M3D – 1
THREE DIMENSIONAL GEOMETRY
C1A Coordinates of a point in space
There are infinite number of points in space. We want to identify each and every point of space with the
help of three mutually perpendicular coordinates axes OX, OY and OZ.
C1B
Vector representation of a point in space
If coordinate of a point P in space is (x, y, z) then the position vector of the point P with respect to the origin
is xî  y ĵ  zk̂ .
C2
Distance formula
Distance between any two points (x1, y1, z1) and (x2, y2, z2) is given as
C3
( x1  x 2 ) 2  ( y 1  y 2 ) 2  ( z 1  z 2 ) 2
Distance of a point P from coordinates axes
Let PA, PB and PC are distances of the point P(x, y, z) from the coordinate axes OX, OY and OZ
respectively then
PA  y 2  z 2 , PB  z 2  x 2 , PC  x 2  y 2
C4
Section Formula
If point P divides the distance between the points A(x1, y1, z1) and B (x2, y2, z2) in the ratio of m : n, then
coordinates of P are given as
 mx 2  nx1 my 2  ny 1 mz 2  nz 1 
,
,


mn
mn 
 mn
 x  x 2 y 1  y 2 z1  z 2 
,
,

Note : Mid point is  1
2
2 
 2
C5
Centroid of a triangle ABC
 x  x 2  x 3 y 1  y 2  y 3 z1  z 2  z 3 
G 1
,
,

3
3
3


C6
Incentre of triangle ABC
 ax 1  bx 2  cx 3 ay 1  by 2  cy 3 az 1  bz 2  cz 3 
,
,


abc
abc
abc


C7
Centroid of a tetrahedron
A (x1, y1, z1) B (x2, y2, z2) C (x3, y3, z3) and D (x4, y4, z4) are the vertices of a tetrahedron then coordinates of


the centroid is 

C8
x , y , z
i
4
i
4
4
i

.


Direction cosines and direction ratios
(i)
Direction cosines : Let , ,  be the angles which a directed line makes with the positive
directions of the axes of x, y and z respectively, then cos , cos , cos  are called the direction
cosines of the line. The direction cosines are usually denoted by (l, m, n).
(ii)
If l, m, n be the direction cosines of a line, then l2 + m2 + n2 = 1.
(iii)
Direction ratios : Let a, b, c be proportional to the direction cosines l, m, n then a, b, c are called
the direction ratios.
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 2
If a, b, c are the direction ratios of any line L then a î  b ĵ  ck̂ will be a vector parallel to the line
L.
If l, m, n are direction cosines of line L then l î  mĵ  nk̂ is a unit vector parallel to the line L.
(iv)
If l, m, n be the direction cosines and a, b, c be the direction ratios of a vector, then

l 



l 


or
(v)
a
a2  b2  c2
a
a2  b2  c2
b
,m 
,m 
a2  b 2  c2
b
a2  b2  c2
,n 
,n 


2
2
2 
a b c 
c


2
2
2 
a b c 
c
If OP = r, when O is the origin and the direction cosines of OP are l, m, n then the coordinates of
P are (lr, mr, nr).
If direction cosines of the line AB are l, m, n, |AB| = r, and the coordinates of A is (x1, y1, z1) then
the coordinates of B is given as (x1 + rl, y1 + rm, z1 + rn)
(vi)
If the coordinates P and Q are (x1, y1, z1) and (x2, y2, z2) then the direction ratios of line PQ are,
a = x2 – x1, b = y2 – y1 & c = z2 – z1 and the direction cosines of the line PQ are
l
(vii)
x 2  x1
y  y1
z  z1
.
,m  2
and n  2
| PQ |
| PQ |
| PQ |
Direction cosines of axes : Since the positive x-axis makes angles 00, 900, 900 with axes of x, y
and z respectively. Therefore
Direction cosines of x-axis are (1, 0, 0)
Direction cosines of y-axis are (0, 1, 0)
Direction cosines of z-axis are (0, 0, 1)
C9
Angle between two line segments :
If two lines have direction ratios a1, b1, c1 and a2, b2, c2 respectively then we can consider two vectors
parallel to the lines as a1i + b1j + c1k and a2i + b2j + c2k and angle between them can be given as
cos  
a 1a 2  b 1b 2  c1c 2
a12
 b 12  c12 a 22  b 22  c 22
(i)
The line will be perpendicular if a1a2 + b1b2 + c1c2 = 0
(ii)
The lines will be parallel if
(iii)
Two parallel lines have same direction cosines i.e. l1 = l2, m1 = m2, n1 = n2
a 1 b 1 c1


a2 b2 c2
Practice Problems :
1.
Find the direction cosines of the line which are connected by the relations
l – 5m + 3n = 0 and 7l2 + 5m2 – 3n2 = 0.
2.
Prove that the straight lines whose direction cosines are given by the relations al + bm + cn = 0 and
fmn + gnl + hlm = 0 are perpendicular to each other if
f g h
   0 , and parallel if
a b c
a2f2 + b2g2 + c2h2 – 2bcgh – 2cahf – 2abfg = 0.
3.
Show that the straight lines whose direction cosines are given by the equations al + bm + cn = 0 and
ul2 + vm2 + wn2 = 0 are mutually perpendicular if a2(v + w) + b2(u + w) + c2(u + v) = 0, and parallel if
a 2 b 2 c2


0.
u
v w
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 3
4.
1
Show that the angle between any two diagonals of a cube is cos 1   .
3
5.
A line makes angles , , ,  with the four diagonals of a cube. Prove that cos2 + cos2 + cos2 +
cos2 =
4
.
3
6.
If a variable line in two adjacent positions has direction cosines (l, m, n) and (l + l, m + m, n + n),
show that the small angle  between the two positions is given by ()2 = (l)2 + (m)2 + (n)2.
7.
If l1, m1, n1 and l2, m2, n2 be the direction cosines of two mutually perpendicular lines, show that the
direction cosines of the line perpendicular to both of them are (m1n2 – m2n1), (n1l2 – n2l1), (l1m2 – l2m1).
 1 1 2 
 1
2
3 
]
 and 
[Answers : (1) 
,
,
,
,


 14 14 14 
 6 6 6
C10
Projection of a line segment on a line
(i)
(ii)
If the coordinates P and Q are (x1, y1, z1) and (x2, y2, z2) then the projection of the line segment
PQ on a line having direction cosines l, m, n is |l(x2 – x1) + m (y2 – y1) + n(z2 – z1)|

   a .b

Vector form : projection of a vector a on another vector b is a.b  
|b|

In the above case we can consider PQ as (x2 – x1) î + (y2 – y1) ĵ + (z2 – z1) k̂ in place of a and

l î  mĵ  nk̂ in place of b
(iii)




l | r |, m | r | & n | r | are the projection of r in OX, OY & OZ axes.
(iv)
 
r | r | ( l î  m ĵ  nk̂ )
A PLANE
If line joining any two points on a surface lies completely on it then the surface is a plane.
OR
If line joining any two points on a surface is perpendicular to some fixed straight line. Then this surface is
called a plane. The fixed line is called the normal to the plane.
C11
Equation of a plane
(i)
Normal form of the equation of a plane is lx + my + nz = p, where l, m, n are the direction
cosines of the normal to the plane and p is the distance of the plane from the origin.
(ii)
General form : ax + by + cz + d = 0 is the equation of a plane, where a, b, c are the direction ratios
of the normal to the plane.
(iii)
The equation of a plane passing through the point (x1, y1, z1) is given by a(x – x1) + b(y – y1) +
c(z – z1) = 0 where a, b, c are the direction ratios of the normal to the plane.
(iv)
Plane through three points : The equation of the plane through three non-collinear points
x
x1
(x1, y1, z1), (x2, y2, z2), (x3, y3, z3) is
x2
x3
(v)
y
y1
y2
y3
z
z1
z2
z3
1
1
0
1
1
Intercept Form : The equation of a plane cutting intercept a, b, c on the axes is
Einstein Classes,
x y z
  1.
a b c
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 4
(vi)

Vector Form : The equation of a plane passing through a point having position vector a and
  
 

normal n is (r  a).n  0 or r .n  a.n
Note :
(a)
Vector equation of a plane normal to unitvector n̂ and at a distance d from the origin is

r .n  d
(b)
Coordinate planes
(c)
(i)
Equation of yz-plane is x = 0
(ii)
Equation of xz-plane is y = 0
(iii)
Equation of xy-plane is z = 0
Planes parallel to the axes :
If a = 0, the plane is parallel to x-axis i.e. equation of the plane parallel to the x-axis is
by + cz + d = 0.
Similarly, equation of planes parallel to y-axis and parallel to z-axis are ax + cz + d = 0 and
ax + by + d = 0 respectively.
(d)
Plane through origin : Equation of plane passing through origin is ax + by + cz = 0.
(e)
Transformation of the equation of a plane to the normal form : To reduce any equation
ax + by + cz – d = 0 to the normal form, first write the constant term on the right hand side and
make it positive, then divide each term by
a 2  b 2  c 2 , where a, b, c are coefficients of
x, y and z respectively e.g.
ax
 a2  b2  c2

by
 a2  b 2  c2

cz
 a2  b 2  c2

d
 a2  b2  c2
where (+) sign is to be taken if d > 0 and (–) sign is to be taken if d < 0.
(f)
Any plane parallel to the given plane ax + by + cz + d = 0 is ax + by + cz +  = 0. Distance
between two parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is given as
| d1  d 2 |
a2  b 2  c2
.
(g)
Equation of a plane passing through a given point & parallel to the given vectors : The

equation of a plane passing through a point having position vector a and parallel to


  

b & c is r  a  b  µc (parametric form) where  & µ are scalars or
     
r .(b  c )  a.(b  c ) (non parametric form)
(h)
A plane ax + by + cz + d = 0 divides the line segment joining (x1, y1, z1) and (x2, y2, z2) in the
 ax  by 1  cz 1  d 
 .
ratio   1
 ax 2  by 2  cz 2  d 
(i)
The xy-plane divides the line segment joining the points (x1, y1, z1) and (x2, y2, z2) in the
ratio
(j)
z1
x
y
. Similarly yz-plane in  1 and zx-plane in  1 .
z2
x2
y2
Coplanarity of four points
The points A(x1 y1 z1), B(x2 y2 z2), C(x3 y3 z3) and D(x4 y4 z4) are coplanar then
Einstein Classes,
x 2  x1
x 3  x1
y 2  y1
y 3  y1
z 2  z1
z 3  z1  0
x 4  x1
y 4  y1
z 4  z1
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 5




Very similar in vector method the point A(r1 ), B(r2 ), C(r3 ) and D(r4 ) are coplanar if
     
[r4  r1 , r4  r2 , r4  r3 ]  0 .
Practice Problems :
1.
A variable plane moves in such a way that the sum of the reciprocals of its intercepts on the
coordinates axes is constant. Show that the plane passes through a fixed point.
2.
Find the vector equation of a plane which is at a distance of 6 units from the origin and which has ĵ
as the unit vector normal to it.
3.

Find the Cartesian equation of a plane whose vector equation is r . ( 2 î  5 ĵ  4k̂ )  3 .
4.
Find the vector equation of a plane whose Cartesian equation is 5x – 7y + 2z = 4.
5.
Reduce the equation of the plane x – 2y + 2z – 9 = 0 to the normal form, and hence find the length of
the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the
plane.
6.
Find the vector equation of the plane through (3, 4, –1), which is parallel to the plane

r . ( 2 î  3 ĵ  5k̂ )  2  0 .
7.
Find the equation of the plane passing through the line of intersection of the planes 2x – 7y + 4z = 3
and 3x – 5y + 4z + 11 = 0 and the point (–2, 1, 3).
8.
Find the equation of the plane which contains the line of intersection of the planes
x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0 and which is perpendicular to the plane 5x + 3y – 6z + 8 = 0.
9.
Find the equation of a plane through the intersection of


r . ( î  3 ĵ  k̂ )  5 and r . ( 2 î  ĵ  k̂ )  3 , and passing through the point (2, 1, –2).
10.
11.
12.
the
planes

Find the vector equation of the plane through the line of intersection of the planes r . ( 2 î  3 ĵ  4k̂ )  1


and r . ( î  ĵ)  4  0 , and perpendicular to the plane r . ( 2 î  ĵ  k̂ )  8  0 .
Find the Cartesian as well as the vector equation of the plane through the intersection of the planes


r . ( 2 î  6 ĵ)  12  0 and r . ( 3 î  ĵ  4k̂ )  0 , which is at a unit distance from the origin.


Find the distance between the parallel planes r .( 2 î  3 ĵ  6k̂ )  5 and r .(6 î  9 ĵ  18k̂ )  20  0 .


1 1 1
1 2 2
[Answers : (1)  , ,  (2) r . ĵ  6 (3) 2x + 5y – 4z = 3 (4) r . (5 î  7 ĵ  2k̂ )  4 (5)
,
,
k
k
k
3 3 3




(6) r . ( 2 î  3 ĵ  5k̂ )  11  0 (7) 15x – 47y + 28z – 7 = 0 (8) 33x + 45y + 50z – 41 = 0 (9) r . ( 3 î  2 ĵ)  8

5
(10) r . ( 5 î  2 ĵ  12k̂ )  47 (11) 2x + y + 2z + 3 = 0 and x – 2y + 2z – 3 = 0 (12) units ]
3
C12
Sides of a plane :
A plane divides thre three dimensional space in two equal parts. Two points A (x1 y1 z1) and B (x2 y2 z2) are
on the same side of the plane ax + by + cz + d = 0 if ax1 + by1 + cz1 + d and ax2 + by2 + cz2 + d are both
positive or both negative and are opposite side of plane if both of these values are in opposite sign.
C13
A plane & a point
(i)
Distance of the point ( x, y , z ) from the plane ax + by + cz + d = 0 is given by
ax  by   cz   d .
a2  b2  c2
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 6
(ii)
(iii)

 
The length of the perpendicular from a point having positive vector a to plane r . n  d is given

| a.n  d |

by p 
.
|n|
The coordinates of the foot of perpendicular from the point (x, y, z) to the plane
ax + by + cz + d = 0 are given by
(iv)
x  x 1 y   y 1 z   z 1
(ax 1  by 1  cz 1  d )
.



a
b
c
a2  b2  c2
To find image of a point w.r.t. a plane
Let P (x1, y1, z1) is given point and ax + by + cz + d = 0 is given plane. Let ( x, y , z  ) is the image
point, then
(a)
(b)
x  x 1   a ,
y   y 1  b ,
z   z 1  c

x  a  x1 ,
y   b  y 1 ,
z   c  z 1
 x  x1 
 y  y 1   z  z1 
a
  b
  c
0
 2 
 2   2 
Solve about to get ( x, y , z ) .
The coordinates of the image of point (x1, y1, z1) w.r.t. the plane ax + by + cz + d = 0 are given by
x  x 1 y   y 1 z   z 1
(ax 1  by 1  cz 1  d )


 2
a
b
c
a2  b2  c2
1.
2.
3.
C14
Practice Problems :
Find the image of the point P(1, 3, 4) in the plane 2x – y + z + 3 = 0.
A variable plane which remains at a constant distance 3p from the origin cuts the coordinates axes at
A, B, C. Show that the locus of the centroid of ABC is x–2 + y–2 + z–2 = p–2.
A variable plane is at a constant distance p from the origin and meets the coordinates axes in A, B, C.
Show that the locus of the centorid of the tetrahedron OABC is x–2 + y–2 + z–2 = 16p–2.
[Answers : (1) Q(–3, 5, 2)]
Angle between two planes :
(i)
Consider two planes ax + by + cz + d = 0 and ax  b y  cz  d  0 . Angle between these
planes is the angle between their normals. Since direction ratios of their normals are (a, b, c) and
(a, b , c) respectively, hence , the angle between them, is given by
cos  
aa  bb  cc
a 2  b 2  c 2 a  2  b  2  c 2
a
b
c


a  b  c
 


n .n
The angle  between the planes r.n1  d1 and r.n2  d2 is given by, cos    1 2 .
| n1 | . | n 2 |
 


Planes are perpendicular if n 1 .n 2  0 & planes are parallel if n 1  n 2 .
Planes are perpendicular if aa  bb   cc  0 and planes are parallel if
(ii)
1.
2.
Practice Problems :
Find the angle between the planes 2x – 3y + 4z = 1 and –x + y = 4.
Find the value of  for which the planes 2x – 4y + 3z = 7 and x + 2y + z = 18 are perpendicular to
each other.
 5 
 (2)  = 2]
[Answers : (1) cos 1 

 58 
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 7
C15
Angle bisectors
The equations of the planes bisecting the angle between two given planes
(i)
a1x + b1y + c1z + d1 = 0 and a2x + b2z + c2z + d2 = 0 are
a 1 x  b 1 y  c1 z  d 1
a12  b 12  c12

a2x  b 2y  c2z  d2
gives the bisector of the angle which contains
a 22  b 22  c 22
the origin.
Equation of bisector of the angle containing origin. First makes both the constant terms positive.
Then the positive sign in
(ii)
a 1 x  b 1 y  c1 z  d 1
a12
 b 12
 c12

a2x  b 2y  c2z  d2
gives the bisector of the angle which
a 22  b 22  c 22
contains the origin.
Bisector of acute/obtuse angle : First make both the constant terms positive. Then
(iii)
C16
a1a2 + b1b2 + c1c2 > 0

origin lies on obtuse angle
a1a2 + b1b2 + c1c2 < 0

origin lies in acute angle
Family of planes
Any plane passing through the line of intersection of non-parallel planes or equation of the
plane through the given line in several form.
(i)
a1x + b1y + c1z + d1 = 0
&
a2x + b2y + c2z + d2 = 0 is
a1x + b1y + c1z + d1 +  (a2x + b2y + c2z + d2) = 0
The equation of plane passing through the intersection of the planes


 

r .n 1  d 1 & r .n 2  d 2 is r . (n 1   n 2 )  d 1  d 2 where  is arbitrary scalar
(ii)
C17
Area of a triangle :
Let A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) be the vertices of a triangle, then area of the triangle is

( 2x

2y

2z )
where  x 
y1
1
y2
2
y3
z1
1
z2 1,y 
z3 1
z1
1
z2
2
z3
x1 1
x1
y1
1
x 2 1 and  z  x 2
y2 1
x3 1
y3 1
x3
Vector Method – From two vectors AB and AC . Then area is given by
i
1
1
| AB  AC | x 2  x1
2
2
x 3  x1
C18
j
k
y 2  y1
z 2  z1
y 3  y1
z 3  z1
Volume of a tetrahedron :
Volume of a tetrahedron with vertices A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) and
x1
x
D(x4, y4, z4) is given by V  1 2
6 x3
x4
y1
y2
z1 1
z2 1
y3
y4
z3 1
z4 1
.
A LINE
C19
Equation of a line
(i)
A straight line in space is characterised by the intersection of two planes which are not
parallel and therefore, the equation of a straight line is a solution of the system constituted by the
equations of the two planes, a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0. This form is also
known as non-symmetrical form.
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 8
The equation of a line passing through the point (x1, y1, z1) and having direction ratios a, b,
(ii)
x  x1 y  y 1 z  z 1


 r . This form is called symmetric form. A general point on
a
b
c
the line is given by (x1 + ar, y1 + br, z1 + cr).
c is
(iii)
Vector equation : Vector equation of a straight line passing through a fixed point with


 

position vector a and parallel to a given vector b is r  a   b where  is a scalar..
(iv)
The equation of the line passing through the points (x1, y1, z1) and (x2, y2, z2) is
x  x1
y  y1
z  z1


x 2  x1 y 2  y 1 z 2  z 1


Vector equation of a straight line passing through two points with position vectors a & b is
 
 
r  a   (b  a )
(v)
Reduction of cartesion form of equation of a line to vector form & vice versa
(vi)
x  x1 y  y 1 z  z 1


a
b
c


r  ( x 1 î  y 1 ĵ  z 1k̂ )   (a î  b ĵ  ck̂ )
Note :
Straight lines parallel to co-ordinates axes :
Straight lines
Equations
Straight lines
Equation
(i) Through origin
y = mx, z = nx
(v) Parallel to x-axis
y = p, z = q
(ii) x-axis
y = 0, z = 0
(vi) Parallel to y-axis
x = h, z = q
(iii) y-axis
x = 0, z = 0
(viii) Parallel to z-axis
x = h, y = p
(iv) z-axis
x = 0, y = 0
Practice Problems :
1.
Find the vector equation of a line passing through a point with the position vector ( 2 î  ĵ  4k̂ ) and
parallel to the line joining the points (  î  4 ĵ  k̂ ) and ( î  2 ĵ  2k̂ ) . Also, find the Cartesian
equivalents of the equation.
x3 y2 z6
. Find the vector equation of the line.


2
5
3
2.
The Cartesian equations of a line are
3.
The Cartesian equations of a line are 6x – 2 = 3y + 1 = 2z – 2. Find (a) the direction ratios of the line,
and (b) the Cartesiam and vector equations of the line parallel to this line and passing through the
point (2, –1, –1).
4.
Find the coordinates of the point where the line through A(3, 4, 1) and B(5, 1, 6) crosses the xy-plane.
5.
Find the coordinates of the point where the line
6.
Find the equations of the line passing through the point (–1, 3, –2) and perpendicular to each of the
line
x1 y  2 z  5
meets the plane 2x + 4y – z = 3.


2
3
4
x y z
x  2 y 1 z 1
.
  and


1 2 3
3
2
5
7.


Show that the lines r  ( î  ĵ  k̂ )   ( 3 î  ĵ) and r  (4 î  k̂ )  µ ( 2 î  3k̂ ) intersect each other. Find
their point of intersection.
8.
Prove that the points A(2, 1, 3), B(5, 0, 5) and C(–4, 3, –1) are collinear.
9.
Find the value of  for which the points A(–1, 3, 2), B(–4, 2, –2) and C(5, 5, ) are collinear.
10.
Show that the points whose position vectors are ( 2 î  3 ĵ  5k̂ ), ( î  2 ĵ  3k̂ ) and (7 î  k̂ ) aree
collinear.
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 9
11.
Using the vector method, find the values of  and µ for which the points A(3, , µ), B(2, 0, –3) and
C(1, –2, –5) are collinear.
[Answers : (1)

x 2 y 1 z 1


(2) r  ( 3 î  2 ĵ  6k̂ )   ( 2 î  5 ĵ  3k̂ )
2
2
1

 13 23 
(3) r  ( 2 î  ĵ  k̂ )   ( î  2 ĵ  3k̂ ) (4)  , , 0  (5) P(3, –1, –1) (6)
 5 5 
(7) (4, 0, –1) (11)  = 2 and µ = –1]
C20
x1 y3 z 2


2
7
4
Reduction of non-symmetrical form to symmetrical form :
Let equation of the line is non-symmetrical form be a1x + b1y + c1z + d1 = 0, a2x + b2y + c3z + d2 = 0. To find
the equation of the line in symmetrical form, we must know (i) its direction ratios (ii) coordinates of any
point on it.
Direction ratios : Let l, m, n be the direction ratios of the line. Since the line lies in both the
planes, it must be perpendicular to normals of both planes. So a1l + b1m + c1n = 0, a2l + b2m
+ c2n = 0. From these equations, proportional values of l, m, n can be found by crossmultiplication as
(i)
l
m
n


b 1c 2  b 2 c 1 c 1a 2  c 2 a 1 a 1b 2  a 2 b 1
Alternative method
i
j
k
The vector a 1
a2
b1
c1 = i(b1c2 – b2c1) + j(c1a2 – c2a1) + k(a1b2 – a2b1) will be parallel to the line of
b2
c2
intersection of the two given plane, hence l : m : n = (b1c2 – b2c1) : (c1a2 – c2a1) : (a1b2 – a2b1)
(ii)
Point of the line – Note that as l, m, n cannot be zero simultaneously, so at least one must be
non-zero. Let a1b2 – a2b1  0, then the line cannot be parallel to xy plane, so it intersect it. Let
it intersect xy-plane in (x1, y1, 0). Then a1x1 + b1y1 + d1 = 0 and a2x1 + b2y1 + d2 = 0. Solving
these, we get a point on the line. Then its equation becomes.
x  x1
y  y1
z0


or
b 1c 2  b 2 c 1 c 1a 2  c 2 a 1 a 1b 2  a 2 b 1
b 1d 2  b 2 d 1
d a  d 2a1
y 1 2
a 1b 2  a 2 b 1
a 1b 2  a 2 b 1
z0


b 1c 2  b 2 c 1
c 1a 2  c 2 a 1
a 1b 2  a 2 b 1
x
Note :
If l  0, take a point on yz-plane as (0, y1, z1) and if m  0, take a point on xz-plane as (x1, 0, z1)
Alternative method
If
a1 b1

Put z = 0 in both the equations and solve the equations a1x + b1y + d1 = 0, a2x + b2y + d2 = 0
a2 b2
otherwise put y = 0 and solve the equations a1x + c1z + d1 = 0 and a2x + c2z + d2 = 0.
C21
Foot, length and equation of perpendicular from a point to a line L
(i)
Cartesian form : Let equation of the line be
xa yb zc


 r (say )
l
m
n
..(i)
and A(, , ) be the point
Any point on line (i) is P (lr + a, mr + b, nr + c)
..(ii)
If it is the foot of the perpendicular from A on the line, then AP is perpendicular to the line, So
l(lr + a – ) + m(mr + b – ) + n (nr + c – ) = 0 i.e. r = n( – a)l + ( – b)m + ( – c)n since
l2 + m2 + n2 = 1. Putting this value of r in (ii), we get the foot of perpendicular from point A on the
given line. Since foot of perpendicular P is known, then the length of perpendicular is given by
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 10
AP  ( lr  a   ) 2  (mr  b   ) 2  (nr  c   ) 2 the equation of perpendicular is given by
x
y 
z


lr  a   mr  b   nr  c  
(ii)

Vector Form : Equation of a line passing through a point having position vector  and


 
 
 
perpendicular so the lines r  a1  b 1 and r  a 2  b 2 is parallel to b1  b 2 . So the vector
 

 

equation of such a line is r  a   (b1  b 2 ) . Position vector  of the image of a point  in a
  


 
  2(a   ).b   
straight line r  a   b is given by   2a  
 2  b   . Position vector of the foot of
 | b |

  
   (a  
).b 
the perpendicular on line is f  a    2  b . The equation of the perpendicular is
 | b | 
  
    (a  
 
).b 
r    µ (a   )  
 2 b 

 |b|
 
C22
To find image of a point w.r.t. a line
Let L 
x  x2 y  y 2 z  z2


is given line
a
b
c
Let ( x, y , z ) is the image of the point P (x1, y1, z1) with respect to the line L. Then
(i)
(ii)
a( x1  x)  b( y 1  y )  c( z 1  z  )  0
x 1  x
y 1  y
z1  z
 x2
 y2
 z2
2
2
2



a
b
c
from (ii) get the value of x, y , z  in terms of  as
x  2a  2x 2  x1 , y   2b  2y 2  y 1 ,
z   2c  2z 2  z 1
now put the value of x, y , z  in (i) get  and resubtitute the value of  to get ( x, y , z  )
Practice Problems :
1.
Find the image of the point (1, 6, 3) in the line
x y 1 z  2
.


1
2
3
[Answers : (1) M(1, 0, 7)]
C23
Angle Between A Plane And A Line :
(i)
If  is the angle between line
x  x1 y  y 1 z  z 1


and the plane ax + by + cz + d = 0, then
l
m
n

al  bm  cn
sin   
2
2
 (a  b  c 2 ) l 2  m 2  n 2

Einstein Classes,




Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 11
(ii)


 b.n 
 

Vector form : If  is the angle between a line r  (a  b ) and r .n  d then sin      
 | b || n | 
(iii)
Condition for perpendicularity
(iv)
l m n  
  , bn  0
a b c
 
Condition for parallel al + bm + cn = 0
b .n  0
Practice Problems :
1.
2.


Find the angle between the line r  ( î  ĵ  3k̂ )   ( 2 î  2 ĵ  k̂ ) and the plane r .(6 î  3 ĵ  2k̂ )  5 .

Find the value of m for which the line r  ( î  2 ĵ  k̂ )   ( 2 î  ĵ  2k̂ ) is parallel to the plane

r .( 3 î  2 ĵ  mk̂ )  12 .
3.


Show that the line r  ( 2 î  2 ĵ  3k̂ )   ( î  ĵ  4k̂ ) is parallel to the plane r .( î  5 ĵ  k̂ )  5 . Also,
find the distance between them.
4.
Find the angle between the line
x 2 y 1 z3
and the plane 3x + 4y + z + 5 = 0.


3
1
2
 7 
10
 8 
]
units (4) sin 1 
[Answers : (1) sin 1   (2) – 2 (3)
 52 
3 3
 21 


C24
C25
Condition For A Line To Lie In A Plane
x  x1 y  y 1 z  z 1


would lie in a plane
l
m
n
(i)
Cartesian form : Line
(ii)
ax + by + cz + d = 0, if ax1 + by2 + cz1 + d = 0 & al + bm + cn = 0


 


Vector form : Line r  a   b would lie in the plane r .n  d if b.n  0 & a.n  d
Coplanar Lines :
(i)
If the given lines are
x y  z 
x y  z 


and


, then condition
l
m
n
l
m
n
          
for intersection/coplanarity is
l
m
n  0 & plane containing the above two
l
m
n
x y z 
lines is
(ii)
l
l
m
m
n 0
n
Condition of coplanarity if both the lines are in general form let the lines be
ax + by + cz + d = 0 = ax  b y  cz  d
&
a b c d
a  b  c d 
0
x + y + z +  = 0 = x  y   z    are coplanar if
   
     
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 12
Practice Problems :
1.
Find the equation of the plane passing through the point (0, 7, –7) and containing the line
x1 y3 z 2
.


3
2
1
2.
Find the equation of the plane passing through the line of intersection of the planes x – 2y + z = 1 and
2x + y + z = 8, and parallel to the line with direction ratios 1, 2, 1. Find also the perpendicular
distance of (1, 1, 1) from this plane.
3.
Find the equation of the plane passing through the line of intersection of the planes 2x + y – z = 3,
5x – 3y + 4z + 9 = 0, and parallel to the line
x1 y  3 z  5
.


2
4
5
4.
Find the equation of the plane passing through (2, 3, –4) and (1, –1, 3), and parallel to the x-axis.
5.


Show that the lines r  ( î  ĵ  k̂ )   ( 3 î  ĵ) and r  (4 î  k̂ )  µ ( 2 î  3k̂ ) are coplanar. Also, find the
equation of the plane containing both these lines.
6.
Show that the lines
7.
Show that the lines
8.
Find
x y2 z3
x2 y6 z3
are coplanar. Also, find the equation


and


1
2
3
2
3
4
of the plane containing these lines.
the
xad ya zad
xbc yb zbc
are coplanar..


and








equation
of
the
plane
which
contains
the
two
parallel
lines
x 3 y  4 z 1
x1 y  2 z


and

 .
3
2
1
3
2
1
[Answers : (1) x + y + z = 0 (2)
13
194
units (3) 7x + 9y – 10z – 27 = 0 (4) 7y + 4z – 5 = 0

(5) r .( 3 î  9 ĵ  2k̂ )  14  0 (6) x – 2y + z + 7 = 0 (8) 8x + y – 26z + 6 = 0]
C26
Skew Lines :
(i)
The straight lines which are not parallel and non-coplanar i.e. non-intersecting
          
are called skew lines. If  
(ii)
l
l
m
m
n
n
 0 , then lines are skew..
Shortest distance : Suppose the equation of the lines are
x y  z 
x   y    z   


and


then shortest distance is given by
l
m
n
l
m
n

(    )(mn  mn )  (   )(nl  n l )  (     )( lm   l m )
 (m n   m  n )
          
l
m
n 
=


l
m
n
Einstein Classes,
2
 (m n   m  n )
2
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 13




Vector Form : For lines a1  b1 & a 2  b 2 to be skew
(iii)
 
 
(b1  b 2 ).(a 2  a1 )  0
or
   
[b1b 2 (a 2  a1 )]  0

 
Shortest distance between the two parallel lines r  a1  b &
(iv)




 
(a 2  a 1 )  b

r  a 2  µb is d 
|b|
Practice Problems :
1.
Find the shortest distance between the lines whose vector equations are

r  (1  t ) î  (t  2) ĵ  ( 3  2t )k̂ and

r  (s  1) î  ( 2s  1) ĵ  ( 2s  1)k̂
2.
Find the length and the equations of the line of shortest distance between the lines
x3 y8
x 3 y7 z6

 z  3 and


3
1
3
2
4
3.
Show that the lines
x1 y  2 z  3
x  4 y 1


and

 z intersect each other. Find their point
2
3
4
5
2
of intersection.
4.
Find the distance of the point (2, 3, 4) from the plane 3x + 2y + 3z + 5 = 0, measured parallel to the
line
5.
x 3 y2 z

 .
3
6
2
Find the distance of the point (–2, 3, –4) from the line
x  2 2y  3 3z  4
measured parallel to


3
4
5
the plane 4x + 12y – 3z + 1 = 0
[Answers : (1)
C27
8 29
z3
units (2)
(3) (–1, –1, –1) (5) 17/2 units]
29
1
Sphere
General equation of a sphere is given by x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0. The centre of the sphere is
(–u, –v, –w) and radius is
Einstein Classes,
u2  v2  w 2  d .
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 14
SINGLE CORRECT CHOICE TYPE
1.
2.
The area of a quadrilateral ABCD, where
A  (0, 4, 1), B  (2, 3, –1), C  (4, 5, 0) and
D  (2, 6, 2) is equal to
(a)
9 sq. units
(b)
18 sq. units
(c)
27 sq. units
(d)
81 sq. units
(c)
4.
1
(a)

2
(b)

6
(d)

3
None of these
8.
(a)
|d| = |dl| + |dm| + |dn|
(b)
(d)2 = (dl)2 + (dm)2 + (dn)2
(c)
d = dl + dm + dn
(d)
None of these
The straight lines
x2 y3 z4


1
1
k
k = {3, –3}
(b)
k = {0, –1}
(c)
k = {–1, 1}
(d)
k = {0, –3}
(d)
d
a b c
4 | abc |
1
c
2


1
a 12
1
a 12


1
b 12
1
b 12


1
c12
1
c12
Let f(x) be a polynomial in x satisfying the
3
None of these
The distance of the point (1, –2, 3) from the plane
x – y + z – 5 = 0, measured parallel to the line
,
1
30
,
30
,
2
30
2
30
1
30
,
30
,
5
3
30
,
1
30
If origin is the centroid of ABC with vertices
A(, 1, 3), B(–2, , – 5) and C(4, 7, ), the values of
,  and  is
(a)
 = 2,  = –8,  = 2
(b)
 = –2,  = 8,  = 2
(c)
 = –2,  = –8,  = 2
(d)
 = –2,  = –8,  = –2
Let f be a one-one function with domain (–2, 1, 0)
and range {1, 2, 3} such that exactly one of the
following statements is true f(–2) = 1, f(1)  1,
f(0)  2 and the remaining two are false. The
distance between points (–2, 1, 0) and (f(–2), f(1),
f(0)) is
(a)
8
(b)
7
(c)
6
(d)
5
(a)
1 unit
(b)
2 units
A(1, 0, 4), B(0, –11, 3), C(2, –3, 1) are three points
and D is the foot of perpendicular from A to BC.
The co-ordinates of D is
(c)
3 units
(d)
None of these
(a)
(4, 5, 5)
(b)
(4, –5, –5)
(c)
(4, 5, –5)
(d)
(–4, –5, –5)
Einstein Classes,
11.
,
30
(d)
2
30
30
1
9.
,
30
(c)
2
x y z 1
 
, is equal to
2 3
6
b

2
c
2
a2 – b2 + c2 = a12 – b12 + c12
and
10.
(c)
1
1
(d)
(b)
d 2 a2  b 2  c2
2 | abc |
2
b

2
a2 + b2 + c2 = a12 + b12 + c12
1
d 2 a2  b 2  c2
| abc |
2
1
(c)
(a)
The plane ax + by + cz = d, meets the coordinate
axes at the points A, B and C respectively. Area of
triangle ABC is equal to
2

1
(a)
(b)
a
2

1
1
condition f ( x )f    f(x) + f   and f(2) = 5. The
x
x
direction cosines of the ray joining origin and point
(f(0), f(1), f(2)) is
If the direction cosines of a variable line in two
adjacent positions be l, m, n and l + dl, m + dm,
n + dn and the small angle between the two
positions is d, then
(a)
6.
a
2
1
(b)
x1 y 4 z  5


, will intersect provided
k
2
1
5.
Two systems of rectangular axes have the same
origin. If a plane cuts them at distances a, b, c and
a1, b1, c1 from the origin, then
The angle between lines whose direction cosines are
given by l + m + n = 0, l2 + m2 – n2 = 0, is
(a)
3.
7.
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
12.
13.
14.
15.
The image of the point (2, –3, 4) with respect to the
plane 4x + 2y – 4z + 3 = 0 is
(a)
 40 16 14 
,
, 

 9 9 9 
(b)
 40 16 14 
 , ,  
9
9
9
(c)
 40 16 14
  , ,  
 9 9 9
(d)
 40 16 14 
 , , 
9 9 
 9
A variable plane passes through a fixed point
(a, b, c) and meets the coordinates axis in A, B, C.
The locus of mid point of the plane common through
A, B, C and parallel to the coordinate planes is
(a)
ax–1 + by–1 + cz–1 = 1
(b)
ax + by + cz = 1
(c)
ax–2 + by–2 + cz–2 = 1
(d)
none of these
M3D – 15
ANSWERS
(SINGLE CORRECT CHOICE TYPE)
1.
a
9.
c
2.
d
10.
d
3.
b
11.
c
4.
d
12.
d
5.
b
13.
a
6.
a
14.
b
7.
a
15.
b
8.
a
The image of the point P(, , ) by the plane
lx + my + nz = 0 is Q (, ,  ) then
(a)
2 +  2 + 2 = l2 + m2 + n2
(b)
2 +  2 + 2 = ( ) 2  ( ) 2  (  ) 2
(c)
        0
(d)
l (     )  m (    )  n (     )  0
A square ABCD of diagonal 4 units is folded along
the diagonal AC so that plane DAC, BAC are at
right angle, the shortest distance between DC and
AB is
(a)
2/3
(b)
4/3
(c)
4
(d)
none of these
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 16
EXCERCISE BASED ON NEW PATTERN
COMREHENSION TYPE
COMPREHENSION-3
COMPREHENSION-1
Consider the two lines L1 and L2 whose equations
A ray of light emancipating from the point source
P(1, –3, 2) and travelling parallel to the line
are
x2 y z1
 
is incident on the plane
1
3
2
x + y – 3z = 0 at the point Q. After reflecting from
the plane the ray travels along the line QR. It is
also known that the incident ray, reflected ray and
the normal to the plane at the point of incidence
are in the same plane.
1.
2.
3.
The point Q is
x2
y
z1
x y 1 z  2


and 

1
4
2
2
3
1
respectively. Consider the plane 4x + 3y + 4z = 1.
7.
8.
(a)
(3, 15, 6)
(b)
(3, 6, 3)
(c)
(–3, –6, –3)
(d)
(–3, –15, –6)
The equation of the line QR are
(a)
x  2 y  22 z  4


15
37
10
(b)
x  3 y  15 z  6


3
7
2
(c)
x3 y6 z3


15
37
10
(d)
x 3 y6 z 3


3
7
2
9.
The equation of the plane PQR is
(a)
5x + 2y – z + 3 = 0
(b)
11x – 5y + 2z = 30
(c)
5x – y – z = 6
(d)
x–y+z=6
Choose the correct statement
(a)
both lines are parallel to the plane
(b)
only L1 is parallel to the plane
(c)
only L2 is parallel to the plane
(d)
none of these
The x-coordinate of point of intersection of one of
above lines with the plane is
(a)

22
13
(b)

(c)

37
33
(d)
37
33
The line which intersect the plane makes an angle
of
(a)
 13 

sin 1 

 574 
(b)
 13 

cos 1 

 574 
(c)
 13 

tan 1 

 574 
(d)
none
COMPREHENSION-4
P(–3, 2, 5) is a point and 2x + 3y – z + 2 = 0 is a
plane.
10.
The length of the perpendicular from P to the plane
is
COMPREHENSION-2
4.
5.
6.
ABC is a triangle where A = (2, 3, 5), B = (–1, 3, 2)
and C = (, 5, µ). The median through A is equally
inclined to the axes.
(a)
The value of ‘’ is
(c)
1
(b)
3
7
(b)
5
(c)
10
(d)
13
11.
(d)
7
(b)
5
(c)
10
(d)
13
The area of triangle ABC is
(a)
2
(b)
8
(c)
2
(d)
32
4
14
The y-coordinate of foot of the perpendicular from
P to the plane is
The value of ‘µ’ is
(a)
2
14
14
14
(a)
Einstein Classes,
20
13
18
7
(a)

(c)
67
14
(b)
37
14
(d)
none
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 17
12.
13.
14.
15.
The ‘z’ coordinate of image of P with respect to the
plane is
(a)
15

7
(b)
23
7
(c)
32
7
(d)
none
COMPREHENSION-5
Consider the points A(1, –4, 5) and B(0, 6, 1) and
the plane 3x – y + 2z = 7.
Choose the correct statement from the following
(a)
both points are on the same side of the
plane
(b)
both points are on the opposite sides of
the plane
(c)
both points are in the plane
(d)
none of these
The ratio in which the line segment AB is divided
by the plane is
(a)
3 : 5 internally
(b)
3 : 5 externally
(c)
10 : 11 internally
(d)
10 : 11 externally
2
If P( + , ,  – 1) is a point on the same side of
the plane as the point A then the set of values of  is
(a)
(b)
(c)


 1  73    1  73
    ,

, 

 

5
5

 

(b)
x
(c)
(d)
18.
(a)
(c)
3 19
2
3 19
Einstein Classes,
(b)
(d)
7
3 19
none
The x-coordinate of the point of intersection of the
above line with the line L1 is
(a)
158
171
(b)
197
171
(c)
235
171
(d)
82
57
Matching-1
Column - A
(A)
Column - B
A line makes angles
(P)
, ,  with the positive
directions of the axes of
reference. The value of
cos 2 + cos 2 + cos 2 is (Q)
(B)
The direction ratios of two
perpendicular lines are
1, –3, 5 and , 1 + , 2 + .
Then  is
(C)
The angle between the
(R)
–7/3
–1
6
lines joining the points
(1, 1, 0), (–3, 3 + 1, 3)
and (0, –1, 0), (–1, 3 – 1, )
1
is cos 1   . If  is an integer
 8
then its value is
(D)
The volume of the
(S)
tetrahedron whose vertices
are (0, 1, 2), (4, 3, 6), (2, 3, 2)
and (3, 0, 1) is (in unit3)
0
Matching-2
5
3 19
82
235
545
z
y
57
171 
171 
7
11
1
MATRIX-MATCH TYPE
The shortest distance between them is
4
82
235
545
z
y
57
171 
171 
1
11
7
x
COMPHENSION-6
Consider the lines L1 and L2
16.
82
235
545
z
y
57
171 
171 
11
1
7
x


 1  73    1  73
    ,

, 

 

6
6

 

x 3 y 4 z1
x 1 y  3 z 1


and


2
1
3
1
3
2
The equations of the line of shortest distance
(a)


 1  73    1  73
    ,

, 




3
3

 



 1  73    1  73
    ,

, 




4
4

 

(d)
17.
(A)
Column - A
Column - B
If A = (1, 2, 3),
B = (–2, 4, ) and
AOB = /2 where O is
the origin then  is
(P)
13
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 18
(B)
If the points A(1, 2, –1), (Q)
B(2, 6, 2) and C(, –2, –4)
are collinear then  is
4
(C)
If A = (1, 1, 1),
(R)
B = (2, –1, 3), C = (0, 4, –2),
D = (1, 2, ) and AB, AC
and AD are coplanar then
 is
0
The projection of a line (S)
segment on the axes of
reference are 3, 4 and 12
respectively. The length
of the line segment is
–2
(D)
(D)
Matching-3
(A)
(C)
1.
Column - A
Column - B
The lines
(P)
6
x 3 y z4
 
and
2
1
3
(B)
(C)
(D)
x y 1
z


aree
  1   2
perpendicular to each
other. Then  is equal to
The shortest distance
between the line
x3 y
z
 
and
3
0 4
the y-axis is
The projection of the
line segment joining
the point (6, –2, 1)
and the origin on the
2.
(Q)
(4, 7, –5)
(c)
(0, –1, 3)
(d)
(–2, –5, 7)
The direction cosines of the line passing through
(R)
–4
11
3.
4.
Column - B
3
,
2
6
11
11
1
,
11
,
3
(d)
12/5
1
,
11
11
1
(c)
3
,
1
(b)
x 2 y 1
z


1
2
1
6
aree
11
1
The distance of the
(P)
point (2, 0, –3) from
the plane 5x – 12y = 0 is
The distance between
(Q)
the planes 4x – 5y + 3z = 5
and 4x – 5y + 3z + 2 = 0 is
Einstein Classes,
(b)
(a)
x  (k  1)
y
z 1


1
1
2
are intersecting, is
Matching-4
(B)
(2, 3, –1)
at cos 1
infinite
xk y 1 z 1


and
4
2
1
(A)
(a)
the origin and cutting the line
x 2 y 1 z 1


line
4
3
0
is
The number of real values (S)
of k for which the lines
Column - A
The shortest distance
(R)
10/13
between the lines
x – y = 0 = 2x + z and
x + y – 12 = 0 = 3x – y + z – 1
is
A variable plane at a
(S)
9
distance of 1 unit from
the origin cuts the
coordinate axes at A, B
and C. If the centroid
D(x, y, z) satisfies the
relation x–2 + y–2 + z–2 = k
then the value of k is
MULTIPLE CORRECT CHOICE TYPE
A point Q at a distance 3 from the point P(1, 1, 1)
lying on the line joining the points A(0, –1, 3) and P,
has the coordinates
,
11
1
6
,
1
11
6
,
1
11
The equation of a plane is 2x – y – 3z = 5 and
A(1, 1, 1), B(2, 1, –3), C(1, –2, –2) and D(–3, 1, 2)
are four points. Which of the following line segments
are intersecting by the plane ?
(a)
AD
(b)
AB
(c)
AC
(d)
BC
Which of the following is/are correct about a
tetrahedron
7/52
(a)
Centroid of a tetrahedron lies on lines
joining any vertex to the centroid of
oppsite face.
1/23
(b)
Centroid of a tetrahedron lies on the lines
joning the mid point of the opposite faces.
(c)
Distance of centroid from all the vertices
are equal
(d)
None of these
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 19
5.
6.
1.
The equation of the line x + y + z – 1 = 0,
4x + y – 2z + 2 = 0 written in the symmetrical form
is
(a)
x1 y 2 z0


1
2
1
(b)
x
y
z 1


1 2
1
(c)
x  1/ 2 y 1 z  1/ 2


1
2
1
(d)
x1 y  2 z  2


2
1
2
2.
x1 y  3 z  2
x 2 y 1 z  3


and


2
1
3
1
3
2
3.
Consider the planes 3x – 6y + 2z + 5 = 0 and
4x – 13y + 3z = 3. The plane 67x – 162y + 47z + 44 =
0 bisects the angle between the planes which
(a)
contains origin (b)
is acute
(c)
is obtuse
(d)
none of these
Assertion-Reason Type
Each question contains STATEMENT-1 (Assertion)
and STATEMENT-2 (Reason). Each question has
4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.
(A)
Statement-1 is True, Statement-2 is True;
Statement-2 is a correct explanation
for Statement-1
(B)
Statement-1 is True, Statement-2 is True;
Statement-2 is NOT a correct
explanation for Statement-1
(C)
Statement-1 is True, Statement-2 is False
(D)
Statement-1 is False, Statement-2 is True
Consider the planes 3x – 6y – 2z = 15 and
2x + y – 2z = 5.
STATEMENT-1 : The parametric equations of the
line of intersection of the given planes are
x = 3 + 14t, y = 1 + 2t, z = 15t.
 

STATEMENT-2 : The vector 14 i  2 j  15k is
The equations of two straight lines are
4.
5.
STATEMENT-1 : The given lines are coplanar.
STATEMENT-2 : The equations
2r – s = 1
r + 3s = 4
3r + 2s = 5 are consistent
STATEMENT-1 : The planes x + y = 0, y + z = 0
and x + z = 0 meet in a unique point
STATEMENT-2 : The above planes have a
common line of intersection.
STATEMENT-1 : Two planes x – 2y + 2z = 6 and
3x – 6y + 6z = 2 are parallel.
STATEMENT-2 : If the two planes are parallel
then they have the same direction rations of
normals.
STATEMENT-1 :

1
2 3
,
1
1
be the
,
2
3
6.
direction cosines of any directed line
STATEMENT-2 : If l, m, n are the direction
cosines of a line, then l2 + m2 + n2 = 1
STATEMENT-1 : There be a unique
equation of the plane passing through the points
(4, 3, 5), (1, 2, 3) and (7, 4, 7).
STATEMENT-2 : The three given points are
collinear.
7.
STATEMENT-1 :
a
b
c


0
xy yz zx
represents a pair of planes
STATEMENT-2 : The general equation of second
degree in x, y, z i.e. equation
ax2 + by2 + cz2 + 2hxy + 3gzx + 3fyz = 0 represents
a pair of planes iff abc + 2fgh – af2 – by2 – ch2 = 0
parallel to the line of intersection of given planes.
(Answers) EXCERCISE BASED ON NEW PATTERN
COMPREHENSION TYPE
1.
d
2.
a
3.
b
4.
a
7.
b
8.
a
9.
a
10.
c
13.
b
14.
c
15.
d
16.
d
MATRIX-MATCH TYPE
1.
[A-Q, B-P, C-S, D-R]
2.
[A-S, B-R, C-R, D-P]
4.
[A-R, B-P, C-Q, D-S]
MULTIPLE CORRECT CHOICE TYPE
1.
a, c
2.
a, d
3.
b, c
4.
a, b
6.
a, b
ASSERTION-REASON TYPE
1.
D
2.
A
3.
D
4.
A
7.
A
Einstein Classes,
5.
11.
17.
c
b
a
6.
12.
18.
3.
[A-R, B-S, C-P, D-Q]
5.
a, b, c, d
5.
D
6.
d
c
c
A
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 20
INITIAL STEP EXERCISE
(SUBJECTIVE)
1.
Planes are drawn parallel to the co-ordinate planes
through the points (1, 2, 3) and (3, –4, –5). Find the
lengths of the edges of the parallelopiped so formed.
17.
Find the equation of the plane perpendicular to the
plane x + 3y + 3z = 8 and passing through the points
(2, 2, 1) and (9, 3, 6).
2.
Show that the points A(2, 3, 4), B(–1, 2, –3) and C
(–4, 1, –10) are collinear. Also find the ratio in which
C divides AB.
18.
If a plane passes through the point (–3, –3, 1) and is
normal to the line joining the points (2, 6, 1) and
(1, 3, 0), find its equation.
3.
The vertices of a triangle are A(5, 4, 6), B(1, –1, 3)
and C(4, 3, 2). The internal bisector of BAC meets
BC in D. Find AD.
19.
Find the equation of the plane passing through the
point (a, b, g) and perpendicular to the planes
a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0.
4.
Find the points of trisection of the line segment
joining the points (2, –2, 7) and (5, 1, –5).
20.
5.
Find the angle between any two diagonals of a cube.
Find the equation of the plane passing through the
line of intersection of the planes 4x – 5y – 4z = 1
and 2x + y + 2z = 8 and the point (2, 1, 3).
6.
If P, Q, R S are (3, 6, 4), (2, 5, 2), (6, 4, 4), (0, 2, 1)
respectively, find the projection of PQ on RS.
21.
The plane x – y – z = 4 is rotated through 900 about
its line of intersection with the plane x + y + 2z = 4.
Find its equation in the new position.
7.
Find the ratio in which the surface x2 + y2 + z2 = 25
divides the line joining (0, 1, 2) and (3, 4, 5).
22.
Find the angle between the plane x + y – 2z + 5 = 0
and the line whose direction cosines are
8.
What are the direction cosines of a line that passes
through the points P(6, –7, –1) and Q(2, –3, 1) and
is so directed that it makes an acute angle  with
the positive direction of x-axis.
9.
If Q be the foot of perpendicular from P(2, 4, 3) on
the line joining the points A(1, 2, 4) and B(3, 4, 5),
find the co-ordinates of Q.
10.
Find the projection of the line joining (1, 2, 3) and
(–1, 4, 2) on the line having direction ratios 2, 3, –6.
2x + y + 2z = 9 and 3x – 4y + 12z + 13 = 0.
11.
Find the locus of a point which moves such that the
sum of its distance from points A(0, 0, –) and
B(0, 0, ) is constant.
Which of these bisector planes bisects the acute
angle between the given planes. Does origin lie in
the acute angle or obtuse angle between the given
planes ?
12.
Determine the values of a and b so that the points
(a, b, 3), (2, 0, –1) and (1, –1, –3) are collinear.
25.
13.
Two vertices of a triangle are A(3, 4, 2) and
B(1, 3, 2). The medians of the triangle intersect at
(2, 4, 3), find the remaining vertex C of the triangle.
If a variable plane forms a tetrahedron of constant
volume 64k3 with the co-ordinate planes, find the
locus of the centroid of the tetrahedron.
26.
Find the locus of a point, the sum of squares of
whose distances from the planes x – z = 0, x – 2y + z
= 0 and x + y + z = 0 is 36.
27.
A variable plane at constant distance p from the
origin meets the co-ordinates axes at P, Q and R.
Find the locus of the point of intersection of planes
drawn through P, Q, R and parallel to the co-ordinates planes.
28.
If a variable plane cuts the co-ordinate axes in A,
B, and C and is at a constant distance p from the
origin, find the locus of the centroid of the
tetrahedron OABC.
14.
If the direction cosines of three coplanar lines be l1,
m1, n1; l2, m2, n2 and l3, m3, n3, prove that
l1 m1 n1
l2 m2 n 2  0
l3 m3 n 3
15.
If the direction cosines of two lines at right angles
be l1, m1, n1 and l2, m2, n2 find the direction cosines
of a line which is perpendicular to both.
16.
Find the equation of the plane through the points
(2, 1, 2) and (1, 3, –2) and parallel to the x-axis.
Einstein Classes,
1 2 1
.
,
,
6 6 6
23.
Find the plane which bisects the obtuse angle
between the planes 4x – 3y + 12z + 13 = 0 and
x + 2y + 2z = 9.
24.
Find the planes bisecting and angles between planes
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 21
29.
Find the equation of the line drawn through point
(1, 0, 2) to meet at right angles the line
40.
the
x 1 y  2 z 1


.
3
2
1
30.
The direction cosine of the shortest distance between
x  2 y  3 z 1


1
4
3
and
1
4
5
x  3 y  4 z 1
,
,


are
.
42 42 42
2
3
2
Find the image of the point P(3, 5, 7) in the plane
2x + y + z = 16.
Find
31.
lines
Find the equation of the line of intersection of planes
(i)
its equation
(ii)
the points where it intersects the
lines.
4x + 4y – 5z = 12, 8x + 12y – 13z = 32
32.
Show that the line represented by equations
x = ay + b, z = cy + d in symmetrical form is
41.
x 2 y2 z6
.


2
3
6
x b y zd
 
a
1
c
33.
Find the point of intersection of the line
x  2 y 1 z  3


3
2
2
and
Find the distance of the point (1, 0, –3) from the
plane x – y – z = 9 measured parallel to the line
the
42.
Find the co-ordinates of the foot of perpendicular
from the point (2, 6, 3) to the line
x y 1 z  2


. Also find the equation of this
2
2
3
plane
2x + y – z = 3.
perpendicular.
34.
Show that the lines
x  3 y 1 z  2


and
2
3
1
43.
x  3 y 1 z  2


.
3
4
2
x 7 y z7
 
are coplanar. Also find the
3
1
2
equation of the plane containing them.
35.
44.
x  x1 y  y 1 z  z 1


l1
m1
n1 and parallel to the line
their point of intersection and equation of the plane
in which they lie.
Find the equation of the line which can be drawn
from the point (1, –1, 0) to intersect the lines
x 2 y 1 z  3
x4 y z1


 
and
2
3
4
4
5
2
orthogonally.
37.
Find the equation of the plane through the line
Are the lines 3x – 2y + z + 5 = 0 = 2x + 3y + 4z – 4
x  4 y  6 z 1


and
coplanar ? If yes, find
3
5
2
36.
Find the equation of the plane passing through
(1, 2, 0) which contains the line
Find the angle between the lines x – 3y – 4 = 0,
4y – z + 5 = 0 and x + 3y – 11 = 0, 2y – z + 6 = 0.
38.
Find the co-ordinates of the point where the line
x – 2y + z = 1, x + 2y – 2z = 5 intersects the plane
x + y – 2z = 7.
39.
Find the length of the shortest distance between the
lines
x y  z 


l2
m2
n2 .
45.
Let A = (1, 2, –3) and B = (4, 1, 5). P is a variable
point such that PAB is an equilateral triangle. Find
the equation of the locus of the point P.
46.
A plane meets the coordinates axes at A, B and C
such that the centroid of the ABC is the point
(a, b, c). Find the equation of the plane ABC.
47.
Let PM be the perpendicular from the point
P(1, 2, 3) to the x-y plane. If OP makes an angle 
with the positive direction of the z-axis and OM
makes an angle  with the positive direction of the
x-axis, where O is the origin, then find  and .
x 1 y  4 z  1
x 4 y3 z 2




and
.
1
3
2
2
3
3
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 22
FINAL STEP EXERCISE
(SUBJECTIVE)
1.
If the points P, Q, R, S are (4, 7, 8), (–1, –2, 1),
(2, 3, 4) and (1, 2, 5) respectively, show that PQ and
RS intersect. Also find the point of intersection.
2.
The direction cosines of three mutually
perpendicular lines are l1, m1, n1; l2, m2, n2 and
l3, m3, n3. Prove that the direction cosines of the line
equally inclined to them are
l1  l 2  l 3 m 1  m 2  m 3 n 1  n 2  n 3
,
,
3
3
3
3.
11.
A triangle is so placed that the middle points of its
sides are on the axes. If a, b, c be the length of its
sides, show that the equation to its plane is
x y z
   1 , where 8x 2 = b 2 + c 2 – a 2 ,
1
x1 y1 z1
8y12 = c2 + a2 – b2 and 8z12 = a2 + b2 – c2.
12.
The plane dx + by = 0 is rotated through an angle 
about its line of intersection with the plane z = 0.
Show that the equation to the plane in new position
is
ax  by  z a 2  b 2 tan   0 .
The direction cosines of two lines are given by the
equations 3m + n + 5l = 0, 6nl – 2lm + 5mn = 0. Find
the angle between them ?
13.
4.
Prove that the line joining the mid-point of
opposite edges of a tetrahedron are concurrent.
Find the reflection of the plane ax + by + cz + d = 0
in the plane ax  b y  cz  d  0 .
14.
5.
Prove that the two lines whose direction cosines are
given by the relation pl + qm + rn = 0 and
al 2 + bm 2 + cn 2 = 0 are perpendicular if
p2(b + c) + q2(c + a) + r2(a + b) = 0 and parallel if
If P be any point on the plane lx + my + nz = p and
Q be a point on the line OP such that OP . OQ = p2,
show that the locus of the point Q is p(lx + my + nz)
= x2 + y2 + z2.
15.
A point P moves on a plane
p2 q2 r2
  0.
a
b c
6.
7.
9.
10.
4.
3
16.
If the edges of a rectangular parallelopiped are
p, q, r show that the angle between four diagonals
are given by
8.
through P and perpendicular to OP meets the
co-ordinate axes in A, B and C. If the planes through
A, B and C parallel to the planes x = 0, y = 0, z = 0
intersect in Q, find the locus of Q.
If a line makes angles , , ,  with the diagonals
of a cube, prove that
cos 2  cos 2  cos 2   cos 2  
 p2  q2  r 2 
.
cos1  2
2
2 
p

q

r


x y z
   1 . A plane
a b c
P is a given point and PM and PN are
perpendicular from P to xz and xy-plane
respectively. If OP makes angles , , ,  with the
plane OMN and the co-ordinates planes
respectively, prove that
cot2 = cot2 + cot2 + cot2 + 2
where O is the origin.
17.
If two pairs of opposite edges of a tetrahedron are
mutually perpendicular, show that the third pair
will also be mutually perpendicular.
Through a point P(h, k, l) a plane is drawn at right
angles to OP to meet the co-ordinates axes in A, B
and C. If OP = p, show that the area of ABC is
p5
.
2hkl
Find the length and direction cosines of a line
segment whose projections on the co-ordinates axes
are 6, –3, 2.
18.
Find the volume of the tetrahedron with vertices
P(2, 3, 2), Q(1, 1, 1), R(3, –2, 1) and S(7, 1, 4).
Find the equation of line of intersection of the planes
19.
Find the equation of the projection of the line

   
  
r .(3 i  j  k )  1 and r .( i  4 j  2k )  2
x 1 y  1 z  3


2
1
4
on
the
x + 2y + z = 9.
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
plane
M3D – 23
20.
Find the image of the point (, , ) with respect to
the plane 2x + y + z = 6. Hence find the image of the
line
23.
a1x + b1y + c1z + d1 = 0 = a2x + b2y + c2z + d2 = 0 and
which is parallel to the line
x 1 y  2 z  3


with respect to the
2
1
4
plane.
21.
x y  z 


.
l
m
n
Find the equation of the line which passes through
the point P(, , ) and is parallel to the line
a1x + b1y + c1z + d1 = 0 = a2x + b2y + c2z + d2 = 0
22.
Find the equation of the plane which passes through
the line
Find the equation of the projection of line
3x – y + 2z – 1 = 0, x + 2y – z – 2 = 0 on the plane
3x + 2y + z = 0.
24.
If the planes x – cy – bz = 0, cx – y + az = 0 and
bx + ay – z = 0 pass through a straight line, then
find the value of a2 + b2 + c2 + 2abc.
25.
If the planes y = az + cx, x = cy + bz and z = bx + ay
meet in a line show that the line of intersection of
x
these planes is
1 a
26.
2

y
1b
2

z
1  c2
Prove that the shortest distance between any two
opposite edges of a tetrahedron formed by the
planes y + z = 0, x + z = 0, x + y = 0, x + y + z = 3a
is 2a.
ANSWERS SUBJECTIVE (INITIAL STEP EXERCISE)
1.
2, 6, 8
2.
C divides AB externally in the
ratio 2 : 1
3.
1530
unit
8
4.
Q  (4, 0, –1)
5.
1
  cos 1  
3
6.
2 units.
7.
AB internally in the ratio
8.
2 2 1
, , 
3 3 3
9.
 19 28 41 
Q , , 
9 9 9
10.
8
7
11.
x 2  y2 z2

 1 (a is constant quantity)
a 2  2 a 2
12.
a = 4, b = 2
13.
C  (2, 5, 5)
15.
(m1n2 – m2n1), (n1l2 – n2l1), (l1m2 – l2m1)
16.
2y + z = 4.
17.
3x + 4y – 5z = 9
19.
(b1c2 – b2c1) (x – ) + (c1a2 – a1c2)(y – ) + (a1b2 – a2b1) (z – ) = 0.
20.
32x – 5y + 8z = 83.
21.
5x + y + 4z = 20
22.
1
90 0  cos 1  
6
23.
x + 35y – 10z = 156
24.
35x + y + 62z = 78, 17x + 25y – 10z = 156
25.
xyz = 6k3
Einstein Classes,
23  2529
23  2529
and externally in the ratio
50
50
18.
26.
.
x + 3y + z + 11 = 0
x2 + y2 + z2 = 36
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
M3D – 24
27.
x–2 + y–2 + z–2 = p–2
28.
x–2 + y–2 + z–2 = 16p–2
29.
x 1 y
z2


1
2
7
30.
(–9, –1, 1)
31.
x1 y  2 z 0


2
3
4
33.
5 2 8
 , , 
 2 3 3
35.
(2, 4, –3)
36.
x1 y 1 z


14
 12 2
37.
900
38.
(1, –2, –4)
39.
61
283
40.
34.
x+y+z=0
(i)
–16x + y + 4z + 33 = 0, –23x + 8y + 11z + 26 = 0
(ii)
1 
1 
 13 11
 7
,
,   and  , 3 , 

3
2 
3 
 6
 3
42.
x 2 y3 z 2


0
3
2
44.
(m1n2 – m2n1) (x – x1) + (l2n1 – l1n2) (y – y1) + (l1m2 – l2m1) (z – z1) = 0
45.
x2 + y2 + z2 – 2x – 4y + 6z – 60 = 0, 3x – y + 8z – 14 = 0
46.
x y z
  3
a b c
43.
47.
41.
7
6x + 2y + 13z – 10 = 0.
 5
,   tan 1 2
  tan 1 
 3 


ANSWERS SUBJECTIVE (FINAL STEP EXERCISE)
1.
3 5 9
A , , 
2 2 2
9.
Length = 7;
13.
2(aa  bb  cc )(ax  by  cz  d )  (a 2  b 2  c 2 )(ax  by  cz  d)
15.
1
1
1
1
1
1

  2 2 2
ax by cz x
y
z
18.
½ unit.
20.
    2  2  12  2  2    6  2    2   6 
,
,


3
3
3


21.
x 
y
z


b1c 2  b 2 c1 c1a 2  c 2 a 1 a1b 2  a 2 b1
22.
x1 y 1 z 1


11
9
 15
23.
24.
(a1x + b1y + c1z + d1) (a2l + b2m + c2n) – (a1l + b1m + c1n) (a2x + b2y + c2z + d2) = 0
1
Einstein Classes,
6 3 2
,
,
7 7 7
3.
1
  cos 1  
6
10.


 



1
r
(106 i  73 j  23k )  t( 2 i  7 j  13k )
222
19.
2 x  1 4 y  17 z


4
 14
5
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111