a2 + b1 b2 + c1 c2
a12 + b12 + c12
a22 + b22 + c22
1⋅3 + 2 ( − 1 ) + ( − 3 ) ( − 2 )
[(1 + 4 + 9 ) ( 9 + 1 + 4 )]
=
3−2+6
14
14
=
7 1
=
14 2
\
P = 60°.
6. Find the angle between the two lines whose direction cosines are given by the equations.
l + m + n = 0; l2 + m2 – n2 = 0.
we have
l+ m+n=0
[1]
l2 + m2 – n2 = 0
From [1],
l = – m – n.
Then from [2], we get
[– m – n]2 + m2 – n2 = 0
fi
2m2 + 2mn = 0
Dharm\C\N-ENGM\EMATH1.PM5
[2]
fi
2m (m + n) = 0
11
Analytical Geometry in Three Dimensions
m=0
m = 0,
m=– n
and if
or
m = – n.
then l = – n
then l = 0.
PART-A
This gives
From [1], if
Hence, DR’s of the line
are
or
(– n, 0, n)
(–1, 0, 1)
\ The actual D.C’s are
 1
1 
,0,
 −

2
2 

Thus, q is given by
and (0, – n, n)
and (0, –1, 1)
and

1
1 
,
 0 , −

2
2 

 1   1  1


 + 
 
cos q =  − 1  (0) + (0)  −


2
2



 2
2

7. Find whether the lines whose direction cosines are given by

1
 =
or q = 60°.
2

2l – m + 2n = 0
and lm + mn + nl = 0 are at right angles.
Solution:
Given: m = 2l + 2n. and therefore the second equation becomes:
l(2l + 2n) + n(2l + 2n) + nl = 0
fi
2l2 + 5ln + 2n2 = 0
(2l + n) (l + 2n) = 0
fi
fi
or
l
n
=
1 −2
or
l
n
=
2 −1
fi
fi
l=–
n
or l = – 2n
2
m = (2) (1) + 2(– 2) = – 2
m = 2(2) + 2 (– 1) = 2
The DR’s of the lines are
and
Hence
(l1 : m1 : n1) = (1, – 2, – 2)
(l2 : m2 : n2) = (2, – 1, 2)
l1l2 + m1m2 + n1n2 = 0
fi
1 ◊ 2 + (– 2) (– 1) + (– 2) (2) = 0.
fi Lines are perpendicular.
8. Show that the lines whose D.C.s satisfy the relations l + m + 4n = 0 and mn + nl + lm = 0
are parallel.
[V.T.U. F/M 2005]
Solution:
l + m + 4n = 0
...[1]
From [1],
mn + nl + lm = 0
...[2]
From [1], l = – (m + 4n), substituting in [2]
mn – n (m + 4n) – (m + 4n) m = 0
i.e.,
mn – mn – 4n2 – m2 – 4mn = 0
⇒
m2 + 4mn + 4n2 = 0
or
m = – 2n
or
(m + 2n)2 = 0
i.e.,
m1 = – 2n1 and m2 = – 2n2
Dharm\C\N-ENGM\EMATH1.PM5
12
Engineering Mathematics – I
m1
or
n1
=
m2
=–2
n2
Also, we have from (1),
l1 + m1 + 4n1 = 0
l1
∴
=
l2
m1
or
m2
and
=
n1
n2
=–2
l2 + m2 + 4n2 = 0
− ( m1 + 4 n1 )
− ( m2 + 4 n2 )
.
But m1 = – 2n1 and m2 = – 2n2
Hence
l1
l2
m1
or
=
=
− ( − 2 n1 + 4 n1 )
− ( − 2 n2 + 4 n2 )
m2
,
n1
=
n2
l1
,
n1
=
⇒
n2
=
m1
m2
=
n1
n2
l1
=
l2
l2
⇒
the lines are parallel.
n1
n2 l1
l2 m1 m2
9. Show that the straight lines whose D.R’s are given by the equations
l + m + n = 0 and 2mn + 3nl – 5lm = 0 are at right angles.
we have
l+m+n=0
fi
l = – (m + n)
Also,
2mn + 3nl –5lm = 0
fi
2mn – 3mn – 3n2 + 5m2 + 5mn = 0.
fi
5m2 + 4mn – 3n2 = 0;
\
Let
and from [1],
m=
m1 =
− 4 n ± 16 n2 + 60 n2
10
−2+
19
n1
5
l1 = –
−3+
19
=
− 2 ± 19
− 4 ± 2 19
.n=
n
5
10
and
n1
[1]
m2 =
and
−2−
l2 = –
5
If ‘q ’ be the angle between two lines, then
19
5
n2
− 3 − 19
5
n2
cos q = l1l2 + m1m2 + n1n2
=
=
fi
q =
9 − 19
25
n1 n2
25
n1n2 +
4 − 19
25
n1n2 + n1n2
(–10 –15 + 25) = 0.
π
2
10. Two lines are connected by the relations l – 5m + 3n = 0, and 7l2 + 5m2 – 3n2 = 0. Find
D.C’s (l : m : n).
Solution:
Given:
l – 5m + 3n = 0
[1]
2
2
2
7l + 5m – 3n = 0
[2]
From [1],
l = 5m – 3n
Dharm\C\N-ENGM\EMATH1.PM5
13
Analytical Geometry in Three Dimensions
Substitute the value of l in [2], we get
n
m
=
= l (say)
3
2
fi
\
\
PART-A
fi
fi
fi
fi
when 3m –2n = 0
7(5m – 3n)2 + 5m2 – 3n2 = 0
180m2 – 210mn + 60n2 = 0
6m2 – 7mn + 2n2 = 0
(3m –2n) (2m – n) = 0
Either 3m – 2n = 0 or 2m – n = 0
m = 2l , n = 3l
From [1],
l – 10l + 9l = 0
fi
l= l
fi
fi
l
=l
−1
m n
l
=
=
2 3
1
fi D.R’s of one line are: (1 : 2 : 3)
\


Actual D.C’s are:  1 , 2 , 3 
 14
14
14 

When
2m – n = 0
m
n
=
= l (say)
fi
1
2
\
m=l
n = 2l
l – 5l + 6l = 0
\ From [1],
l=–l
fi
m n
l
=
=
1 2
−1
Hence D.R’s of 2nd line are: (–1 : 1 : 2)
\ The D.C’s of 2nd line are:
\
 −1 1

:
: 2 .

6
6 
 6
11. Show that the straight lines whose D.C’s are given by the equations:
ul + vm + wn = 0, al2 + bm2 + cn2 = 0
are parallel if
u2
v2
w2
+
+
= 0.
a
b
c
and ^r if u2 (b + c) + v2 (c + a) + w2 (a + b) = 0.
Dharm\C\N-ENGM\EMATH1.PM5
l
=l
1
14
Engineering Mathematics – I
Solution: D.C’s of the lines are given by
ul + vm + wn = 0
al2 + bm2 + cn2 = 0
Eliminating ‘n’ from (i) and (ii),
(i)
(ii)
 ul + vm 
by putting n = – 
 from (i) in equation (ii), we have
 w

2
 ul + vm 
al + bm + c 
 =0
 w

2 2
2 2
2 2
aw l + bw m + c(u l + v2m2 + 2uvlm) = 0
2
fi
or
2
l2(aw2 + cu2) + 2uvclm + m2 (bw2 + cv2) = 0
Now, dividing both sides by m2, we have
l2
l
2
2
2
2
[3]
2 (aw + cu ) + 2uvc m + (bw + cv ) = 0
m
which is quadratic in (l/m).
If the lines are parallel, then their direction cosines are equal. Hence the roots of [3]
must be equal \ the lines will be parallel
if
4c2u2v2 = 4(aw2 + cu2) (bw2 + cv2)
if
c2u2v2 – (abw4 + acw2v2 + bcu2w2 + c2u2v2) = 0
if
– w2(abw2 + bcu2 + cav2) = 0
if
abw2 + bcu2 + acv2 = 0
if
u2
v2
w2
+
+
=0
a
b
c
Again, if (l1 : m1 : n1) and (l2 : m2 : n2) are
l1
l2
The D.C’s of two lines, then
are the roots of [3].
,
m1 m2
\ From [3],
Product of roots =
l1 l2
m1 m2
ab =
Q
or
if
l1 l2
=
=
bw2 + cv2
aw2 + cu2
c
a
m1 m2
=
n1 n 2
= k (say)
+ bu 2
bw 2 + cv2
aw 2 + cu2
(By symmetry, changing a, b, c and u, v, w in cyclic order).
If the lines are perpendicular, then, l1l2 + m1m2 + n1n2 = 0
av 2
k(bw2 + cv2) + k (aw2 + cu2) + k (av2 + bu2) = 0
or
if
Dharm\C\N-ENGM\EMATH1.PM5
bw2 + cv2 + aw2 + cu2 + av2s + bu2 = 0
(cancelling k)
(b + c) u2 + (c + a) v2 + (a + b) w2 = 0
12. Show that the straight lines whose direction cosines are given by al + bm + cn = 0 and fmn
f
g h
+ +
+ gnl + hlm = 0 are perpendicular if (i)
= 0 and parallel if af ± bg ± ch = 0.
a b c
Solution: We have
al + bm + cn = 0;
i.e.
al + bm
c
n=–
and
fmn + gnl + hlm = 0.
– fm ◊
fi
al + bm
al + bm
– gl ◊
+ hlm = 0
c
c
agl2 + lm (af + bg – ch) + bfm2 = 0
fi
ag ◊
fi
l1
Let
m1
and
l2
 l 
 
2 + (af + bg – ch)  m  + bf = 0
m
l2
[1]
be the roots of this equation.
m2
 l1  l2 
bf

 =
\ product of roots = 
,
m
m
ag
 1  2 
l1 l2
or
=
bf
l1 l2
|||ly
=
m1 m2
l1 l2
or
ag
n1 n2
f
a
l1 l2
or
f /a
h
f
c
a
the lines are perpendicular if (i) l1l2 + m1m2 + n1n2 = 0
i.e.
f
=
=
m1 m2
g
b
m1 m2
g/b
=
n1 n2
h/c
g
h
+
=0
a b c
Again, the lines are parallel
+
if
l1
l2
l1
taking first two,
l2
=
=
m1
m2
m1
m2
=
n1
n2
.
.
Hence, Eqn. (1) has equal roots
\
(af + bg – ch)2 = 4ag ◊ bf
af + bg – ch = ± 2 abgf
or
af + bg m 2 abgf = ch
or
\
af m
Dharm\C\N-ENGM\EMATH1.PM5
bg = ±
or
ch ;
or
( af m
af m
bg +
bg )2 = ( ch )2
ch = 0
PART-A
15
Analytical Geometry in Three Dimensions
16
Engineering Mathematics – I
13. Find the angle between two diagonals of a cube.
[VTU, Aug., 1999 ; VTU J/A, 2003 ; VTU, J/F, 2003, 2004]
Z
Solution: Take OP and MM′ be any two diagonals of
the cube, where M(a, 0, 0), P(a, a, a) and M′ (0, a, a).
The DC’s of OP and MM′ are
1 , 1 , 1 and – 1 , 1 , 1
3
3
3
3
3
3
If q is the angle between these diagonals OP
and MM¢ then
1  1 
1 . 1
1 . 1
1
+
= .
 −
 +
3 
3
3
3
3
3 3
14. If a straight line makes angle a , b , g and d with
the four diagonals of a cube, then show that
4
(i) cos2 a + cos2b + cos2g + cos2d = .
3
8
2
2
2
2
(ii) sin a + sin b + sin g + sin d =
3
cos q =
C(0, 0, a)
M¢(0, a, a)
P(a, a, a)
O
Y
A(a, 0, 0)
M(a, 0, 0)
X
Fig. 1.9
[VTU, Aug., 2001]
[VTU, Mar., 2001]
Solution: Let the origin be one of the corner of the cube and three edges OA, OB and OC be
along the coordinate axes. Let OA = OB = OC = a. Then
O = (0, 0, 0); P = (a, a, a)
A = (a, 0, 0); L = (0, a, a)
B = (0, a, 0); M = (a, 0, a)
C = (0, 0, a); N = (a, a, 0)
Let a , b , g , d be the angle between the diagonals, OP, AL, BM and CN respectively
with the given line having D.C’s (l : m : n), then
al + am + an
l+m+n
=
cos a =
3
a2 + a2 + a2
Z
C
M
P
L
O
X
A
N
B
Y
Fig. 1.10
|||ly, cos b =
Dharm\C\N-ENGM\EMATH1.PM5
−l+m+n
3
, cos g =
l+m−n
3
17
Analytical Geometry in Three Dimensions
l−m+n
3
we get, cos2a + cos2b + cos2 g + cos2d
, squaring and adding,
PART-A
cos d =
( l + m + n )2 ( − l + m + n )2 + ( l + m − n )2 + ( l − m + n )2
3
2
2
2
4 (l + m + n ) 4
= .
=
3
3
=
since l2 + m2 + n2 = 1.
(i) cos2 a + cos2b + cos2 g + cos2 d =
4
.
3
Also, (1 – sin2 a) + (1 – sin2 b) + [1 – sin2 g) + [1 – sin2 d ) =
4
3
4
.
3
8
fi
(sin2a + sin2 b + sin2 g + sin2 d ) =
, which proves (ii).
3
15. If the edges of a rectangular parallelopiped are of lengths a, b, c. Show that the angles
4 – (sin2 a + sin2 b + sin2 g + sin2 d ) =
 ± a 2 ± b2 ± c 2 
between four diagonals will be given by cos–1  2
2
2 

 a +b +c 
Solution: The diagonals are OP, AL, BM and CN respectively, then
O (0, 0, 0); P (a, b, c)
A (a, 0, 0); L (o, b, c)
B (o, b, o); M (a, o, c)
C (0, 0, c); N (a, b, o)
D.R’s of OP are: (a – o) : (b – o) : (c – o)
=a: b: c
D.R’s of AL are: (o – a) : (b – o) : (c – o)
=– a:b:c
D.R’s of BM are: (a – o) : (o – b) : (c – o)
= a:– b:c
D.R’s of CN are: (a – o) : (b – o) : (o – c)
= a:b:– c
The angle between OP and BM is q,
where
a ⋅ a + b( − b) + c( c )
cos q =
a 2 + b2 + c2
a 2 + b2 + c 2
=
Z
C
M
L
P
O
B
X
A
Y
N
Fig. 1.11
a 2 − b2 + c 2
( a2 + b2 + c2 )
Similarly, by taking other pair of diagonals AL and CN, we can prove the other
results, namely, cos q =
a 2 + b2 − c 2
( a2 + b2 + c2 )
Dharm\C\N-ENGM\EMATH1.PM5
and cos q =
− a 2 + b2 + c 2
( a2 + b2 + c2 )
.
18
Engineering Mathematics – I
All the results are consolidated in the single statement as:
cos q =
± a2 ± b2 ± c2
( a2 + b2 + c2 )
 ± a2 ± b2 ± c2 
q = cos–1  2
.
2
2 
 a + b + c 
16. Prove that the lines whose D.C’s are given by the relations
al + bm + cn = 0 and mn + nl + lm = 0 are
1 1 1
+ + =0
a b c
(i) Perpendicular if
and
(ii) Parallel if
a+ b+ c =0
Solution:
Let (l1 : m1 : n1) and (l2 : m2 : n2) be the D.C’s of the two lines.
Eliminating ‘n’ between the given equations, we get
 − ( al + bm) 
 − ( al + bm) 
m 
+l 
 + lm = 0
c
c




– alm – bm2 – al2 – blm + clm = 0
or
∏ m2, we have
2
 l 
 l 
a   + (a + b – c)   + b = 0
m
m
If roots are
[1]
l
l1
and 2 ,
m2
m1
 l   l 
b
then  1  ⋅  2  = product of the roots =
a
m
m
 1  2
fi
l1l2
mm
= 1 2
b
a
fi
l1l2
mm
nn
= 1 2 = 1 2
1 
1 
1 
a
b
c 
 
 
 
(By symmetry)
If the lines are perpendicular, then
l1l2 + m1m2 + n1n2 = 0
1 1 1
+ +
fi
= 0. If the lines are parallel, then their D.C’s must be same i.e., the
a b c
roots of eqn (i) must be equal.
\
or
or
(a + b + c)2 = 4ab,
(a + b) – c = ± 2
(a + b m 2 ab ) = c
or
(Q b2 = 4ac)
a b , (using equation [1])
or
( a ± b )2 = ( c )2
or
a+ b =± c
a + b + c = 0.
17. Show that the lines whose D.C’s are given by the equations
l + m + n = 0 and al2 + bm2 + cn2 = 0
Dharm\C\N-ENGM\EMATH1.PM5
19
Analytical Geometry in Three Dimensions
(i) Perpendicular if a + b + c = 0
1 1 1
+ + =0
a b c
Solution: We have,
l+m+n=0
al2 + bm2 + cn2 = 0
Eliminating n from (i) and (ii), we get
(ii) Parallel if
(i)
(ii)
al2 + bm2 + c (– l – m)2 = 0
(a + c) l2 + 2clm + (b + c) m2 = 0
i.e.
2
 l 
 l 
(a + c)   + 2c   + (b + c) = 0
m
m
which is quadratic in l/m. Let l1/m1 and l2/m2 be its roots.
i.e.
(iii)
l1l2
b+c
=
m1m2
a+c
Then
(iv)
|||ly by eliminating m from (i) and (ii) we can prove that
l1l2
b+c
=
n1n2
a+b
(v)
From (iv) and (v), we get
n1 n2
l1l2
mm
= 1 2 =
=l
b+c
a+c
a +b
(i) The lines whose D.C’s are
(l1 : m1 : n1) and [(l2 : m2 : n2) are ^r if l1l2 + m1m2 + n1n2 = 0.
i.e.
or
(say)
l (b + c) + l (a + c) + l (a + b) = 0
a + b + c = 0.
(ii) The lines are parallel if
m1
l1
n
=
= 1
m2
l2
n2
This will hold good if the roots of
(iii) are equal
fi
discriminant = 0
i.e.
ab + bc + ca = 0
fi
4c2 – 4(a + c) (b + c) = 0
1 1 1
+ + =0
a b c
is the condition for the two lines to be parallel.
18. If P (4, 2, 3) and Q (1, –3, 4), then find the projection of PQ on a line which makes 30°,
120° and 90° with coordinate axes.
The actual D.C’s of the line are:
(∏ abc), we get
cos 30°, cos 120°, cos 90° the D.C’s of the line are
Dharm\C\N-ENGM\EMATH1.PM5
3 −1
,
, 0.
2 2
PART-A
are
20
Engineering Mathematics – I
3 −1
,
, 0.
2 2
= l(x2 – x1) + m(y2 – y1) + n(z2 – z1)
The projection of PQ on the line whose D.C’s are
3
 −1
(1 – 4) + 
 (– 3 – 2) + 0(4 – 3)
2
 2 
=
5
1
−3 3
+
+0=
(5 – 3 3 )
2
2
2
19. If the actual D.C’s of two lines are connected by l – 3m + n = 0, l2 – 3m2 – 2n2 = 0. Find
the D.C’s.
=
Solution: Given:
l – 3m + n = 0
l2 – 3m2 – 2n2 = 0
∏ n and n2, and put
l
m
= p,
=q
n
n
m
l
–3◊
+1=0
n
n
we get
l2
m2
–2=0
n2
n2
fi
P – 3q + 1 = 0
and
p2 – 3q2 – 2 = 0
From the first, p = 3q – 1, Substituting in the second,
we get
–3◊
(3q – 1)2 – 3q2 – 2 = 0
6q2 – 6q – 1 = 0
or
6 ± 36 + 24
3 ± 15
–1=
6
2⋅6
Solving,
q=
But
p = 3q – 1
\
 3 ± 15 
9 ± 3 15 − 6
3 ± 3 15
p = 3 
=
 – 1 =
6
6
6


fi
l
3 ± 15
3 ± 3 15 m
=
,
=
n
n
6
6
(l : m : n) are proportional to 3 ± 3 15 , 3 ±
(3 ± 3 15)2 + (3 ± 15)2 + 62 =
\ Actual D.C’s are
But
3 ± 3 15
204 ± 24 15
,
15 , 6
204 ± 24 15
3 ± 15
204 ± 24 15
,
6
204 ± 24 15
20. Find the area of DPQR where P = (1, 0, 0), Q = (0, 2, 0), R = (0, 0, 3) using PQ ◊ PR sin
Q P$ R.
D.R’s of
D.R’s of
PQ = (0 – 1) : (2 – 0) ; (0 – 0) = – 1 : 2 : 0
PR = (0 – 1) : (0 – 0) : (3 – 0) = – 1 : 0 : 3
Dharm\C\N-ENGM\EMATH1.PM5
21
Analytical Geometry in Three Dimensions
1
=
Now,
sin q =
Also
PQ =
(0 − 1)2 + (2 − 0)2 + (0 − 0)2 =
1+4 = 5
PR =
(0 − 1)2 + (0 − 0)2 + (3 − 0)2 =
1 + 9 = 10
Area of
DPQR =
1+4+0 1+0+9
1 − cos2 θ =
1−
=
1+0+0
cos q =
5 10
50
1
7
.
=
50
50
1
PQ ◊ PR ◊ sin Q P$ R
2
1
7
7
50 7
⋅ ( 5)( 10)
=
= ⋅
Sq. units
2
2
50
50 2
21. Find the direction ratios of the bisectors of the angle between the lines whose direction
cosines are (l1 : m1 : n1) and (l2 : m2 : n2).
=
Solution: Let L1, L2 be the lines with D.C’s (l1 : m1 : n1) and (l2 : m2 : n2) respectively.
C!
B
A
C
q
1
O
1
A!
Fig. 1.12
Let OA, OB be lines thro’ the origin such that OA is in direction O + L1 and OB is in
direction of L2.
If q is the angle between L1 and L2 then A O$ B = q
Let OA = OB = 1 unit.
Then A = (l1, m1, n1), B = (l2, m2, n2)
Let A be the point on AO produced such that OA¢ = 1
then
A¢ = (– l1, – m1, – n1)
Now, if C and C¢ are the mid points of AB and A¢B then OC and OC¢ are along the
internal bisector and external bisector of the angle A O$ B = q.
Now,
 l + l m + m2 n1 + n2 
,
C= 1 2 , 1
2
2 
 2
 l − l m − m1 n2 − n1 
,
C1 =  2 1 , 2
2
2 
 2
\ D.R’s of the internal bisector is the D.R’s of the line joining O and C.
(l1 + l2, m1 + m2, n1 + n2)
Dharm\C\N-ENGM\EMATH1.PM5
PART-A
( − 1) ( − 1) + (2)(0) + (0)(3)
\
22
Engineering Mathematics – I
|||ly DR’s of external bisector are
(l2 – l1, m2 – m1, n2 – n1)
22. It A = (1, 3, 5), B = (6, 4, 3), C = (2, –1, 4) and D = (0, 1, 5) find the projection of AB on
CD.
Solution: The D.R’s of CD are (– 2, 2, 1)
[Using the formula (x2 – x1) : (y2 – y1) : (z2 – z1)]
Now, the D.Cs of CD are


2
1
−2
 − 2 2 1
,
,
, , 

 = 
4 + 4 +1
4 + 4 + 1 
 3 3 3
 4 + 4 + 1
\ the projection of AB on CD
=–
2
2
1
10 2 2
10
+ − =−
(6 – 1) +
(4 – 3) +
(3 – 5) = –
.
3
3
3
3
3 3
3
Projection: l(x2 – x1) + m[y2 – y1) + n[z2 – z1).
23. Can the numbers
your answer.
1
,
1
1
,
2
2 2 2
3
2 2
2
be the D.C’s of any directed line? Give reasons for
Solution: If (l : m : n) are D.C’s of any line, then l2 + m2 + n2 = 1
Here l =
1
1 ,n= 1
,m=
2 2
8 2
4 2
Since l2 + m2 + n2 =
1
2
(2 2)
+
1
2
(4 2)
+
1
(8 2)2
1
1
1
16 + 4 + 1
+
+
=
π 1.
8 32 128
128
So, the given numbers are not the D.C’s of the directed line.
24. Find the angle between the lines whose direction cosines are proportional to 1, 2, 1 and
2, –3, 6.
=
Solution: Have
a1 = 1, b1 = 2, c1 = 1
a2 = 2, b2 = –3, c2 = 6
If ‘q ’ be the required angle
\
cos q =
=
\
a1a2 + b1b2 + c1c2
a12
+ b12 + c12
1 × 2 + 2 ( − 3) + 1 × 6
12 + 22 + 12
 2 
q = cos–1 

7 2 
Dharm\C\N-ENGM\EMATH1.PM5
a22 + b22 + c22
22 + ( − 3)2 + 62
=
2−6+6
6
49
=
2
( 6 )7
=
2
7( 6 )
25. A line makes an angle of 45° with OX and 60° with OY. What angle does it make with
OZ?
Solution: If the line makes an angle g with the positive z-axis, Then since the line makes
angles 45°, 60° and g with the axes:,
D.C’s are : (cos 45° : cos 60° : cos g)
 1 1

: : cos γ 

 2 2

or
But we know l2 + m2 + n2 = 1
cos2g = 1 –
fi
1 1
+
+ cos2 g = 1
2 4
3 1
=
4 4
1
2
\
g = 60°;
or
120°,
26. What are the D.C’s of the line equally included to the axes? How many such lines are
there?
Solution:
Since
a = b = g;
(Given)
fi
cos a = cos b = cos g
or
l=m=n
Since l2 + m2 + n2 = 1
\
l2 + l2 + l2 = 1
cos2 g = ±
\
3l2 = 1
 ± 1 ± 1 ± 1
:
:
\ D.C’s are 

3
3
 3
Now consider
+
+
+
–
+
–
–
–
Q (l : m : n), (– l : –m : – n)
equal angle with the axis.
LM
MN
l=±
\
+
+
–
+
+
–
+
+
–
–
(or
–
+
(or
+
–
(or
–
–
(or
are same. So there can
27. If a straight line makes an angle of
3
+
+)
+
–)
–
+)
+
+)
different lines which make
π
with each of x-axis, y-axis, then what angle does
4
it make with the z-axis?
Solution: Let g be the angle it makes with z-axis, since
cos2 a + cos2 b + cos2 g = 1
π
π
\
cos2
+ cos2
+ cos2 g = 1
4
4
Dharm\C\N-ENGM\EMATH1.PM5
–
+
+
+
be 4
1
PART-A
23
Analytical Geometry in Three Dimensions
24
Engineering Mathematics – I
1 1
+
+ cos2g = 1
2 2
cos2 g = 0
cos g = 0 = cos p/2
g = p/2.
28. (l1 : m1 : n1), (l2 : m2 : n2) are the D.C’s of two mutually ^r lines, show that the D.C’s of the
line ^ to both of them are (m1n2 – n1m2), (n1l2 – n2l1), (l1m2 – m1l2).
fi
Solution: If (l : m : n) be the D.C’s of the line ^r two given lines, then
ll1 + mm1 + nn1 = 0
ll2 + mm2 + nn2 = 0
\
l
m
n
=
=
m1n2 − n1m2
n1l2 − l1n2 l1m2 − m1l2
=
Σl 2
Σ(m1n1 − n1m2 )2
(By the principle of algebra)
1
1
=
=1
sin θ sin 90o
where q = 90° is the angle between the given lines
=
l = n2m1 – n1m2,
m = n1l2 – l1n2
n = l1m2 – m1l2
i.e. the D.C’s are (m1n1 – n1m2), (n1l2 – n2l1), (l1m2 – m1l2).
29. Show that the pair of lines whose D.C’s satisfy the equations,
l + m + n = 0, 2l2 + 2m2 – n2 = 0 are parallel.
Solution: Let (l1 : m1 : n1), (l2 : m2 : n2) be D.C’s of the given lines then these satisfy the given
equations
l+m+n=0
[1]
2l2 + 2m2 – n2 = 0
[2]
Eliminating ‘n’ from these eqn, we get 2l2 + 2m2 – (l + m)2 = 0
2l2 + 2m2 – (l + m)2 = 0
or
l=m
or
(l – m)2 = 0
Thus, we have
l1 = m1, l2 = m2
\
so that
l1 m1
=
=1
l2 m2
Now using [1] and [3], we get
n1
− (l1 + m1 )
=
=1
n2
− (l2 + m2 )
Expressions [3] and [4], give
l1 = l2, m1 = m2, n1 = n2.
fi lines are parallel.
Dharm\C\N-ENGM\EMATH1.PM5
[3]
30. The coordinates of the angular points A, B, C, D of a tetrahedron are (–2, 1, 3) (3, –1, 2),
(2, 4, –1) and (1, 2, 3) respectively. Find the angle between the edges AC and BD.
Solution: The D.R’s of AC are (4 : 3 : – 4).

−

Actual D.C’s are
4
3
:
41
41
:
−4 

41 
3
1 
 −2
:
:


14
14 
 14
If q$ is the angle between AC and BD, then
and the Actual D.C’s of BD are
−8+9−4
−3
 −3 
θ = cos–1 

14 × 41
574
 574 
31. Find the angle between the straight line which intersection of the planes 2x + 2y – z + 15
= 0 = 4y + z + 29 and the line
x + 4 y −3 z +1
.
[VTU J/A, 2003]
=
=
4
1
−3
cos q =
=
;
⇒
Solution. Let l, m, n be the D.C’s of the line of intersection of the planes 2x + 2y – z + 15 = 0.
l
m n
=
=
3 −1 4
D.C’s of the second line are 4, – 3, 1. If θ is the angle between these two lines, then
4y + z + 29 = 0. Then
cos θ =
3(4) + ( − 1)( − 3) + (4)(1)
9 + 1 + 16 16 + 9 + 1
=
19
26
or
 19 
θ = cos–1   .
 26 
EXERCISES
π

 Ans : θ = 2 


2. A, B, C are the points (1, 4, 2), (– 2, 1, 2), (2, – 3, 4). Find the angles of the triangle ABC.
1. Find the angle between the lines whose D.R’s are (2 : 3 : 4) and (1 : – 2 : 1)

 2 
1
 Ans : 90° , cos−1
, cos−1 
 


3
 3 

3. Find the D.C’s of the line which is ^r to the lines with D.C’s proportional to (1 : –2 : –2) and

 2 −1 2 
, 
 Ans : Actual D.C’s are  ,
 3 3 3 

(O : 2 : 1)
4. If (l1 : m1 : n1), (l2 : m2 : n2) and (l3 : m3 : n3) are the D.C’s of 3 mutually ^r lines, then show
that the line whose D.C’s are l1 + l2 + l3 , m1 + m2 + m3 , and n1 + n2 + n3 makes equal angles
3
3
3
with them.
5. If (l1 : m1 : n1), (l2 : m2 : n2) be two directions inclined as an angle q, show that actual D.C’s of
the direction bisecting them are
1
1
1
(l + l2) sec q/2;
(m1 + m2) sec q/2;
(n1 + n2) sec q /2.
2 1
2
2
Dharm\C\N-ENGM\EMATH1.PM5
PART-A
25
Analytical Geometry in Three Dimensions
26
Engineering Mathematics – I
6. If P (1, 2, 3), Q (4, 5, 7), R (– 4, 3, – 6) and S (2, 9, 2).
7.
8.
9.
10.
S.T. PQ and RS are parallel.
If A (4, 7, –2), B (2, 0, 3), C(– 5, – 8, – 9), D (– 9, – 4, – 5). S.T. AB is ⊥r to CD.
Find the D.C’s of a st. line equally inclined to the coordinate axes.
If P(2, 3, 5), Q (3, 5, 2), R(5, 2, 3) are the vertices of a triangle find the angle P.
A straight line is inclined with the axes of y and z at angles of 45° θ and 60°. Find the
inclination of the line to the x-axis.
1.4
PLANES
A plane is a surface such that a straight line joining any two of its points lies entirely in the
surface.
Equations of Planes
An equation of first degree: Ax + By + Cz + D = 0 (1) in x, y, z always represents a plane.
This equation is called the General equation of a plane.
1. Any plane through (x1, y1, z1)
(one point form)
Let
Ax + By + Cz + D = 0
[1]
Let it pass through P (x1, y1, z1).
Then, we have Ax1 + By1 + Cz1 + D = 0
[2]
Subtracting [2] from [1], we get
A (x – x1) + B( y – y1) + C (z – z1) + D – D = 0
or
A (x – x1) + B( y – y1) + C (z – z1) = 0
is the equation of the plane through (x1, y1, z1).
2. Normal Form
[VTU, Aug., 1999]
Z
(x ,
z
y,
)
P
N
p
a
O
Y
X
Fig. 1.13
Let a be a plane.
Let ON be a normal to plane
Let ON = p and (l, m, n) be D.C’s of the normal to the plane.
Then, the equation of the plane is
lx + my + nz = p
or
x cosa + y cos b + z cos g = p
where (l, m, n) are D.C’s of the normal to the plane, p, is the length of the ^r from the origin
to the plane and p is always positive.
Dharm\C\N-ENGM\EMATH1.PM5
27
Analytical Geometry in Three Dimensions
[VTU, Aug., 2001]
PART-A
3. Intercept Form
Z
C
D
E
N
O
Q
B
A
F
Y
X
Fig. 1.14
We have the equation of plane in normal form:
x cos a + y cos b + z cos g = 0
We find from the figure,
ONA, ONB and on ONC are right angled Ds
ON = p; 

OA = f 

ONA = a ; 
|||ly, we get the relations
[1]
 p
fi cos a =   ;
f
fi
f = p sec a
cos a = (p/f), cosb = (p/g), cos g = ( p/h)
fi
f = p seca , g = p sec b , h = p secg
Substituting [2] in [1], we get
x y z
p
p
p
+ +
+y
+z
=p;
=1
h
f g h
g
f
4. Reduction of the general equation
Ax + By + Cz + D = 0 to the normal form
Let Ax + By + Cz + D = 0 be the eqn. to a given plane.
x
(
)
2
Ax + By + Cz = – D, ÷ ± ∑ A , we get
or,
±
A
∑ A2
⋅x ±
B
∑ A2
⋅y ±
C
∑ A2
⋅z = ±
Since in the normal form P is always positive
\
±
D
∑ A2
Dharm\C\N-ENGM\EMATH1A.PM5
should be positive.
D
∑ A2
[2]
28
Engineering Mathematics – I
5. Plane through three given points (three point form)
Let
[V.T.U. F/M. 2005]
Let the plane be
P, lies on it
\
P = (x1, y1, z1)
Q = (x2, y2, z2)
R = (x3, y3, z3)
Ax + By + Cz + D = 0
Ax1 + By1 + Cz1 + D = 0
[1]
[2]
Q lies on it
\
Ax2 + By2 + Cz2 + D = 0
[3]
R lies on it
\
Ax3 + By3 + Cz3 + D = 0
[4]
Solving for A : B : C : D from [2], [3], [4]; and substitute in [1], OR, eliminate A : B : C : D
from [1] to [4],
x
y z 1
x1 y1 z1 1
= 0.
We get, the three point form
x 2 y2 z2 1
x 3 y3 z3 1
6. Deduction of equation of the plane in the intercept form, from the 3 point form.
[V.T.U. F/M 2005]
If we choose in the equation of the plane in three point form,
(x1 = f, y1 = 0, z1 = 0);
(x2 = 0, y2 = g, z2 = 0)
(x3 = 0, y3 = 0, z3 = h).
Then we get the equation of the plane passing through the points (f, 0, 0); (0, g, h) and
(0, 0, h), i.e. a plane making intercepts f, g and h, on the three axes. The equation of the
plane then becomes:
x
y
z 1
f
0
0 1
0
g 0 1
0 0
or,
or,
h 1
0
0 1
z 1
+ fg
+0=0
0 h
h 1
h 1
xgh – fy (–h) + fg (z – h) = 0
(∏ fgh)
x
\
= 0;
y z 1
0 0 1
x g 0 1 –f g 0 1 =0
0 h 1
0 h 1
g
– fy
x y z
+ +
+ (– 1) = 0
fi
f
g h
7. Condition for four points to be coplanar
x y z
+ + =1
f g h
Let
A = (x1, y1, z1), B = (x2, y2, z2)
C = (x3, y3, z3) and D = (x4, y4, z4)
be any four points. Then the equation of the plane passing through three points A, B, C is
x − x1
x2 − x1
x3 − x1
Dharm\C\N-ENGM\EMATH1A.PM5
y − y1
y2 − y1
y3 − y1
z − z1
z2 − z1
z3 − z1
=0
[1]
29
Analytical Geometry in Three Dimensions
Since D = (x4, y4, z4) also lies on [1]
y4 − y1
y2 − y1
y3 − y1
z4 − z1
z2 − z1
z3 − z1
=0=
x2 − x1
x3 − x1
x4 − x1
y2 − y1
y3 − y1
y4 − y1
z2 − z1
z3 − z1
z4 − z1
PART-A
\
x4 − x1
x2 − x1
x3 − x1
is the condition for A, B, C, D to be coplanar.
WORKED EXAMPLES
31. Find the intercepts made on the coordinate axes by
plane x + 2y – 2z = 9. Find the length of the normal from
origin to the plane x + 2y – 2z = 9 and also the D.R’s of
normal.
Solution: To find the intercepts on the axes, we write
equation in the form:
x y z
+ +
=1
f
g h
x + 2y – 2z = 9
Given plane:
i.e.,
the
the
the
Z
C
the
[1]
x
+
2y
O
(0,0,0)
x 2 y 2z
+
−
=1
9
9
9
x
y
z
+
−
=1
9 9  9
−
2  2
  

9 −9
fi the intercepts are 9,
,
, the given plane is
2
2
x + 2y – 2z = 9
DR’s of normal to plane (1 : 2 : –2).
\ The actual D.C’s
1
2
2
1 + 2 ( − 2)
2
,
2
2
2
1 + 2 + ( − 2)
−2
2
2
1 + 22 + ( − 2)2
1 2 −2
.
, ,
3 3 3
We now write the equations of the given plane in the normal form
lx + my + nz = p
\ The eqn. of the plane:
x + 2y – 2z = 9
can be written as:
9
x 2 y 2z
+
−
=
;
3
3
3
3
2z
1
2
x+
y–
= 3.
i.e.,
3
3
3
Hence the length of the normal from the origin to the plane is 3.
fi
Dharm\C\N-ENGM\EMATH1A.PM5
–
2z
=
9
B
A
Fig. 1.13
Y
X
30
Engineering Mathematics – I
32. Find the equation of the plane having y intercept 10, z interecept-4 and perpendicular to
the plane 7x + y + 13z – 17 = 0.
[V.T.U. F/M. 2005]
Solution: Given y-intercept 10 ; z-intercept–4
fi Plane passes through (0, 10, 0) and (0, 0, – 4).
\
a (x – 0) + b(y – 10) + c(z – 0) = 0
[1]
Hence,
i.e.,
a . 0 + b (– 10) + c (– 4) = 0
b −2
5b + 2c = 0 fi
or b : c = – 2 : 5
=
5
c
Further the plane given by (1) is perpendicular to the plane
7x + y + 13z – 17 = 0
\
7a + b + 13c = 0. But b : c = – 2 : 5
Hence we get 7a – 2 + 65 = 0
or
7a + 63 = 0
\
a=–9
Substituting the values of a, b, c in (1), we get
– 9x – 2 (y – 10) + 5z = 0
or
9x + 2y – 5z – 20 = 0 is the required equation.
33. Find the image of the point (1, 2, 3) in the plane: x + y + z = 9
Solution: Let P (1, 2, 3) and Q (x1, y1, z1) be the image of the point P in the given plane
x + y + z = 9.
Let R be the point on the plane where the line PQ cuts the plane. Then R is the mid
point of PQ.
 x + 1 y1 + 2 z1 + 3 
,
,
R=  1

2
2 
 2
(mid point formula)
This point lies on the plane
x+y+z=9
\
x1 + 1 y1 + 2 z1 + 3
+
+
=9
2
2
2
fi
x1 + y1 + z1 = 12
Now D.R’s of PQ are
fi
[1]
[(x1 – 1) : (y1 – 2) : (z1 – 3)]
Now PQ is ^r to the plane and D.R’s of Normal to the plane are (1 : 1 : 1)
Thus, we have
P
x1 − 1 y1 − 2 z1 − 3
=
=
=k
1
1
1
R
fi
x1 = k + 1, 
y1 = k + 2, 
(2)

z1 = k + 3 
Q
\ (1) fi k + 1 + k + 2 + k + 3 = 12
fi
3k = 6
k=2
Dharm\C\N-ENGM\EMATH1A.PM5
Fig. 1.16(a)
31
Analytical Geometry in Three Dimensions
Q = (3, 4, 5)
33(a). Find the image of the point (1, – 1, 2) in the plane
2x + 2y + z = 11
[V.T.U. F/M, 2005]
Solution: Let P(1, – 1, 2) and Q (x1, y1, z1) be the image of the point P in the given plane
2x + 2y + z = 11
Let R be the point on the plane where the line PQ cuts the plane. Then R is the mid
point of PQ.
P(1,–1,2)
 x + 1 y1 − 1 z1 + 2 
=
,
R=  1

2
2 
 2
\
lies in the plane :
fi
fi
R
2x + 2y + z = 11.
2.
x1 + 1
y − 1 z1 + 2
= 11.
+2. 1
+
2
2
2
2x1 + 2y1 + z1 = 20
Q(x1,y1,z1)
(1)
Now, DR’s of PQ are [(x1 – 1) : (y1 + 1) : (z1 – 2)].
and PQ is ^r to the plane. DR’s of normal to the plane are (2 : 2 : 1)
fi
x1 − 1 y1 + 1 z1 − 2
=
=
= k (say),
2
2
1
(x1 = 2k + 1, y1 = 2k – 1, z1 = k + 1)
Equation (1) become
fi
4k + 2 + 4k – 2 + k + 2 = 20
fi
9k = 18
\
or
(2)
k=2
Substituting k = 2 in (2), we get
x1 = 5, y1 = 3, z1 = 4.
therefore, the required image = Q = (5, 3, 4).
33 (b). Find the equation of the plane passing through the points (9, 0, 6), (2, 1, 1) and
perpendicular to the plane 2x + 6y + 6z = 9.
[V.T.U. J/A, 2003]
Solution: Equation of plane be
A(x – x1) + B(y – y1) + C(z – z1) = 0
Equation passes through (9, 0, 6) so
A(x – 9) + B(y – 0) + c(z – 6) = 0
Equating passes through (2, 1, 1) so
– 7A + B – 5C = 0
(1)
r
Since equation is ^ to the plane 2x + 6y + 6z = 9.
We have
2A + 6B + 6C = 0
(2)
Solving (1) and (2)
A
B
C
=
=
− 36
− 32 44
or
a b
c
= =
9 8 − 11
The required equation of plane is
9(x – 9) + 8y – 11 (z – 6) = 0
Dharm\C\N-ENGM\EMATH1A.PM5
or
9x + 8y – 11z = 15.
PART-A
Now by putting k = 2 in [2], we get the image point
32
Engineering Mathematics – I
34. Find the D.C’s of the normal to, and the length of the perpendicular from the origin on,
the plane whose equation is Ax + By + Cz + D = 0.
Solution:
Let Ax + By + Cz + D = 0 be the equation of the plane
[1]
Equation in Normal form is:
x cosa + y cos b + z cos g = p
[2]
If [1] is put in the normal form, [1] and [2] represent the same plane, and so the two
equations are equivalent. Hence the coefficients are proportional
cos β cos γ
−p
cos α
=
=
=
B
C
D
A
[1] \ cosa : cosb : cosg = A : B : C
i.e D.C’s of the normal to the plane
[3]
\
= (cosa : cosb : cosg )
= A :B :C
= (x-coeff : y-coeff : z-coeff)
[2] Also,
−p
cos α cos β cos γ
=
=
=
A
B
C
D
=
\
[4]
−p
=
D
(cos2 α + cos2 β + cos2 γ )
2
2
A + B +C
1
2
2
A +B +C
=
2
p=
\
2
+1
2
A + B2 + C 2
−D
2
A + B2 + C 2
Note:
(a) D.C’s of Normal:
= coeff x : coeff y : coeff z
(b) The position is specified by p =
(c) ⊥r from the origin =
=
−D
2
A + B2 + C 2
−D
2
2
A +B +C
2
|− D |
2
2
2
.
. The ⊥r in magnitude
=
D
2
A +B +C
A + B2 + C2
35. Find the length of the perpendicular from the point (x1, y1, z1) on the plane.
(ax + by + cz + d) = 0
We know that:


|− D |

p= 
 A2 + B2 + C 2 


where p = length of the ^r from the origin O (0, 0, 0).
Hence, we require the ^r from P (x1, y1, z1) to the plane
(ax + by + cz + d) = 0
Dharm\C\N-ENGM\EMATH1A.PM5
[1]
[2]
33
PART-A
Analytical Geometry in Three Dimensions
z-axis
N
(x1, y1, z1)
P
O
xax
is
y-axis
Fig. 1.16(b)
Shifting the origin to the point
(x1, y1, z1), i.e. put
x = (X + x1)
y = (Y + y1)
z = (Z + z1)






[3]
We get a(X + x1) + b(Y + y1) + c(Z + z1) = d = 0
fi
aX + bY + cZ + (ax1 + by1 + cz1 + d) = 0
aX + bY + cZ + k = 0
k = (ax1 + by1 + cz1 + d)
[4]
The perpendicular distance of p from the plane [2] is the same as the perpendicular
from the new origin to the plane [4].
Hence apply Eqn. [1], we get
fi
p=
−k
2
2
a +b +c
2
=
− ( ax1 + by1 + cz1 + d )
a 2 + b2 + c2
Note: The perpendicular distance is, in magnitude equal to the numerical value
ax1 + by1 + cz1 + d .
a2 + b2 + c2
36. Find the D.C’s of the normal to the plane
3x + 2y + 6z + 14 = 0.
Also obtain the perpendicular distance of the origin from this plane.
Solution: D.C’s of the Normal
= coeff x : coeff y : coeff z
= 3:2:6
p=
−D
2
2
A + B +C
2
=
− 14
2
2
2
3 +2 +6
=
− 14
7
= 2.
37. Find the equation of the plane through the points (2, 2, 1) and (9, 3, 6) and perpendicular
to the plane 2x + 6y + 6z = 9.
[V.T.U. J/F, 2004]
Dharm\C\N-ENGM\EMATH1A.PM5
34
Engineering Mathematics – I
Solution: The equation of a plane through a point (2, 2, 1) is
a (x – 2) + b(y – 2) + c (z – 1) = 0
A point (9, 3, 6) also lies on (1), then
a (9 – 2) + b(3 – 2) + c (6 – 1) = 0
fi
2x + 6y + 6z = 9
Plane (1) is perpendicular to plane (3), then
2a + 6b + 6c = 0
On solving (2) and (4), we get
...(1)
7a + b + 5c = 0
...(2)
...(3)
...(4)
a
b
c
a b
c
or = =
=
=
6 − 30 10 − 42 42 − 2
3 4 −5
On substituting the value of a, b, c in (1) we get
3 (x – 2) + 4 (y – 2) – 5 (z – 1) = 0 or 3x + 4y – 5z = 9.
38. Prove that the equation of the plane through the points (1, – 2, 4) and (3,– 4, 5) and
perpendicular and (3, – 4, 5) and perpendicular to yz-plane or parallel to x-axis is y + 2z = 6.
Solution: Any plane through (1, – 2, 4) is
A(x – 1) + B(y + 2) + C(z – 4) = 0
[1]
where A : B : C are D.R’s of Normal. Since [1] passes through (3, – 4, 5)
\
A(3 – 1) + B(– 4 + 2) + C(5 – 4) = 0
or
A ◊ 2 + B ◊ (– 2) + C ◊ 1 = 0
[2]
Again, Since [1] is parallel to x-axis (DC’s 1 : 0 : 0)
\ Normal to [1] is perpendicular to x–axis
\ A◊1 + B◊0 + C◊0 = 0
[3]
Eliminating (A : B : C) from [1], [2] and [3], determinantally,
we get
x −1 y + 2 z − 4
2
1
−2
1
0
0
=0
fi
1(y + 2) + 2(z – 4) = 0
fi
y + 2z = 6 Hence proved.
39. Find the equation of the plane through the point P(2, 3, – 1) and perpendicular to the
line OP, where O is the origin.
Solution:
O = (0, 0, 0)
P = (2, 3, –1)
\ DR’s of OP = (2 – 0) : (3 – 0) : (– 1 – 0) = (2 : 3 : – 1)
Any plane perpendicular OP
= 2x + 3y + (– 1) z = k
P lies on it.
\ 2 ◊ 2 + 3 ◊ (3) + (– 1) (– 1) = k = 14
plane:
(2x + 3y – z) = 14
40. Find the plane through the point (a , b , g )
Ax + By +
Solution: Any plane parallel to
Ax + By +
is
Ax + By +
Dharm\C\N-ENGM\EMATH1A.PM5
and parallel parallel to
Cz + D = 0
Cz + D = 0
Cz + k = 0
35
Analytical Geometry in Three Dimensions
(a , b , g ) lies on it).
Aa + Bb + Cg + k = 0
PART-A
\
Subtracting we get
A(x – a ) + B( y – b ) + C(z – g ) = 0
41. Find the plane through the point (3, 2, 5) and parallel to the plane
2x + 4y + 3z – 7 = 0.
Solution: Any plane parallel to 2x + 4y + 3z –7 = 0 has the same normal.
Hence D.R’s of Normal = (2 : 4 : 3)
Any plane parallel to the above
= 2x + 4y + 3z + k = 0
(3, 2, 5) lies on it
\
\
2◊3 + 4◊2 + 3◊5 + k = 0
k = – 29
\ the plane is : 2x + 4y + 3z – 29 = 0
42. Find the equation of the plane which passes through the point (1, – 2, 1) and is ^r to each
of the planes
3x + y + z – 2 = 0 and
x – 2y + z + 4 = 0.
Solution: Let
Ax + By + Cz + D = 0
be the equation of the plane. It passes through (1, – 2, 1). So, we get
[1]
A – 2B + C + D = 0
Since [1] is perpendicular to the planes
we have
\
3x + y + z – 2 = 0
3A + B + C = 0
A – 2B + C = 0
and
[2]
x – 2y + z + 4 = 0
[3]
[4]
B
C
A
=
=
= k (say)
2
7
−
−
3
\
A = 3k; B = – 2k; C = –7k.
Substituting in [2], we get
D=0
\ Equation of the required plane is 3x –2y –7z = 0.
43. Find the distance between the parallel planes
2x + 3y – z + 4 = 0
Solution: The given planes are
and
4x + 6y – 2z – 5 = 0
2x + 3y – z + 4 = 0
4x + 6y – 2z – 5 = 0
We shall find any one point on one plane say [1]
To this end put x = 0, y = 0 is [1], we get z = 4
\ one point on the plane [1] is (0, 0, 4)
Dharm\C\N-ENGM\EMATH1A.PM5
[1]
P1
[2]
P2
Fig. 1.16(c)
36
Engineering Mathematics – I
Distance between the parallel planes = perpendicular distance of (0, 0, 4) from plane [2]
4(0) + 6(0) − 2(4) − 5
=
16 + 36 + 4
=
13
56
.
44. If the foot of the perpendicular from the origin to the plane is (2, – 1, 2). Find the
equation of the plane.
Solution: Let
Ax + By + Cz + D = 0
[1]
be the equation of the required plane. Since D.R’s of the normal to [1] is (A : B : C), also OP
is a normal to that plane
\ the D.R’s of normal to the plane are (2 : –1 : 2)
The plane becomes
2x – y + 2z + D = 0
(2, – 1, 2) lies on it \ 4 + 1 + 4 + D = 0
fi
D=–9
\
2x – y + 2z – 9 = 0
is the required plane.
45. Prove that the four points (– 6, 3, 2), (3, – 2, 4) (5, 7, 3) and (– 13, 17, – 1) lie in one plane.
Find the equation of the plane containing them.
Solution: Let
A (– 6, 3, 2)
B (3, – 2, 4)
C (5, 7, 3)
and
D (– 13, 17, – 1)
The equation to the plane containing the points A, B and C is
x +6
y−3 z −2
3 +6 −2 −3 4 −2
5+6
7 −3 3 −2
=0
fi
x +6 y−3 z −2
9
2
−5
11
4
1
=0
fi
(x + 6) (– 5 – 8) – (4 – 3) (9 – 22) + (z – 2) (36 + 55) = 0
fi
– 13(x + 6) + 13(4 – 3) + 91(z – 2) = 0
fi
– 13x + 13y + 91z – 299 = 0
fi
x – y – 7z + 23 = 0
Put x = – 13, y = 17, z = – 1 in the LHS of this equation,
we get
– 13 – 17 + 7 + 23 = – 30 + 30
= 0 = RHS
\ D(– 13, 17, – 1) lies on this plane
Thus, the 4 points lie on the same plane whose eqn is
x – y – 7z + 23 = 0
46. Find the equation of the plane through (2, 3, – 4) and (1, – 1, 3) and parallel to the x-axis
Solution: Let Ax + By + Cz + D = 0 be the plane
[1]
If crosses through (2, 3, – 4) and (1, – 1, 3)
\
and
From (2) – (3),
Dharm\C\N-ENGM\EMATH1A.PM5
2A + 3B – 4C + D = 0
[2]
A – B + 3C + D = 0
[3]
A + 4B – 7C = 0
[4]
. P(2, 3, – 4)
P : plane
. Q(1, – 1, 3)
Since [1] is parallel to the x-axis whose
D.C’s are (1 : 0 : 0)
\
A◊1 + B◊0 + C◊0 = 0
Putting this value in [4],
we get,
O(0, 0, 0)
A=0
x-axis
Fig. 1.16(d)
4B = 7C
B=
From [2],
fi
D.C’s (1 : 0 : 0)
2◊0 +
7
C
4
21
C – 4C + D = 0
4
fi
D=–
Putting this values in [1], we get
5
C
4
5
7
Cy + Cz –
C=0
4
4
fi
7y + 4z – 5 = 0 is the required plane
47. Find the equation of the plane through the points (1, – 2, 4) and (3, – 4, 5) and
perpendicular to yz-plane
Solution: The equation of the given yz-plane is x = 0.
[1]
PLANE
Let the equation of the plane ^r to [1] be
. Q(3, –4, 5)
Ax + By + Cz + D = 0
[2]
. P(1, –2, 4)
Then
A◊ 1 + B◊ 0 + C◊ 0 = 0
i.e.
A=0
[3]
(Q aa1 + bb1 + cc1 = 0)
[3]
By data, [2] passes thro’
(1, –2, 4) and (3, – 4, 5)
A – 2B + 4C + D = 0
–2B + 4C + D = 0
[4]
Using (3) 3A – 4B + 5C + D = 0
– 4B + 5C + D = 0
[5)
Using [3]
From [4] and [5], we get by Cross-multiplication
C
D
B
=
=
− 4 + 2 − 10 + 10
4−5
\
i.e.
i.e.
fi
yz-plane, or x = 0
Fig. 1.16(e)
C
D
B
=
=
= k (say)
−2 6
−1
B = – k, C = – 2k, D = + 6k
\
putting these values is [2],
– ky – 2k z + 6k = 0
i.e.
y + 2z – 6 = 0, is the desired eqn of the plane.
48. A plane meets the coordinate axes at A, B, C such that the centroid of the triangle DABC
is the point (a , b , g ). Show that the equation of the plane is
Dharm\C\N-ENGM\EMATH1A.PM5
x
y z
+ +
= 3.
α β γ
PART-A
37
Analytical Geometry in Three Dimensions
38
Engineering Mathematics – I
Solution: Let
A (a, 0, 0)
B (0, b, 0)
C (0, 0, c)
equation of the plane is
then the
x y z
+ +
=1
a b c
a b c
Since the centroid of the triangle is  , ,  but it is given to be (a , b , g )
 3 3 3
a
b
c
We have
= a,
= b,
= g.
3
3
3
\
a = 3a , b = 3b , c = 3g
x
y
z
+
+
and [1] becomes
=1
3α 3 β 3γ
[1]
x
y z
+ + =3
α β γ
49. Find the length of the perpendicular from the points (1, 2, 3) and (2, – 1, 2) from the plane
3x – 4y + 5z = 12.
Are these points on the same side of the plane?
i.e.
Solution: Length of ^r from the point (1, 2, 3) to the plane:
p=
3 − 4 × 2 + 5 × 3 − 12
=
9 + 16 + 25
1 − 21
50
=
2
5
r
and length of ^ from the point (2, –1, 2) to the plane:

8
2⋅2⋅2 4 2
4
=
=
Q

2.
5 
50
2 ⋅ (5)
50

50 5
Putting down (1, 2, 3) in the eqn of plane we find –ve value and putting (2, –1, 2) in
the eqn of the plane we find +ve value. Thus, they are on opposite sides of the plane.
50. Find the distance between the parallel planes
p=
3 × 2 − 4 × − 1 + 5 × 2 − 12
=
8
=
2x + 3y –z + 4 = 0
and
4x + 6y – 2z – 5 = 0
Solution: The given planes are:
2x + 3y –z + 4 = 0
[1]
4x + 6y – 2z – 5 = 0
[2]
We shall find any one point on one plane say [1]. To find the same we put x = 0, y = 0
in [1], we get z = 4
\ (0, 0, 4) is a point on plane
\ Distance between the parallel plane
= ^r distance of (0, 0, 4) from the plane
=
Dharm\C\N-ENGM\EMATH1A.PM5
4(0) + 6(0) − 2(4) − 5
16 + 36 + 4
=
13
56
[1]
[2]
51. Show that the two points P = (1, – 1, 3) and Q = (3, 3, 3) are equally distant from the
plane (5x + 2y – 7z + 9) = 0 and are on opposite sides of the plane.
Solution: Given plane:
(5x + 2y – 7z + 9) = 0
[1]
P = (1, – 1, 3)
Q = (3, 3, 3)
Let PM and QN be the perpendiculars from P and Q on the plane [1].
PM =
QN =
− ( Aα + B β + Cγ + D )
2
2
A + B +C
2
=
− [5 ⋅1 + 2( − 1) + ( − 7)(3) + 9]
− [5 ⋅ 3 + 2 ⋅ 4 + ( − 7)(3) + 9]
2
2
5 + 2 + ( − 7)
=
2
=
9
78
−9
78
5 + 2 + ( − 7)
Now, PM and QN are equal in magnitude, hence, the two points are equally distant
from the plane.
Also PM and QN have opposite signs, and hence the two points lie on opposite sides of
the plane.
2
2
2
EXERCISES
1. Find the equation of the plane
(i) thro’ the point (1, – 1, 2) and parallel to the xy-plane
(Ans. z – 2 = 0)
(ii) thro’ the point (2, – 1, 6), (1, – 2, 4) and perpendicular to the plane x – 2y – 2z + 9 = 0
(Ans. 2x + 4y – 3z + 18 = 0)
2. Find the image of the line
x −1 y − 2 z − 3
in the plane 2x + y + z = –2.
=
=
2
1
4

x + 5 y +1 z − 0
=
=
 Ans.
 (each = k)
1 
−4
−2

3. Find the distance between the parallel planes 2x –2y + 2 + 3 = 0 and 4x – 4y + 2z – 5 = 0.
4. Show that the form points
11 

 Ans. perpendicular to distance = 6 


(0, – 1, 0), (2, 1, – 1), (1, 1, 1)
and (3, 3, 0) are coplanar. Find the equation of the plane thro’ them.
(Ans. 4x – 3y + 2z = 3)
5. Find the distances of the points (2, 3, – 5), (3, 4, 7) from the plane x + 2y – 2z = 9, are the
points on the same side of the plane?
(Ans. 3, – 4) Since the
signs are opposite, the two points are on opposite sides of the plane.
6. If the plane ax + by + cz + d = 0 is parallel to the line with D.R’s (l : m : n). Show that
al + bm + cn = 0.
7. Find the equation of the plane thro’ the points (2, 2, 1) (1, –2, 3) and parallel to x–axis
(Ans. y + 2z – 4 = 0)
Any Plane Through the Line of Intersection two Planes: (Coaxal Planes)
If P1 = 0 and P2 = 0 are any two given planes then the equation to any plane passing
through the intersection of the planes
Dharm\C\N-ENGM\EMATH1A.PM5
PART-A
39
Analytical Geometry in Three Dimensions
40
Engineering Mathematics – I
P1 = 0 and P2 = 0 is of the form:
P1 + k (P2) = 0
fi one plane + k (other plane) = 0.
If
P1: A1x + B1 y + C1z + D1 = 0
and
P2: A2x + B2 y + C2z + D2 = 0 are the equations of two
planes, then
A1x + B1y + C1z + D1 + k (A2x + B2 y + C2z + D2) = 0 (1)
is the plane passes thro’, the intersection of the given planes.
P1 = 0
P1 + KP2 = 0
P2 = 0
Fig. 1.16(f)
WORKED EXAMPLES
52. Find the equation of the plane passing through the line of intersection of
P1: 3x + 4y – 2z + 6 = 0
[1]
P2: x + 2y + z –2 = 0
[2]
and passing thro’ the origin.
Solution: Any plane through the point of intersection of [1] and [2] is of the form
(P1 + kP2 = 0)
i.e.
(3x + 4y – 2z + 6) + K (x + 2y + z – 2) = 0
(0, 0, 0) lies on it.
\
(0 + 0 + 0 + 6) + k (0 + 0 + 0 – 2) = 0
\
6 – 2k = 0
\
K = 3.
Hence the plane is:
(3x + 4y – 2z + 6) + 3 (x + 2y + z – 2) = 0
fi
6x + 10y + z = 0
53. Find the plane passing through the line of intersection of 3x + 4y – 2z + 6 = 0 and x + 2y
+ z – 2 = 0 and passing through (3, – 4, 2).
Solution: Any plane through the intersection of [1] and [2] is of the form
P1 + kP2 = 0
(3x + 4y – 2z + 6) + k(x + 2y + z –2) = 0
(3, – 4, 2) lies on it
\
9 – 16 – 4 + 6 + k (3 – 8 + 2 – 2) = 0
so
k = – 1.
\ The plane:
(2x + 2y – 3z + 8) = 0.
54. Find the plane through the same line given Eqn (2) and parallel to the z-axis.
(3x + 4y – 2z + 6) + k (x + 2y + z – 2) = 0
x(3 + k) + y(4 + 2k) + z(–2 + k) + (6 – 2k) = 0
This is parallel to Z-axis
\ coeff of z is zero
or
(– 2 + k) = 0
Dharm\C\N-ENGM\EMATH1A.PM5
41
Analytical Geometry in Three Dimensions
fi
5x + 8y + 2 = 0
55. Find the equation to the plane containing the line 2x – 5y + 2z = 6, 2x + 3y – z = 5 such
that it is
x
y
z
= .
(i) parallel to the line
=
1
−6 7
(ii) perpendicular to the plane x + y – z = 7.
Solution: Any plane through the intersection of the given planes is
2x –5y + 2z – 6 + k (2x + 3y – z – 5) = 0
fi
(2 + 2k) x + (3k – 5) y + (2 – k) z –6 –5k = 0
[1] is parallel to the line
[1]
y
z
x
=
=
1
−6 7
fi
1(2 + 2k) – 6(3x – 5) + 7(2 – k) = 0
fi
k=2
Putting down for k in [1], we get 6x + y – 16 = 0.
Again [1] is perpendicular to plane x + y – z = 7
fi
1 (2 + 2k) +1 (3k – 5) – (2 – k) = 0
fi
k = 1.
Putting down for k in eqn [1], we get
4x – 2y + z – 11 = 0
56. Find the equation of the plane thro’ the line of intersection of the planes
3x + 2y + 4z + 6 = 0
and
(2x – y + z + 2) = 0,
so as to be parallel to the y-axis.
Solution:
P1 = (3x + 2y + 4z + 6) = 0
[1]
P2 = (2x –y + z + 2) = 0
Any plane thro’ the line of intersection of the two planes above, is of the form P1 + kP2 = 0.
UV
W
i.e.
(3x + 2y + 4z + 6) + k (2x – y + z + 2) = 0
fi
x(3 + 2k) + y (2 – k) + z (4 + k) + (6 + 2k) = 0
which is parallel to the y-axis (Given)
Hence coeff of y should be zero
\
i.e. (2 – k) = 0
fi
k=2
Substitute in [3], we get the equation of the required plane:
x (3 + 2 ◊ 2) + y (2 – 2) + z(4 + 2) + (6 + 2 ◊ 2) = 0
fi
(7x + 6z + 10) = 0
57. Find the plane thro’ the line of intersection of
(3x + 2y + 4z – 5) = 0,
and
(2x – y + 2z + 3) = 0,
so as to be perpendicular to the latter plane.
Dharm\C\N-ENGM\EMATH1A.PM5
[2]
[3]
[4]
PART-A
So the plane: x (3 + 2) + y (4 + 4) + z (–2 + 2) + (6 – 4) = 0
42
Engineering Mathematics – I
P1 = (3x + 2y + 4z – 5) = 0
P2 = (2x – y + 2z + 3) = 0
Any plane thro’ the line of intersection of the two planes above, is of the form
(P1 + kP2) = 0
UV
W
[1]
(3x + 2y + 4z – 5) + k (2x – y + 2z + 3) = 0
[2]
x (3 + 2k) + y(2 – k) + z (4 + 2k) + (– 5 + 3k) = 0
This is required to be perpendicular to the latter plane of the two P1 = 0 and
P2 = 0, namely P2 = 0
[3]
Solution:
fi
fi
\
fi
2x – y + 2z + 3 = 0
[4]
(3 + 2k) (2) + (2 – k) (– 1) + (4 + 2k) (2)
k(4 + 1 + 4) + (6 – 2 + 8) = 0
fi
Substituting for k, in [2], we get
[5]
k = – 4/3
[6]
(3x + 2y + 4z – 5) – (4/3) (2x – y + 2z + 3) = 0
fi
x + 10y + 4z – 27 = 0
58. Find, the equation of the plane passing thro’ the intersection of the planes x + 3y – z = 4
and 2x + 2y + 2z = 1 and perpendicular to plane.
x + y – 4z = 0
Solution: The eqn of plane:
x + 3y –z – 4 + k (2x + 2y + 2z – 1) = 0
fi
(1 + 2k) x + (3 + 2k)y + (–1 + 2k)
Since it is perpendicular to x + y – 4z = 0,
we have
(1 + 2k) 1 + (3 + 2k) 1 + (–1 + 2k) (–4) = 0
fi k=2
\ The required plane is
fi
Z
–4=0
x + 3y – z – 4 + 2 (2x + 2y + 2z –1) = 0
5x + 7y + 3z – 6 = 0.
EXERCISES
1. Obtain the equation of the line of intersection of the planes
4x + 4y – 5z = 12 and 8x + 12y –13z = 32
x −1 y − 2 z 

 Ans : 2 = 3 = 4 


2. Find the equation of the line passing thro, the point (1, 2, 3) and parallel to the line of
intersection of the planes
x – y + 2z = 5 and 3x + y + z = 6.
3. Show that the straight lines
3x + 2y + z –5 = 0 = x + y – 2z – 3
and
8x – 4y –4z = 0 = 7x + 10y – 8z
in the symmetric form.
are at right angles.
Dharm\C\N-ENGM\EMATH1A.PM5
43
Analytical Geometry in Three Dimensions
3x –y + 2z – 4 = 0
and
x+ y+ z–2=0
and passing thro’ (2, 2, 1).
(Ans: 7x – 5y + 4z – 8 = 0)
5. Find the equation of the plane passing thro’ the intersection of the planes
x + 3y – z = 4
and
2x + 2y + 2z = 1
and perpendicular to the plane
x + y – 4z = 0
(Ans: 5x + 7y + 3z – 6 = 0)
1.5
STRAIGHT LINES
A straight line may be regarded as the common line of intersection of the planes
A1x + B1y + C1z + D1 = 0
and
A2x + B2y + C2Z + D2 = 0.
Equations of Straight Lines
1. Symmetrical form of equation of a line
To find the equation of the straight line passing through the point (a , b, g ) and having
direction cosines (l : m : n)
(a, b, g)
(l : m : n)
A
(x, y, z)
P
Fig. 1.16(g)
line.
i.e.
or,
Let A = (a , b , g ) be the given point. Consider any point P = (x, y, z) on the straight
Then D.R’s AP = (x – a ) : (y – b ) : (z – g )
But it is given that the D.C’s are (l : m : n).
Evidently these two are equivalent.
fi
(x – a ) : (y – b ) : (z – g ) = l : m : n
x −α y − β z −γ .
fi
=
=
l
m
n
The above equation is called the ‘Symmetric Form’ of the equations to a line.
2. Parametric form
x −α
y − β z −γ
=
We know that
=
l
m
n
Put each ratio equal to r ,
y− β z −γ
x −α
=
\
=
=r
m
n
l
(x – a ) = lr, (y – b ) = mr, z – g = nr
[x = (a + lr),
y = (b + mr)
z = (g + nr)]
Any point on the straight line can be written in the form
[(a + lr, b + mr, g + nr)].
For different values of r , we get different points on the line.
Dharm\C\N-ENGM\EMATH1A.PM5
[1]
[2]
[3]
PART-A
4. Find the equation of the plane thro’ the intersection of the planes
44
Engineering Mathematics – I
y − y1
z − z1
x − x1
=
=
.
y2 − y1 z2 − z1
x2 − x1
If a line passes through the points (x1, y1, z1) and (x2, y2, z2), its direction ratios are
given by
[(x2 – x1) : (y2 – y1) : (z2 – z1)]
Hence if we choose the points A and B and D.R’s as a, b, c of the line then, the
‘Symmetric form’ of the equation of the line:
3. Two point form of a straight line:
x − x1
x − x1
y − y1
y − y1
z − z1
z − z1
=
=
=
; becomes
=
.
b
a
x2 − x1
y2 − y1 z2 − z1
c
(a : b : c)
A(x1, y1, z1)
P(x2, y2, z2)
Fig. 1.16(h)
Note: In particular, equation of a line passing through the origin and having the
direction cosines (l: m : n) are given by
x −0
x
y
z
y−0 z−0
= .
=
=
or
=
l
l
m n
m
n
59. Find the equations of the line x = ay + b, z = cy + d in the symmetrical form.
Solution: Equations of the given line are:
x = ay + b
[1]
z = cy + d
[2]
From [1], we get
ay = x – b
y
x −b
=
1
a
\
[3]
From [2], we get
cy = z – d
\
\
y
z −d
=
1
c
From [3] and [4], we get
x −b
y z −d
=
=
1
a
c
is the required symmetrical form of the given equations.
60. Find the equations of the lines x + 5y – z – 7 = 0;
2x – 5y + 3z + 1 = 0
in symmetrical form.
Solution: Given:
x + 5y – z – 7 = 0
2x – 5y + 3z + 1 = 0
If (a : b : c) be the DR’s of the required line.
Since the line is perpendicular to the normal of [1] and [2] we have
Dharm\C\N-ENGM\EMATH1A.PM5
[4]
[1]
[2]
a + 5b – c = 0
[3]
2a – 5b + 3c = 0
[4]
\
45
b
c
a
=
=
10
− 5 − 15
\ D.R’s are (2 : – 1 : – 3)
Let us find one point on the line. This can be done by taking z = 0 i.e. a the point (x, y,
0) where (x, y).
Satisfy:
x + 5y = 7
2x – 5y = – 1
Solving [5] and [6], we get
x = 2, y = 1
\ A point on the line is (2, 1, 0).
\ Equations of the line are
y −1
z
x−2
=
=
2
−1
−3
61. Put the equations
4x + 4y – 5z = 12,
8x + 12y – 13z = 32
of a straight line in symmetrical form
Solution: If (l : m : n) be the direction cosines of the line, then
4l + 4m – 5n = 0
8l + 12m – 13n = 0
\
l
m
n
=
=
− 40 + 52 48 − 32
− 52 + 60
m
n
l
=
=
8
12 16
l
m n
= .
fi
=
3 4
2
Thus D.R’s of the line are proportional to (2 : 3 : 4).
Next, put z = 0 in the given
eqns.
4x + 4y – 12 = 0;
x+ y–3=0
8x + 12y – 32 = 0; 2x + 3y – 8 = 0
y
1
x
=
\
=
−6+8 3−2
−8+9
fi
x = 1, y = 2.
i.e.
x −1 y − 2 z
=
= .
2
3
4
62. Find the equation of the line joining the points (3, 0, 2) and (1, –2, 3).
The eqn. of the line thro’ the points (x1, y1, z1) and (x2, y2, z2) is given by
x − x1
y − y1
z − z1
=
=
x1 − x2 y1 − y2 z1 − z2
x −3 y−0 z −2
Hence the required eqn is
=
=
3 −1 0 + 2 z − 3
y z−2
x −3
=
i.e.
=
.
2
2
−1
Hence, required eqns of the line in symmetrical form are
Dharm\C\N-ENGM\EMATH1A.PM5
[5]
[6]
PART-A
Analytical Geometry in Three Dimensions
46
Engineering Mathematics – I
63. Find the line thro’ (1, 2, 3) and ^r to both the lines
x y z
x −3 y−4 z −6
and
= =
=
=
4 5 7
2
3
4
Let the D.C’s of the required line (l : m : n)
It is perpendicular to both the lines
2l + 3m + 4n = 0 
4l + 5m + 7n = 0 
\
(l : m : n) = 1 : 2 : 2
\ Straight lines
x −1
y−2 z −3
=
\
=
1
2
−2
point: (1, 2, 3)
EXERCISES
1. While down the equations of the line thro’ the point (2, – 3, – 7) and having D.R.’s (3 : – 4 : 5).

x − 2 y + 3 z + 7
=
=
 Ans :

3
5 
−4

2. Find the equations of a line through (1, 4, – 2) and parallel to the planes
6x + 2y + 2z + 3 = 0
and
x + 2y – 6z + 4 = 0.
3. Obtain the equations of the line of intersection of the planes
x + y – 2z = 8 and 3x – y + 4z = 12
in the symmetric form.
4. Write the equations
x + 2y + 4z = 0 = 2x + 4y – z
of a line in the symmetric form
5. Prove that the points (3, 2, 4), (4, 5, 2) and (5, 8, 0) are collinear.
Find the equations of the line thro’ them.

x −1 y − 4 z + 2 
=
=
 Ans :

19
5 
−8


x −5 y−3
z 
=
=
 Ans :

1
−5
− 2


x
y
z
= 
 Ans : =
2
1
0
−


x − 3 y − 2 z − 4

 Ans : 1 = 3 = 2 


1.6
ANGLE BETWEEN PLANES/STRAIGHT LINES
The angle between two planes, is equal to the angle between their normals.
Let the equations to the two planes be:
Ax + By + Cz + D = 0
and
A¢x + B¢y + C¢z + D¢ = 0
The D.R’s of the normal to the first plane is
(A : B : C) and the D.R’s of the normal to the 2nd plane is (A¢ : B¢ : C¢). Hence, if
the angle between the two planes, then it is the same as the angle between (A : B : C)
(A¢ : B¢ : C¢)
( AA ′ + BB ′ + CC ′
\
cos q =
( ∑ A 2 ) ( ∑ B2 )
Dharm\C\N-ENGM\EMATH1A.PM5
[1]
[2]
q is
and
47
Analytical Geometry in Three Dimensions
(A) The Plane and the Straight Line
x −α y − β z −γ
=
=
will lie in the plane
l
m
n
PART-A
The line
Ax + By + Cz + D = 0
Aa + Bb + Cg + D = 0
Al + Bm + Cn = 0
if
and
\
A (x – a) + B(y – b) + C(z – g) = 0
is the plane through the line if Al + Bm + Cn = 0
(B) Plane Through the Lines
z −γ
x −α y − β
=
=
n1
l1
m1
We have,
x − α1 y − β 1
z − γ1
=
=
l2
m2
n2
and
x −α
l1
l2
is
y−β
m1
m2
z −γ
n1
n2
=0
(C) Angle Between the Line and the Plane
To find the expression for the angle between:
x −α y − β
z −γ
=
=
l
m
n
line:
[1]
and plane: Ax + By + Cz + D = 0
The actual D.C’s of the normal plane Ax + By + Cz + D
= 0 are proportional to A, B, C.
The actual D.C’s of the line are proportional to l, m, n
The angle between the normal to the plane and the line
is given by
cos (90° – q) =
Al + Bm + Cn
A2 + B2 + C2
[3]
l2 + m 2 + n 2
[2]
Normal
line
q
Fig. 1.17
as the angle between the line and the plane is (90° – q)
\
Al + Bm + Cn
sin q =
A2 + B2 + C2
fi
–1
q = sin
l 2 + m2 + n2

 Al + Bm + Cn

 ( ∑ A 2 ) ∑ l2

( )





1.7
COPLANAR LINES (TWO INTERSECTING STRAIGHT LINES)
Two straight lines may be coplanar (or intersect) i.e. they lie in the same plane.
Condition for the two lines to intersect (or to be coplanar)
Dharm\C\N-ENGM\EMATH1B.PM5
48
Engineering Mathematics – I
Let
and
z −γ
x −α y − β
=
=
n
l
m
[1]
z −γ′
x − α′ y − β′
=
=
′
′
n′
l
m
Since the lines [1] and [2] intersect, if some point of [1] lies on [2].
Any point on [1] can be
[2]
written:
(a + lr, b + mr, g + nr)
Substitute and test in [2],
[3]
α + lr − α ′ β + mr − β ′ γ + nr − γ ′
=
=
=k
l′
m′
n′
[4]
a + lr – a ¢ = l¢ k
(a – a ¢ ) + lr – l¢k = 0 

(b – b¢ ) + mr – m¢k = 0 
|||ly, we get
(g – g ¢ ) + nr – n¢k = 0 
Eliminating r and k, we get
fi
fi
α − α ′ l l′
β − β ′ m m′
γ − γ ′ n n′
in [3].
=0
or
[5]
α − α′ β − β′ γ − γ ′
l
m
n
l′
m′
n′
=0
[6]
Note: to get the point of intersection, substitute the common value of r satisfying [4],
To find the plane containing them
Let
and
is
z −γ
x −α y − β
=
=
n
l
m
[1]
a − α′ y − β ′
z −γ′
=
=
l′
m′
n′
be the equations of two lines.
The equation of any plane containing the line [1]
A(x – a) + B(y – b) + C(z – g) = 0
[2]
[3]
where
Al + Bm + Cn = 0
[4]
The plane [3] contains the lines [2] if it is parallel to the line [2] and a point of [2]
namely (a ¢, b ¢, g ¢ ) lies on the plane [3].
\
Al¢ + bm¢ + cn¢ = 0
[5]
and
A(a¢ – a) + B(b¢ – b) + C(g ¢ - g) = 0
[6]
Eliminating
A : B : C from [4], [5], [6], we get
α′ − α
l
l′
β′ − β γ ′ − γ
m
n
m′
n′
=0
Which is the condition for the coplanarity of the lines [1] and [2].
Dharm\C\N-ENGM\EMATH1B.PM5
[7]
49
Analytical Geometry in Three Dimensions
x −α
l
l′
y−β
m
m′
z −γ
n
n′
PART-A
Also Eliminate A : B : C from [3], [4], [5], to get
= 0 is the reqd plane containing the lines [1] and [2].
WORKED EXAMPLES
64. Find the angle between the planes 2x – y + z = 6 and x + y + 2z = 3.
Solution: Here
A = 2, B = – 1, C = 1, D = – 6
A¢ = 1, B¢ = 1, C¢ = 2, D¢ = – 3
If q be the angle between the planes, then
Cos q =
2(1) + ( − 1)(1) + 1(2)
2
2
2
2 + ( − 1) + 1
2
2
1 +1 + 2
=
2
3
6
6
=
1
2
\
q = 60°
65. Find the equation of the plane through (0, 1, – 2) parallel to the plane 2x – 3y + 4z = 0.
Solution: Any plane parallel to the given plane is
2x – 3y + 4z + l = 0
Since [1] passes through (0, 1, – 2)
[1]
\
2 (0) – 3 (1) + 4 (–2) + l = 0
\
l = 11
required plane is
2x –3y + 4z + 11 = 0
66. Find the equation of the plane through (– 1, 1, 1) and (1, – 1, 1) perpendicular to the
plane
x + 2y + 2z = 5.
Solution: Let the equation of the plane be Ax + By + Cz + D = 0
Since [1] is ^r to the plane
x + 2y + 2z = 5,
A + 2B + 2C = 0
Since (1) passes through (– 1, 1, 1) and (1, –1, 1),
[3] – [4] gives
From [2], we get
–A+B+C+D=0
A– B+ C+ D=0
– 2A + 2B = 0
fi
3B = – 2C
\ [1] becomes
B=
− 2C
3
\
A=
− 2C
3
\
D=
− 2C 2C
+
−C = – C
3
3
Dharm\C\N-ENGM\EMATH1B.PM5
[1]
[2]
[3]
[4]
A=B
50
Engineering Mathematics – I
\ [1] becomes
–
i.e.
2C
2C
y + Cz – C = 0
x−
3
3
2x + 2y – 3z + 3 = 0
x +1 y − 3 z + 2
and the plane
=
=
2
1
−3
x + 2y + 3z + 4 = 0
67. Find the angle between the line
Solution: Here
\
A = 1, B = 2, C = 3, D = 4
l = – 3, m = 2, n = 1
a = –1, b = 3, g = 2

Al + β m + Cn
q = sin–1 
2
 A + B 2 + C 2 l2 + m2 + n2





1( − 3) + 2(2) + (3)(1)
= sin–1 

 12 + 22 + 32 (− 3)2 + 22 + 12 
68. Show that the lines


4
−1
= sin–1 
 = sin
 14 14 
1
1
(x + 3) =
(y + 5)
3
2
2
7 
 
=–
1
(z – 7)
3
1
1
(n + 1) =
(y + 1) = – (z + 1)
5
4
are coplanar. Find the equation of the plane containing them.
Solution: The given lines are:
and
and
z −7
x+3 y+5
=
=
−3
2
3
[1]
z +1
x +1 y +1
=
=
4
5
−1
[2]
The equation of the plane which contain the lines [1] and is parallel to [2] is
x +3 y +5 z −7
2
3
−3
4
5
−1
=0
= (x + 3) (–3 + 15) – ( y + 5) (–2 + 12) + (z – 7) (10 – 12) = 0
12x + 36 – 10y – 50 – 2z + 14 = 0
6x – 5y – z = 0
[3]
The plane [3] passes thro’ (–1, –1, –1) a point on the line [2], if
–6+5+1=0
or,
0 = 0 which is true.
Hence the two lines are coplanar and the equations of the plane containing the given
two lines is 6x – 5y – z = 0.
Dharm\C\N-ENGM\EMATH1B.PM5
69. Find the equation of the plane which contains the line
the plane x + 2y + z = 12.
x −1 y +1 z − 3
and is ^r to
=
=
2
4
−1
z −3
x −1 y +1
=
=
4
2
−1
Any plane through the given line is of the form
A(x – 1) + B(y + 1) + C(z – 3) = 0
where
2A – B + 4C = 0
r
Again, by data [1] must be ^ to the given plane
x + 2y + z = 12
\
A◊1 + B◊2 + C◊1 = 0
(using a1a2 + b1b2 + c1c2 = 0)
Solution: Given:
[1]
or,
[3]
A + 2B + C = 0
Solving [2] and [3] by the method of cross multiplication, we have
C
A
B
=
=
4 +1
−1 − 8 4 − 2
A
B
C
=
=
5
−9 2
fi
[1]
[2]
Putting these values in [1], we get
– 9 (x –1) + 2 (y + 1) + 3 (2 – 3) = 0
fi
9x –2y – 5z + 4 = 0
This is the eqn of the required plane.
70. Find the equation of the plane through the points (1, 0, –1), (3, 2, 2) and parallel to the
line x – 1 =
1− y z −2
=
.
2
3
Solution: Equation of the plane through (1, 0, –1) is
A(x – 1) + B(y – 0) + C(z + 1) = 0
Since this passes through (3, 2, 2),
we have
2A + 2B + 3C = 0
[1]
[2]
The plane [1] is parallel to the line x − 1 = y − 1 = z − 2 . Normal to this plane is ^r to
1
−2
+3
this line.
Hence
A – 2B + 3C = 0
Solving [2] and [3], we get
A
B
C
=
=
4 −1
−2
From [1] required eqn of the plane is,
4(x – 1) – 1( y – 0) – 2(z + 1) = 0
fi
Dharm\C\N-ENGM\EMATH1B.PM5
4x – y – 2z – 6 = 0
[3]
PART-A
51
Analytical Geometry in Three Dimensions
52
Engineering Mathematics – I
z −3
x y−2
x −2 y−6
z −3
=
=
=
and
=
are coplanar and find
4
2
3
3
1
2
the equation of the plane in which they lie and the point of intersection.
Solution: Condition of coplanarity is
71. Prove that the lines
α′ − α
l
l′
β′ − β γ ′ − γ
m
n
m′
n′
=0
By data, the lines are
z −3
x −0 y−2
=
=
3
1
2
(a, b, g ) = (0, 2, 3)
\
and
z −3
x −2 y−6
=
=
4
2
3
(a ¢, b¢ , g ¢ ) = (2, 6, 3)
(l, m, n) = (1, 2, 3)
(l¢, m¢, n¢ ) = (2, 3, 4)
Substituting, we get
−2 −4 0
1
2 3
2
3 4
= – 2(8 – 9) – (– 4) (4 – 6) π 0
= – 2 (–1) + 4 (– 2) = 2 – 8
=–6π0
fi Condition of coplanarity is not satisfied.
\ the lines are not coplanar and hence not intersecting also.
72. Find the coordinates of a point of intersection of the line
with the plane
x + 2 y −1
z−4
=
=
1
3
−2
x + 5y + 4z – 7 = 0
z−y
x + 2 y −1
=
=
=r
1
3
−2
fi
(3r –2, –2r + 1, r + 4) = P
a parametric point on the line.
then ‘P’ lies in the given plane x + 5y + 4z – 7 = 0
Solution: Let
fi
(3r – 2) + 5 (– 2r + 1) + 4(r + 9) – 4 = 0
fi
– 3r + 12 = 0
fi
r=4
\ P = (10, – 7, 8) are the coordinates of a pt. of intersection.
73. Find the condition that the three straight lines given below are coplanar:
x
y
z x
y z
=
= ;
=
= .
aα bβ cγ l
m n
Dharm\C\N-ENGM\EMATH1B.PM5
x
y z
= = ;
α β γ
Solution: Clearly, origin (0, 0, 0) is a point on all the 3 straight lines. If the 3 are to be
coplanar, they have a common normal.
Let its D.C’s be (A : B : C)
\
Aa + Bb + Cg = 0
Aaa + Bbb + Ccg = 0
Al + Bm + Cn = 0
α
β
aα bβ
l
m
or
γ
cγ
n
=0
fi
aa (bn – gm) – bb (an – gl) + Cg (am – bl) = 0.
74. Show that the two straight lines
x −a y−b z−c
x − a ′ y − b′
z − c′
=
=
=
;
=
a′
b′
c′
a
b
c
are coplanar. Find the point of intersection and the plane containing them.
Solution: We know that two straight lines intersect if we can find some point that lies on
both the given lines.
Consider the point
P = [(a + a¢), (b + b¢), (c + c¢)]
[1]
Examine it with reference to the two lines,
z −c
x − a y −b
=
=
[2]
c′
a′
b′
x − a ′ y − b′ z − c ′
=
=
[3]
a
b
c
( a + a ′ − a ) b + b′ − b c + c ′ − c
=
=
=1
a′
b′
c′
a + a ′ − a ′ b + b′ − b′ c + c ′ − c ′
=
=
=1
and
a
b
c
Hence, P lies on the lines [2] and [3]. Thus, the two lines intersect, and the point of
intersection is
[(a + a¢), (b + b¢), (c + c¢)]
plane containing the lines
x −a
a′
is:
a
y −b z −c
b′
c′
b
c
Apply R + R
1
2

→
fi
=0
x y z
a ′ b′ c ′ = 0
a b c
x (bc¢ – b¢c) + y (ca¢ – c¢a) + z (ab¢ – a¢b) = 0.
75. Find the angle between the line
Dharm\C\N-ENGM\EMATH1B.PM5
x +1 y −3 z + 2
and the plane x + 2y + 3z + 4 = 0.
=
=
2
1
−3
PART-A
53
Analytical Geometry in Three Dimensions
54
Engineering Mathematics – I
Solution: A = 1, B = 2, C = 3, D = 4, l = – 3, m = 2, n = 1, a = –1, b = 3, g = –2
\
 Al + Bm + Cn 


–1 1( − 3) + 2(2) + (3)(1) 
q = sin–1 

 = sin 
2
2
14 14


∑ l 
 ∑ A
4


2 
= sin–1 
 = sin–1   .
 14 14 
7 
x +1 y +1 z +1
76. Show that the lines
and x + 2y + 3z – 8 = 0 = 2x + 3y + 4z – 11
=
=
1
2
3
intersect. Find their point of intersection.
[V.T.U. F/M 2005]
x +1 y +1 z +1
=
=
= k (say), any point on the plane is therefore is x = k – 1,
1
2
3
y = 2k – 1, z = 3k – 1 which lie on the line of intersection of the two planes :
Solution: Let
 (k – 1) + 2 (2k – 1) + 3(3k – 1) – 8 = 0 and 
 2 (k – 1) + 3 (2k – 1) + 4(3k – 1) – 11 = 0 


 14k – 14 = 0 and 20k – 20 = 0



Thus k = 1 and k = 1.
Since the value of k is equal we conclude that the lines intersect and the point of
intersection from (1) is given by x = 0, y = 1, z = 2
\ (0, 1, 2) is the point of intersection.
77. Show that the two straight lines
(3x + 2y – 3z + 2) = 0 = (x + 2y – 2z + 1)
and
(9x – 2y – 3z – 34) = 0 = (2x – z – 9)
are coplanar and find the plane containing them.
Solution: D.C’s of the lines:
Line [1]:
3l + 2m − 3n = 0 


 l + 2m − 2n = 0 
fi
l: m: n=2:3:4
Line [2]:
9l − 2m − 3n = 0 


 2l + 0m − n = 0 
fi
(l : m : n) = 2 : 3 : 4
The D.C’s are same. So the 2 lines are parallel. Hence, they are evidently coplanar.
The plane containing them:
Any plane thro’ to line [2] is of the trun:
7x – 10y + 4z + 1 = 0
Dharm\C\N-ENGM\EMATH1B.PM5
55
Analytical Geometry in Three Dimensions
1. Find the angle between the line
PART-A
EXERCISES
x +1 y z −3
and the plane
= =
2
3
6


–1  18 
 
 Ans: α = sin 
 7 14  

3x + 2y + z = 7
2. Find the angle between planes 2x – y + z = 6 and x + y + 2z = 7
(Ans: a = 90°, plane perpendicular)
3. Find the angle between the lines
x −7 y +3 z −4
and 6x + 4y –5z –4 = 0 = x – 5y + 2z – 1z
=
=
2
1
−1
1


–1  
 Ans: θ = cos  2  = 60° 
4. Prove that the lines
x − 1 y + 1 z + 10
intersect, and find the
=
=
2
−3
8
x − 4 y + 3 z +1
and
=
=
1
7
−4
coordinates of their point of intersection.
5. Show that the lines x + 4 = y + 6 = z − 1 and 3x – 2y + z + 5 = 0 = 2x + 3y + 4z – 4 are
−2
3
5
coplanar. Find their point of intersection and the plane in which they lie.
1.8
SHORTEST DISTANCE BETWEEN SKEW LINES
(Non-intersecting, non-parallel lines in space)
Two lines are skew if and only if they do not lie in a common plane.
To show that the length of the line intercepted between two lines which is perpendicular to both is the shortest distance between them.
Abbre: (shortest distance is usually written as S.D.)
Let
z −γ
x −α y − β
=
=
l
m
n
[1]
z −γ′
x − α′ y − β′
=
=
[2]
n′
l′
m′
Lines lying in two different planes are said to be skew. The shortest distance line is
the line which is perpendicular to both [1] and [2]. It is also equal to the perpendicular
distance from any point on [1], say (a, b, g ) = A, upon the line through [2] parallel to [1].
(see figure)
and
l:m
P
:n
(l : m :
Q
L
B
n)
A
(a, b, g)
(l¢ : m
¢ : n¢)
(a ¢, b ¢, g ¢)
Fig. 1.18
Dharm\C\N-ENGM\EMATH1B.PM5
M
56
Engineering Mathematics – I
Let (l : m : n) be the D.R’s of the line of shortest distance (S.D.). This is perpendicular
to [1] and [2]
(l : m : n ) ^ (l : m : n)
(l : m : n ) ^ (l1 : m1 : n1)
ll + mm + n n = 0 

ll¢ + mm¢ + nn¢ = 0 
\
\
[3]
λ
µ
ν
=
=
(mn′ − m′n ) (nl′ − n′l )
(lm′ − l′m)
1
Putting each equal to   , we get
k
mn¢ – m¢n = lk
nl¢ – n¢l = mk
lm¢ – l¢m = kn





[4]
[5]
Again, the plane through the line [2], parallel to [1] is given by
x − α′ y − β ′ z − γ ′
l′
m′
n′
l
m
n
=0
[6]
[x – a ¢] [m¢n – mn¢] – [y – b¢] [l¢n – ln¢] + [z – g ¢] [l¢m – lm¢] = 0
[x – a ¢] [mn¢ – m¢n] + [y – b ¢] [l ¢n – ln¢] + [z – g ¢] [l¢m – lm¢] = 0
i.e.
[x – a¢] lk + [y – b¢]mk + [z – g ¢] nk = 0
fi
l [x – a ¢] + m [y – b ¢] + g [z – g ¢] = 0
Shortest distance (S.D.)
= PQ =
λ[α − α ′] + µ[ β − β ′] + ν [γ − γ ′]
[λ 2 + µ 2 + ν 2 ]
[7]
[8]
To find the equations of the S.D. line
The S.D. line PQ may be regarded as the common line of intersection of the planes
APQ and BPQ.
The plane APQ passes through AP (i.e. [1]) and is parallel to PQ [i.e. to the direction
(l : m : n)]
Hence its equation is given by
x −α
l
y−β
m
z −γ
n
λ
µ
ν
=0
[9]
Similarly, the plane BPQ, passes through the line [2] is parallel to the direction
[l : m : n]
\ Its equation is
n − α′ y − β ′ z − γ ′
l′
m′
n′
=0
λ
µ
ν
Dharm\C\N-ENGM\EMATH1B.PM5
[10]
WORKED EXAMPLES
78. Find the magnitude and equations to the line of shortest distance between the lines
x
y+9 z−2
x −6 y−7 z −4
and
.
[V.T.U. J/A 2003]
=
=
=
=
2
4
−3
3
1
−1
Solution. Let l, m, n be the DC’s of the line of shortest distance between the two lines.
Since S.D. is perpendicular to both the lines, we have
3l – m + n = 0
– 3l + 2m + 4n = 0
l m
n
1
1
Solving,
=
=
=
=
2 5 −1
30
22 + 52 + 12
Length of S.D.
= (6 – 0)
2
+ (7 + 9) –
30
Equation of the line of shortest distance is
x −6 y −7 z −4
3
1
1
2
5
−1
x
and
−3
2
y+9 z−2
2
4
5
−1
5

1 
+ (4 – 2)  −
 = 3 30
30
30 

= 4x – 5y + 17z + 79 = 0
= 22x – 5y + 19z – 7 = 0.
79. Find the shortest distance between the lines.
x −3
y−2
=
=
3
4
P1 = (3x + 5y
P2 = x – 2y +
Any plane through [2] is:
 z −1


 −2 
+ 4z + 7) = 0
8z + 2 = 0
[1]
[2]
P1 + kP2 = 0
fi x(3 + k) + y (5 – 2k) + 2(4 + 8k) + (7 + 2k) = 0
[3] is parallel to [1], if normal is perpendicular to [1]
i.e.
fi
Plane:
fi
of
(3 + k) 3 + (5 – 2k) 4 + (4 + 8k) (– 2) = 0
k=1
x (3 + 1) + y (5 – 2) + z (4 + 8) + (7 + 2) = 0
4x + 3y + 12z + 9 = 0
S.D. = perpendicular from any point of [1] i.e.
(3, 2, 1) on [5]
=
Dharm\C\N-ENGM\EMATH1B.PM5
4 ⋅ 3 + 3 ⋅ 2 + 12 ⋅ 1 + 9
2
2
2
4 + 3 + 12
=
39
=3
13
[3]
[4]
[5]
PART-A
57
Analytical Geometry in Three Dimensions
58
Engineering Mathematics – I
80. Find the length and equation of line of shortest distance between the lines
3x + 2y + 4z – 8 = 0 

x – 4y + z + 2 = 0 
[1]
y − 4 z −1
x −5
=
[2]
=
−2
+2
3
Solution: Here, the length as well as the equation is required. So, we reduce the eqn [1] also
to the symmetrical form, first: Line [1]
and
3x + 2y + 4z – 8 = 0 = x – 4y + z + 2
D.C’s: 3l + 2m + 4n = 0
l – 4m + ln = 0
l : m : n = 18 : 1 : –14
[1]
[2]
To find point
Put z = 0:
3x + 2y – 8 = 0
x – 4y + 2 = 0
Solve
x = 2, y = 1, \ point [2, 1, 0].
x−2
y −1 z − 0
=
=
St. line [1]:
18
1
− 14
[3]
[4]
y − 4 z −1
x −5
=
=
2
−2
3
St. line (2):
[5]
Length of S.D.
Let D.C’s be (l : m : n)
\
18l + m – 14 n = 0
3l – 2m + 2n = 0
l : m : n = – 26: – 78: – 39 = 2 : 6 : 3
S.D =
=
[6]
λ[α − α ′] + µ[ β − β ′] + [ν − ν ′]
λ2 + µ2 + ν 2
2(2 − 5) + 6(1 − 4) + 3(0 − 1)
22 + 62 + 32
= 21/7 = 3 units
Eqn:
x −α
l
y−β
m
z −γ
n
λ
µ
ν
x − 2 y −1 z
18
1
−14
2
6
3
=0=
( x − α ′) ( y − β ′) ( z − γ ′)
l′
m′
n′
λ
=0=
µ
x − 5 y − 4 z −1
3
2
−2
2
6
3
fi
87x – 82y + 106z – 92 = 0 = 11x + 5y – 22z – 18
81. Find the shortest distance between the axis of z and the line
x + 2y + z + 1 = 0 = 2x + 2y – z + 9
Dharm\C\N-ENGM\EMATH1B.PM5
ν
Solution: The z-axis passes thro’ the origin (0, 0, 0); and has D.C’s (0 : 0 : 1).
Hence, it can be written
The second st. line is
y−0 z −0
x −0
=
=
0
0
1
P1 = x + 2y + z + 1 = 0 
[1]
[2]

P2 = 2x + 2y – z + 9 = 0 
The shortest distance between the two lines [1] and [2] is the length of the ^r from
any point on [1] say (0, 0, 0) on the plane thro’ – [2], parallel to [1]
Any plane thro’ [2] is of the form P1 + kP2 = 0
(x + 2y + z + 1) + k (2x + 2y – z + 9) = 0
is
fi
x (1 + 2k) + y(2 + 2k) + z(1 – k) + (1 + 9k) = 0
This is to be parallel to the line [1]; hence the normal to [3] is ^r to [1].
fi
(1 + 2k) : (2 + 2k) : (1 – k)
^r to (0 : 0 : 1)
fi
(1 + 2k) 0 + (2 + 2k) 0 + (1 – k) (1) = 0
fi
k=1
Hence, the plane is
[from [3] and [4]]
x (1 + 2 ◊ 1) + y(2 + 2 ◊ 1) + 2(1 – 1) + (1 + 9 ◊ 1) = 0
fi
3x + 4y + 10 = 0
The S.D. = perp. from (0, 0, 0) on [5]
=
D
A2 + B2 + C2
=
− 10
32 + 4 2 + 0 2
=
− 10
5
[3]
[4]
[5]
= 2 units
82. Find the coordinates of the point of intersection of the line of S.D. with the lines
x +3 y−6 z −3
x y−6
z
and
hence find its length.
=
=
=
=
2
3
2
2
−2
−1
Solution: Suppose
and
\
x +3 y−6 z −3
=
=
= p (say)
2
3
−2
[V.T.U. F/M 2005]
L1
2 : 3 : –2
P
x y−6
z
=
=
= q (say)
2
2
−1
x = 2p – 3, y = 3p + 6,
z = – 2p + 3
and any point on L1, L2 and x = 2q, y = 2q + 6, z = – q
SD = PQ
...(1)
...(2)
Q
L2
Fig. 1.19
If P and Q are on L1, L2 respectively, then,
\ DR’s of PQ = 2q – 2p + 3, 2q – 3p, – q + 2p – 3
PQ being the line of S.D. is perpendicular to L1 whose D.R.s are (2 : 3 : – 2) and also to
L2 whose d.r.s are (2 : 2 : – 1)
\
and
 2 (2q – 2p + 3) + 3 (2q – 3p) – 2 (– q + 2p – 3) = 0 


 2 (2q – 2p + 3) + 2 (2q – 3p) – 1 (– q + 2p – 3) = 0 
Dharm\C\N-ENGM\EMATH1B.PM5
PART-A
59
Analytical Geometry in Three Dimensions
60
Engineering Mathematics – I
or
12q – 17p + 12 = 0 and 9q – 12p + 9 = 0
on solving, we get p = 0 and q = – 1
Substituting the value of p in (1) we get P = (– 3, 6, 3) and substituting the value of q
in (2) we get Q = (– 2, 4, 1).
\
S.D. = Distance between P and Q (nearest to each other)
( − 2 + 3)2 + (4 − 6)2 + (1 − 3)2 =
83. If 2d is the shortest distance between the lines
=
9 = 3 units.
y/b + z/c = 1, x = 0 and
x/a – z/c = 1, y = 0
P.T.
1
+
1
2
+
1
2
=
1
b
c
d2
a
Solution: The symmetric form of the eqns of the lines are:
2
x
y −b
z 
=
=
0
a
− c 

y z
x−a

=
=

0 c
a
and
and
[1]
\ if (l : m : n) be the D.C’s of the shortest distance, then
l ◊ 0 + m ◊ b + n (– c) = 0
l◊a+m◊0+ n◊c=0
n
−m
l
=
=
\
ac
− ab
bc
We find from [1], that B(0, b, 0), A (a, 0, 0) lie on the two lines.
Hence S.D. between the two lines
= projection of BA on the line (l : m : n)
( a − 0) bc + (0 − b) ( − ac) + (0 − 0) ( − ab)
=
(b2c 2 + c 2 a 2 + a 2b2 )
2abc
=
b2c 2 + c2 a 2 + a 2b2
This is given to be = 2d.
\
fi
1
d
2
=
b2 c2 + c2 a2 + a2b2
2 2 2
a b c
=
1
a
2
+
1
2
b
+
1
c2
1
1
1
1
+
+
= 2
a 2 b2 c 2
d
EXERCISES
1. Find the S.D. between the lines
y−7 z −3
x −5
=
=
7
3
− 16
Dharm\C\N-ENGM\EMATH1B.PM5
and
y − 13 z − 15
x−9
=
=
3
8
−5
(Ans: 14)
61
Analytical Geometry in Three Dimensions
2x – 2y + 3z – 12 = 0 = 2x + 2y + z
2x – z = 0 = 5x – 2y + 9
(Ans: 6.)
3. Show that the S.D. between any two opposite edges of the tetrahedron formed by the planes
y + z = 0; z + x = 0; x + y = 0; x + y + z = a is
2a
6
.
4. Find the length and equations of the line of S.D. between the lines
x − 15 y − 29 z − 5
=
=
.
3
8
−5
(Ans: 14; 117x + 4y – 41z – 490 = 0 = 9x – 4y – z – 14)
5. Find the length and equations of S.D. between the two lines
and
x − 8 y + 9 z − 10
and
=
=
3
7
− 16
x −1 y − 3
=
= z+2
2
4
8


; x − y + 2z + 6 = 0 = 19x 
 Ans.
14


3x – y – 2z + 4 = 0 = 2x + y + z + 1
1.9
RIGHT CIRCULAR CONE AND RIGHT CIRCULAR CYLINDER
(A) Right Circular Cone
A right circular cone is the surface generated by a straight line revolving about another line
which is fixed, the two lines intersecting at a constant angle q. The fixed line is called the
‘axis’ of the cone; the point of intersection is the ‘vertex’ of the cone; the angle between the
two lines is ‘semi-vertical angle’ of the cone. The straight line which revolves and generates
the cone is called the ‘generator’ of the cone.
B
,z )p
(x 1, y 1 1
A
(a, b, g)
q
M l:m:n C
P¢
B¢
Fig. 1.19
To find equations of a right circular cone
x −α y − β z −γ
[1]
=
=
l
m
n
be the axis, the axis has the D.C’s (l : m : n) of the cone.
Let p(x1, y1, z1) be any point on the cone; and AP makes an angle ‘q ’ with the axis [1].
Let A (a, b, g) be the vertex of the cone,
D.R’s of AP = (x1 – a) : (y1 – b) : (z1 – g)
D.R’s of Axis = (l : m : n)
The two are inclined at an angle q,
where
cos q =
\
cos2q =
( x1 − α )l + ( y1 − β )m + ( z1 − γ )n
2
(l + m2 + n2 ) [( x1 − α )2 + ( y1 − β )2 + ( z1 − γ )2 ]
2
[( x1 − α )l + ( y1 − β )m + ( z1 − γ )n ]
(l + m 2 + n2 ) [( x1 − α )2 + ( y1 − β )2 + ( z1 − γ )2 ]
Dharm\C\N-ENGM\EMATH1B.PM5
PART-A
2. Find the shortest distance between the lines
62
Engineering Mathematics – I
or dropping the suffixes, we get
[l (x – a) + m (y – b) + n [z – g )]2 = cos2 q ( p2 + m2 + n2) [(x – a)2 + (y – b)2 + (l – g)2]
To find the equation of the right circular cone, whose verZ
tex is at the origin, and axis along a line of (z–axis), D.R’s
(l : m : n), and whose semivertical angle is equal to a
P(x, y, z)
Vertex = (0, 0, 0)
P = (x, y, z), any point on RCC
a
\
ZO$ P = a.
D.R’s of OP are:
O(0, 0, 0)
X
[(x – 0) : (y – 0) : (z – 0)] = (x : y : z)
D.C’s of the axis (z-axis) of the cone are (0 : 0 : 1)
\
or
cos a =
cosa =
Y
Fig. 1.20

Q cos θ = a1 a2 + b1b2 + c1 c2

∑ a12 ∑ a22

x (0) + y(0) + z(1)
x 2 + y2 + z 2
z
(x2 + y2 + z2) cos2a = z2
fi
2
x + y2 + z 2




z2
x2 + y2 + z2 =
= z2 sec2a
cos2 α
x2 + y2 = z2 sec2a – z2 = z2 (sec2a – 1)
\
\
x2 + y2 = z2 tan2 a
or
is the equation of the cone.
WORKED EXAMPLES
84. Find the equation of the right circular cone which passes through the point (2, 1, 3), has
its vertex at (1, 1, 2) and axis parallel to the line.
[V.T.U. J/A 2003]
x − 2 y −1 z + 2
=
=
2
3
−4
Solution: DR’s of axis are 2, – 4, 3. DR’s of the
generator AB are 1, 0, 1. If a is the semi-vertical
angle between the axis and AB, then
coa a =
=
2 . 1 + ( − 4) . 0 + 3 . 1
P(x, y, z)
A
(1, 1, 2)
axis
a
1 + 0 + 1 4 + 16 + 9
B(2, 1, 3)
5
(1)
2 29
If P(x, y, z) is any point on the cone, then DR’s of AP are x – 1, y – 1, z – 2. So
cos a =
( x − 1)2 + ( y − 1)( − 4) + ( z − 2)3
2
2
2
( x − 1) + ( y − 1) + ( z − 2)
29
=
5
2
29
from (1)
Squaring, the required equation of cone is
2[2(x – 1) – 4(y – 1) + 3(z – 2)]2 = 25[(x – 1)2 + (y – 1)2 + (z – 2)2].
Dharm\C\N-ENGM\EMATH1B.PM5
Analytical Geometry in Three Dimensions
63
85. Find the equation of a right circular cone whose vertex is at the point (1, 2, 3) and whose
x −1 y − 2 z − 3
, given that its semivertical angle is 30°.
=
=
2
3
4
PART-A
axis is the straight line
Vertex = (1, 2, 3);
x −1 y − 2 z − 3
[1]
=
=
2
3
4
Let P = (x, y, z) be any point on the cone. Then
D.R’s of PA = (x – 1) : (y – 2) : (z – 3)
[2]
D.R’s of Axis = 2 : 3 : 4
[3]
For any point P (x, y, z) on the cone. PA makes an angle 30°, with the axis of the cone.
Axis:
3
= cos 30°
2
\
=
( x − 1) 2 + ( y − 2)3 + ( z − 3) 4
2
(2 + 32 + 42 [( x − 1)2 + ( y − 2)2 + ( z − 3)2
(2x + 3 y + 4z − 20)2
3
=
29 [( x − 1)2 + ( y − 2)2 + ( z − 3)2 ]
4
87 [(x – 1)2 + ( y – 2)2 + (z – 3)2]
fi
= 4 (2x + 3y + 4z – 20)2
86. Find the equation of the right circular cone of semi vertical angle 30°, whose vertex is at
the point (1, 2, 3) and whose axis is parallel to the line (x = y = z).
We know that the equation of the cone with vertex (a, b, g) axis with D.C’s (l : m : n)
and semivertical angle q,
[l (x – a) + m (y – b) + n (z – g)]2
Here vertex
\
Axis has
\
= cos2 q (l2 + m2 + n2) [(x – a)2 + ( y – b)2 + [z – g)2]
(a, b, g) = (1, 2, 3).
[1]
(a = 1, b = 2, g = 3)
D.C’s = (1 : 1 : 1)
q = 30°
[1(x – 1) + 1 ( y – 2) + 1 (z – 3)]2
= (cos2 30°) (12 + 12 + 12) [(x – 1)2 + ( y – 2)2 + (z – 3)2]
3
(x + y + z – 6) 2 =   ◊ 3 [(x – 1)2 + ( y – 2)2 + (z – 3)2]
4
2
4 (x + y + z – 6) = 9[(x – 1)2 + ( y – 2)2 + (z – 3)2]
87. Find the equation of the right circular cone whose vertex is (3, 1, 2) axis has D.C’s
i.e.
 1 
1 : 2 : 3 and the semi-vertical angle is cos–1 
.
 7
Solution: V = Vertex = (3, 1, 2).
Let P(x, y, z) be any general point on the surface of the cone. The D.R’s of VP are
(x – 3) : ( y – 1) : (z – 2).
Dharm\C\N-ENGM\EMATH1C.PM5
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Engineering Mathematics – I
D.R’s of axis are given to be
(1 : 2 : 3)
Angle between them = semi-vertical angle q
1
= cos–1
.
7
1( x − 3) + 2 ( y − 1) + 3 ( z − 2)
cos q =
q = cos–1
But
\
Hence
12 + 22 + 33
7
7
q
v
P(x, y, z)
V(3, 1, 2)
Fig. 1.21
( x − 3)2 + ( y − 1)2 + ( z − 2)2
;
1
cos q =
1
1
A(1, 2, 3)
7
x + 2 y + 3z − 11
=
14 ⋅ ( x − 3)2 + ( y − 1)2 + ( z − 1)2
Squaring, cancelling 7 and cross-multiplying,
2 [(x – 3)2 + (y – 1)2 + (z – 2)2] = 17y + 20z + 2 = (x + 2y + 3z – 11)2
is the required eqn of cone.
88. Find the equation of the right circular cone with origin as vertex, and the axis of
coordinates as three generators.
Solution: Given that the axis coordinates are 3 generators, the axis of the cone has to make
equal angles with the coordinate axes.
1
\ we can take the D.R’s of axis as 1 : 1 : 1. It makes angles cos–1
with x, y, z axes.
3
1
\ Semivertical angle q = cos–1
3
1
i.e.
cosq =
3
Let P(x, y, z) be any general point on the cone.
D.R’s of the line joining P to the vertex V (0, 0, 0) are
(x – 0) : (y – 0) : (z – 0) = x : y : z D.R’s of axis are (1 : 1 : 1)
\
cos q =
But
cosq =
1
3
Simplifying
fi
fi
=
(1) ( x ) + (1) ( y ) + (1) ( z )
12 + 12 + 12
x 2 + y2 + z 2
1
3
x+ y+z
3 ⋅ x 2 + y2 + z 2
x2 + y2 + z2 = (x + y + z)2
2xy + 2yz + 2zx = 0
xy + yz + zx = 0 is the eqn of the cone.
Dharm\C\N-ENGM\EMATH1C.PM5
65
89. Find the equation of the right circular cone whose vertex is at the origin, whose axis is
the line x = – 2y = z and which has semivertical angle 45° V = (0, 0, 0) is the vertex.
P(x, y, z) be any general point on the surface of the cone. Then VP is generator and if
makes the angle 45° with the axis.
or
D.R’s of VP = (x – 0) : (y – 0) : (z – 0) = x : y : z
Eqns. of axis are
x = – 2y = z
y
z
x
=
=
(on dividing by –2)
1 −2
−2
\ D.R’s it axis are –2 : 1 : – 2
Angle between them = 45°
\
fi
cos 45° =
− (2x − y + 2z )
9 ⋅ x 2 + y2 + z 2
9(x2 + y2 + 22) = 2(2x – y + 2z)2 is the equation of the cone.
π
, and vertex (0, 0, 1)
4
given that the axis of the cone is parallel to the x-axis. Also, show that, the section of the
above cone by the plane z = 0 is a rectangular Hyperbola.
90. Find the equation of the right circular cone of semi-vertical angle
Solution: Axis is parallel to x-axis.
\
D.C’s = (1 : 0 : 0)
In the eqn of the general cone we put (a : b : g ) = (0 : 0 : 1)
and
i.e.
1 

π
θ = 4 , cos θ =

2

(l : m : n) = (1 : 0 : 0)
[Sl(x – a)]2 = cos2q S(l 2) [S(x – a)2]
[1]
 1  2
2
2
2
2+
2
[1(x – 0) + 0 + 0]2 = 
 (1 + 0 + 0 ) [(x – 0) + ( y – 0) (z – 0) ]
 2
fi
2x2 = x2 + y2 + (z – 1)2 i.e. x2 – y2 – (z – 1)2 = 0
To find the section by XOY plane, put z = 0 in [2], we get
x2 – y2 – (0 – 1)2 = 0
i.e.
x2 – y2 = 1
(a rectangular Hyperbola)
91. Find the cone got by rotating the straight line (z = 0, y = 2x) about the x–axis.
Let us put the given straight lines the Std. symmetrical form.
x−0 y−0 z−0
.
=
=
1
2
0
fi (0, 0, 0) is a point on the line as well as on the x-axis (axis of rotation).
Hence the vertex is at the origin.
Axis of the cone = x–axis;
Solution: Given: (z = 0, y = 2x);
has D.C’s = (1 : 0 : 0)
Dharm\C\N-ENGM\EMATH1C.PM5
[2]
PART-A
Analytical Geometry in Three Dimensions
66
Engineering Mathematics – I
Semi-vertical angle between
(1 : 2 : 0) and (1 : 0 : 0)
1 ⋅1 + 2 ⋅ 0 + 0 ⋅ 0
cosq =
12 + 02 + 02
12 + 22 + 52
=
1
5
cos2q = (Sl2) [S(x – a)2] = [Sl (x – a)]2
\
(1/5) (12 + 02 + 02) [x2 + y2 + z2] = [1 (x – 0) + 0(y – 0) + 0 (z – 0)]2
\
(x2 + y2 + z2) = 5(x2)
4x2 – y2 – z2 = 0
92. The axis of right circular cone, vertex 0, makes equal angles with the coordinate axes,
and the cone passes through the line drawn through 0, with D.C’s proportional to (3 : 4 : 5).
Find the eqn of the cone
D.C’s of axis = (1 : 1 : 1)
Vertex = (0, 0, 0).
Semivertical angle is the same as that between (3 : 4 : 5) and (1 : 1 : 1).
\
cos q =
3 ⋅1 + 4 ⋅1 + 5 ⋅ 1
2
2
2
1 +1 +1
2
2
2
(3 + 4 + 5 )
=
12
150
=
2 6
5
Eqn of cone is
cos2 q (l2 + m2 + n2) [S(x – a)2] = [Sl(x – a)]2
(24/25) [1 + 1 + 1] [x2 + y2 + z2] = [1 (x – 0) + 1 (y – 0) + 1[z – 0)]2
72 (x2 + y2 + z2 ) = 25 (x + y + z)2
93. Find the equation of the right circular cone with vertex (2, – 3, – 4), semivertical angle
30° and whose axis is equally inclined to the coordinate axes.
[V.T.U. F/M 2005]
Solution: It is given that the axis of the cone is equally inclined with the coordinate axes
we have basically a = b = g fi cos a (l) = cos b (m) = cos g (n)
Since l2 + m2 + n2 = 1
fi
we get 3l2 = 1 or l = 1/ 3
1
l= m= n=
Hence, DR’s of the axis of cone are 1, 1, 1.
3
By question, V = (2, – 3, – 4) is the vertex of the cone and let P (x, y, z) be any point on
the cone.
\ DR’s of VP = (x – 2 : y + 3 : z + 4).
Also D.R.’s of the axis = 1, 1, 1.
Since 30° is the angle between these two lines we have,
cos 30° =
3
=
2
i.e.,
or
1( x − 2) + 1( y + 3) + 1( z + 4)
1 + 1 + 1 ( x − 2)2 + ( y + 3)2 + ( z + 4)2
x + y+z+5
3
( x − 2)2 + ( y + 3)2 + ( z + 4)2
9 [(x – 2)2 + (y + 3)2 + (z + 4)2] = 4 (x + y + z + 5)2
Dharm\C\N-ENGM\EMATH1C.PM5
94. If a is the semi-vertical angle of the right circular cone which passes thro’ the lines OX,
OY, x = y = z, show that
cos a = (9 – 4 3 )–1/2
Solution: If (l : m : n) are the D.C’s of the axis of the cone, since the axis makes the same
angle a with each of the lines OX (D.C’s 1, 0, 0), OY (D.C’s 0, 1, 0) and x = y = z (D.C’s
proportional to 1 : 1 : 1)
\
cos a = l(1) + m (0) + n (0) = l (0) + m (1) + n(0)
=
1 +1 +1
l = cos a, m = cos a
fi
and
l(1) + m(1) + n(1)
l+ m+ n=
3 cosa
or
n=
3 cosa – m – l
= 3 cos a – 2 cosa = ( 3 – 2) cos a
But l2 + m2 + n2 = 1.
\
cos2a + cos2a + ( 3 – 2)2 cos2a = 1
or
[(1 + 1 + ( 3 – 2)2] cos2a = 1
or
(2 + 3 + 4 – 4 3 ) cos2a = 1
or
(9 – 4 3 ) cos2a = 1
or
cos2a =
cosa =
\
Hence the result.
1
(9 − 4 3 )1 / 2
1
9−4 3
= (9 – 4 3 )–1/2
EXERCISES
Find the equation of the right circular cone for which:
x y z
= = ; semivertical angle = 30°.
4 2 3
(Ans: 4(4x + 2y + 3z)2 = 87 (x2 + y2 + z2)
(ii) Vertex = (0, 0, 0), axis: 2x = 3y = 4z; semivertical angle = 45°.
(Ans: 2(6x + 4y + 3z)2 = 61(x2 + y2 + z2)
(iii) Vertex = (2, 3, 5), axis makes equal angles with the coordinate axis and the semivertical
(i) Vertex = (0, 0, 0), axis:
angle is cos–1
2
.
3
(Ans: 3(x + y + z – 10)2 = 2[(x – 2)2 + (y – 3)2 + (z – 5)2])
(B) RIGHT CIRCULAR CYLINDER
Definition: A right circular cylinder is the surface generated by a line (generator) revolves
about a fixed line (axis), parallel to it, at a constant distance from it.
Thus any plane perpendicular to the axis of a right circular cylinder, cuts the cylinder
in a circle.
Equations of a Right Circular Cylinder (R.C.C.)
In order to find the equation of R.C.C. we must know the equations of the axis, a point
on the axis, D.R’s of the axis and base radius r of the cylinder.
Dharm\C\N-ENGM\EMATH1C.PM5
PART-A
67
Analytical Geometry in Three Dimensions
68
Engineering Mathematics – I
Let A(a, b, g) be a point on the axis of the cylinder whose equation is
x −α y − β z −γ
=
=
=
l
m
n
[1]
P(x, y, z)
A
(a, b, g)
y–b
z–g
x–a
=
=
m
n
l
B
M
Fig. 1.22
Let P(x, y, z) be any point on the cylinder.
Draw PM^r on the axis AB.
y − β z −γ
x −α
=
=
l
m
n
A(a, b, g ) is a point on the line [1]
$ is st. angle
In DAPM, M
AP2 = AM2 + MP2
But
AP =
( x − α )2 + ( y − β )2 + ( z − γ )2
AM = projection of AP on AB
=
and
(x − α ) l + ( y − β ) m + (z − γ ) n
l 2 + m2 + n 2
MP = r
\ From [2],
2
2
 ( x − α )2 + ( y − β )2 + ( z − γ )2  =  ( x − α ) l + ( y − β ) m + ( z − γ ) n  + r2


l 2 + m 2 + n2


fi
2
2
2
(x – a) + (y – b) + (z – g) =
[( x − α ) l + ( y − β ) m + ( z − γ ) n]2
l2 + m2 + n2
which is the required equation of the right-circular cylinder.
+ r2
WORKED EXAMPLES
95. Find the equation of the right circular cylinder of radius 2 whose axis is
x −1
y z −3
=
=
3
1
2
Solution: Axis:
x −1
y z −3
=
=
3
1
2
passes thro: A (1, 0, 3)
D.R’s: (2 : 3 : 1)
If P(x, y, z) be any point on the surface of the cylinder and PM be drawn perpendicular to the axis.
Dharm\C\N-ENGM\EMATH1C.PM5
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Analytical Geometry in Three Dimensions
( x − 1)2 + ( y − 0)2 + ( z − 3)2
\
AM = projection of line joining
A (1, 0, 3) to P (x, y, z) on the axis
D.R’s are (2 : 3 : 1)
AM =
\
=
PART-A
AP =
l( x 2 − x1 ) + m ( y2 − y1 ) + n ( z2 − z1 )
l 2 + m2 + n 2
2( x − 1) + 3 ( y − 0) + 1 ( z − 3)
2
2
2
2 + 3 +1
=
2x + 3 y + z − 5
14
Also MP = radius of the cylinder = 2.
By pythogorean theorem
AP2 = AM2 + MP2
2
 2x + 3 y + z − 5 
2
fi
(x – 1) + (y – 0) + (z – 3) = 
 +2
14


14 [x2 + y2 + z2 – 2x – 6z + 10] = (2x + 3y + z – 5)2 + 56
fi
10x2 + 5y2 + 13z2 – 6yz – 4zx –12xy – 8x + 30y – 74z + 59 = 0
96. Find the equation of the right circular cylinder whose axis passes through the origin and
2
generator is the line
2
2
x −3 y z −4
.
= =
2
3
6
[V.T.U. J/A 2003]
Solution: The generator passes through the point A(3, 0, 4)
2 3 6
, , . The radius of the
and has D.R’s. 2, 3, 6 or D.C.’s
7 7 7
cylinder r = AO is the perpendicular distance from O(0, 0, 0) to
the line AP. Thus r2 = (x1 – a)2 + (y1 – b)2 + (z1 – g)2 – [l(x1 – a)
+ m(y – b) + n(z – g)]2. Here a = 0, b = 0, g = 0, x1 = 3, y1 = 0,
z1 = 4, l =
A(3, 0, 4)
P(x, y, z)
r
C
(0, 0, 0)
2
2
2
3
6
24 
6
 30 
,m=
, n = . So r2 = 32 + 02 + 42 –  + 0 +
or r2 = 25 –   .

7
7
7
7 
7
 7 
Equation of cylinder of radius r is
(x – a)2 + (y – b)2 + (z – g2)
=
l( x − α ) + m( y − β ) + n( z − γ )2
l2 + m2 + n2
2
+ r2
3
6 
2
2
 x + y + z
7
7
7

 + 25 −  30 
x2 + y2 + z2 =
 
1
 7 
45x2 + 40y2 + 13z2 – 12xy – 36yz – 24xz = 325.
97. Find the radius of cross section of the Right Circular cylinder passing through the point
(5, 4, – 1) with the axis along the line
x −1
y−0 z
= .
=
9
5
2
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Engineering Mathematics – I
Solution: The axis is given by
\
M
x −1 y − 0 z − 0
=
=
2
9
5
A(1, 0, 0) is a point on the axis & D.R’s are (2 : 9 : 5)
Now, AP =
(5 − 1)2 + (4 − 0)2 + (1 − 0)2 =
33
=
P(5, 4, – 1)
90°
A(1, 0, 0)
AM = projection of the line joining (1, 0, 0) to P(5, 4,
– 1) on the axis whose D.R’s (2 : 9 : 5)
=
M
l( x 2 − x1 ) + m( y2 − y1 ) + n( z2 − z1 )
Fig. 1.23
l 2 + m2 + n 2
2(5 − 1) + 9(4 − 0) + 5( − 1 − 0)
2
2
2 +9 +5
2
=
39
110
By pythagoras, theorem:
PM2 = AP2 – AM2
=
PM =
\
(
33
)
2
2
2109
 39 
−
 = 110
 110 
2109
is the reqd. radius
110
EXERCISES
Find the equation of the right
(i) radius is 2; axis passes thro
(ii) radius is 2, axis passes thro
[Ans:
circular cylinder for which :
(1, 2, 3) has D.R’s (2 : – 3 : 6)
(1, –3, 2) and was D.R’s (2 : –1 : 5).
(i) 49 [(x – 1)2 + ( y – 2)2 + (z – 3)2] – (2x – 3y + 5z – 14)2 = 196
(ii) 30 [(x – 1)2 + ( y + 3)2 + [z – 2)2] – (2x – y + 5z – 15)2 = 120
(iii) Find the equations of the right circular cylinder whose axis passes thro’ the origin, and
which has
x −3 y z −4
= =
as a generator.
2
3
6
[Ans: 49 (x2 + y2 + z2) – (2x + 3y + 6z)2 = 325]
(iv) Find the R.C.C for which
(a) r = 1,
x −1 y − 2 z − 3
x −1 y z − 3
=
=
= =
(b) r = 2
.
2
1
2
2
3
1
[Ans: (a) 10x2 + 5y2 + 13z2 – 6yz – 4zx – 12 xy – 8x + 30y – 74z + 101 = 0
(b) 5x2 + 8y2 – 4xy – 4yz –8zx + 22x – 16y – 14z – 10 = 0]
Dharm\C\N-ENGM\EMATH1C.PM5