LINE
Vector equation of a line:
Equation of a line through a given point and parallel to the given vector:
Let 𝑎̅ be the position vector of the given point A.
Z
𝑏̅
Let 𝑙 be the line which passes through the
point A and is parallel to a given vector 𝑏̅.
A P
l
Let 𝑟̅ be the position vector of any
point P on the line 𝑙
𝑎̅ 𝑟̅
̅
̅̅̅̅
Then 𝐴𝑃 is parallel to the vector 𝑏
Y
̅̅̅̅ = 𝜆𝑏̅, where 𝜆 is a scalar.
i.e. 𝐴𝑃
̅̅̅̅ – ̅̅̅̅
But ̅̅̅̅
𝐴𝑃 = 𝑂𝑃
𝑂𝐴
X
O
̅
∴ 𝜆𝑏 = 𝑟̅ − 𝑎̅
Hence, the vector equation of the line is given by
𝑟̅ = 𝑎̅ + 𝜆 𝑏̅
…………..(1)
This equation is referred as parametric form of the line.
Remark:
(i)
If 𝑏̅ = 𝑎1 𝑖̂ + 𝑏1 𝑗̂ + 𝑐1 𝑘̂, then 𝑎1 , 𝑏1 , 𝑐1 are direction ratios of the line
and conversely, if 𝑎1 , 𝑏1 , 𝑐1 are direction ratios of a line, then 𝑏̅ = 𝑎1 𝑖̂ +
𝑏1 𝑗̂ + 𝑐1 𝑘̂ will be parallel to the line.
(ii) Cartesian form of an equation of a line. Let A ≡ (𝑥1 , 𝑦1 , 𝑧1 ) be a point on
the line and 𝑎, 𝑏, 𝑐 be direction ratios of the line. Consider any point
P(𝑥, 𝑦, 𝑧) on the line.
Then 𝑟̅ = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂
𝑎̅ = 𝑥1 𝑖̂ + 𝑦1 𝑗̂ + 𝑧1 𝑘̂
and
𝑏̅ = 𝑎1 𝑖̂ + 𝑏1 𝑗̂ + 𝑐1 𝑘̂
substituting these values in equation (1) and equating the coefficients of
𝑖̂, 𝑗̂, 𝑘̂, we get
𝑥 = 𝑥1 + 𝜆𝑎1 , 𝑦 = 𝑦1 + 𝜆𝑏1 , 𝑧 = 𝑧1 + 𝜆𝑐1
……… (2)
These are the parametric equations of the line. Eliminating the parameter
𝜆 from (2), we get
𝑥−𝑥1
𝑦−𝑦1
𝑧−𝑧1
=
=
…………..(3)
𝑎1
(iii)
(iv)
(v)
𝑏1
𝑐1
These are the Cartesian equations of the line.
If 𝑙, 𝑚. 𝑛 are direction cosines of the line, the equations of the line are
𝑥−𝑥1
𝑦−𝑦1
𝑧−𝑧1
=
=
…………..(4)
𝑙
𝑚
𝑛
Vector equations of a line passing through the origin and parallel to the
vector 𝑏̅ is
𝑟̅ = 𝜆 𝑏̅
(since 𝑎̅ = 0)
Vector equation of a line passing through the point A(𝑎̅) and parallel to the
given vector 𝑏̅ can also be expressed as
(𝑟̅ − 𝑎̅) × 𝑏̅ = 0.
Page 1 of 11
(vi)
The coordinates of any point on the line
𝑥−𝑥1
𝑦−𝑦1
𝑧−𝑧1
=
=
are (𝑥1 + 𝜆𝑎1 , 𝑦1 + 𝜆𝑏1 , 𝑧1 + 𝜆𝑐1 ),
𝑎1
𝑏1
𝑐1
where 𝜆𝜖𝑅.
Vector equation of the line passing through two gives points:
Let A(𝑥1 , 𝑦1 , 𝑧1 ) and B(𝑥2 , 𝑦2 , 𝑧2 )
Z
l
be any two points on line 𝑙.
A
P
B
Let 𝑎̅ and 𝑏̅ be the position
vectors of two points A and
𝑎̅.
𝑟̅ . 𝑏̅.
B respectively.
Let 𝑟̅ be the position vector of any
O
Y
point P (𝑥, 𝑦, 𝑧) on the line 𝑙.
Then ̅̅̅̅
𝐴𝑃 = 𝑟̅ − 𝑎̅ and
X
̅̅̅̅ = 𝑏̅ − 𝑎̅.
𝐴𝐵
The vectors ̅̅̅̅
𝐴𝑃 and ̅̅̅̅
𝐴𝐵 are collinear
Therefore
̅̅̅̅
̅̅̅̅ )
𝐴𝑃 = 𝜆 (𝐴𝐵
∴
𝑟̅ − 𝑎̅ = 𝜆 (𝑏̅ − 𝑎̅)
∴
𝑟̅ = 𝑎̅ + 𝜆 (𝑏̅-𝑎̅)
i.e. 𝑟̅ = (1 − 𝜆)𝑎̅ + 𝜆 𝑏̅
………….(1)
This is the vector equation of the line passing through the points A(𝑎̅) and B(𝑏̅).
Remark:
(i)
If 𝑟̅ = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂, 𝑎̅ = 𝑥1 𝑖̂ + 𝑦1 𝑗̂ + 𝑧1 𝑘̂ and
𝑏̅ = 𝑥2 𝑖̂ + 𝑦2 𝑗̂ + 𝑧2 𝑘̂
from (1), we get
𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂ = 𝑥1 𝑖̂ + 𝑦1 𝑗̂ + 𝑧1 𝑘̂ + 𝜆 [(𝑥2 − 𝑥1 )𝑖̂ + (𝑦2 − 𝑦1 )𝑗̂ + (𝑧2 − 𝑧1 )𝑘̂]
Equating the coefficients of 𝑖̂, 𝑗̂, 𝑘̂ we get,
𝑥 = 𝑥1 + 𝜆(𝑥2 − 𝑥1 ), 𝑦 = 𝑦1 + 𝜆(𝑦2 − 𝑦1 ), 𝑧 = 𝑧1 + 𝜆(𝑧2 − 𝑧1 ),
on eliminating 𝜆, we obtain
𝑥 − 𝑥1
𝑦 − 𝑦1
𝑧 − 𝑧1
=
=
𝑥2 − 𝑥1 𝑦2 − 𝑦1 𝑧2 − 𝑧1
Which are the equations of the line in Cartesian form passing through two
points A (𝑥1 , 𝑦1 , 𝑧1 ) and B(𝑥2 , 𝑦2 , 𝑧2 ).
(ii) The equation 𝑟̅ = (1 − 𝜆)𝑎̅ + 𝜆 𝑏̅ is called the parametric vector equation of a
line passing through two points, where 𝜆 is parameter.
(iii) The vector equation of a line passing through two points A(𝑎̅) and B(𝑏̅) is also
given by (𝑟̅ − 𝑎̅) × (𝑏̅ − 𝑎̅) = 0. This equation is also referred as nonparametric equation of the line.
(iv) If 𝑟̅ = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂
𝑎̅ = 𝑎1 𝑖̂ + 𝑏1 𝑗̂ + 𝑐1 𝑘̂ and 𝑏̅ = 𝑎2 𝑖̂ + 𝑏2 𝑗̂ + 𝑐2 𝑘̂
Page 2 of 11
Then 𝑟̅ = (1 − 𝜆)𝑎̅ + 𝜆 𝑏̅ gives
𝑥−𝑎1
𝑎2 −𝑎1
=
𝑦−𝑏1
𝑏2 −𝑏1
=
𝑧−𝑐1
𝑐2 −𝑐1
=𝜆
This is the symmetric form of equation of a line.
Examples
1) Find the vector equation of the line through A(3, 4, -7) and
B (6, -1, 1). Also find the Cartesian form.
Sol: The line pass through the pt. A(3, 4, -7) and B(6, -1, 1)
Let 𝑎̅ and 𝑏̅ be the position vector of the points A and B.
𝑎̅ = 3𝑖̂ + 4𝑗̂ − 7𝑘̂ and 𝑏̅ = 6𝑖̂ − 𝑗̂ + 𝑘̂
𝑏̅ – 𝑎̅ = (6𝑖̂ − 𝑗̂ + 𝑘̂) – (3𝑖̂ + 4𝑗̂ − 7𝑘̂)
𝑏̅ – 𝑎̅ = 3𝑖̂ − 5𝑗̂ + 8𝑘̂
The vector equation of the line passing through the points A(𝑎̅) and B(𝑏̅) is
𝑟̅ = 𝑎̅ + 𝜆( 𝑏̅ – 𝑎̅)
𝑟̅ = (3𝑖̂ + 4𝑗̂ − 7𝑘̂) + λ (3𝑖̂ − 5𝑗̂ + 8𝑘̂)
The Cartesian form of the equation of the line passing through
(𝑥1 , 𝑦1 , 𝑧1 ) and (𝑥2 , 𝑦2 , 𝑧2 ) is
𝑥 − 𝑥1
𝑦 − 𝑦1
𝑧 − 𝑧1
=
=
𝑥2 − 𝑥1 𝑦2 − 𝑦1 𝑧2 − 𝑧1
∴
𝑥−3
6−3
i.e.
=
𝑥−3
3
𝑦−4
−1−4
=
𝑥−4
−5
=
=
𝑧−(−7)
1−(−7)
𝑧+7
8
2) Find the equation of the line in vector form passing through the point (4, -2, 5)
and parallel to the vector 3𝑖̂ − 𝑗̂ + 2𝑘̂. Hence, find the equation in Cartesian form.
Sol: Given that the line pass through the point A(4, -2, 5) and
parallel to the vector 𝑏̅ = 3𝑖̂ − 𝑗̂ + 2𝑘̂.
Let a be the position vector of the point A
∴ 𝑎̅ = 4𝑖̂ − 2𝑗̂ + 5𝑘̂
∴ The vector equation of line passing through the pt. A(𝑎̅) and
parallel to vector 𝑏̅ is
𝑟̅ = 𝑎̅ + 𝜆𝑏̅ where λ is scalar
𝑟̅ = (4𝑖̂ − 2𝑗̂ + 5𝑘̂) + λ (3𝑖̂ − 𝑗̂ + 2𝑘̂.)
If 𝑟̅ = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂ then
𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂ = (4𝑖̂ − 2𝑗̂ + 5𝑘̂) + λ (3𝑖̂ − 𝑗̂ + 2𝑘̂)
(𝑥 − 4)𝑖̂ + (𝑦 + 2)𝑗̂ + (𝑧 − 5)𝑘̂ = 3𝜆𝑖̂ − 𝜆𝑗̂ + 2𝜆𝑘̂
Comparing the coefficients of 𝑖̂, 𝑗̂, 𝑘̂ on both sides.
𝑥 − 4 = 3𝜆, 𝑦 + 2 = −𝜆, 𝑧 − 5 = 2𝜆
Page 3 of 11
𝑥−4
3
=
𝑦+2
−1
=
𝑧−5
2
=λ
∴ Cartesian form of the equation of line is
𝑥−4
3
=
𝑦+2
−1
=
𝑧−5
2
3) Find the equation of line passing through the point (5, 4, 3) and having direction
ratios -3, 4, 2.
Sol: Given that the line pass through the point A(5, 4, 3) and having
direction ratios -3, 4, 2
Let 𝑎̅ be the position vector of pt. A
𝑎̅ = 5𝑖̂ + 4𝑗̂ + 3𝑘̂
Let 𝑏̅ be the vector parallel to the line whose direction ratios are
-3, 4, 2
𝑏̅ = −3𝑖̂ + 4𝑗̂ + 2𝑘̂
∴ The vector equation of the line passing through A(𝑎̅) and parallel
to 𝑏̅ is
𝑟̅ = 𝑎̅ + 𝜆𝑏̅
𝑟̅ = (5𝑖̂ + 4𝑗̂ + 3𝑘̂) + λ (−3𝑖̂ + 4𝑗̂ + 2𝑘̂)
𝑥−6
𝑦+4
𝑧−5
4) The Cartesian equation of line is
=
=
find the vector equation of the
2
7
3
line. (March 14)
Sol: The Cartesian equation of line is
𝑥−6
𝑦+4
𝑧−5
=
=
2
7
3
∴ The line is passing through the point
A(6, -4, 5) and having direction ratios 2, 7, 3
Let 𝑎̅ be the position vector of the point A and 𝑏̅ be the vector
parallel to the line.
∴ 𝑎̅ = 6𝑖̂ − 4𝑗̂ + 5𝑘̂, 𝑏̅ = 2𝑖̂ + 7𝑗̂ + 3𝑘̂
The vector equation of line is
𝑟̅ = 𝑎̅ + 𝜆𝑏̅
𝑟̅ = (6𝑖̂ − 4𝑗̂ + 5𝑘̂) + λ (2𝑖̂ + 7𝑗̂ + 3𝑘̂)
5) Find the vector equation of a line passing through the point (𝑖̂ + 2𝑗̂ + 3𝑘̂) and
perpendicular to the vector (𝑖̂ + 𝑗̂ + 𝑘̂) and (2𝑖̂ − 𝑗̂ + 𝑘̂).
Sol: Let 𝑎̅ = 𝑖̂ + 2𝑗̂ + 3𝑘̂, ̅𝑏 = 𝑖̂ + 𝑗̂ + 𝑘̂, ̅𝑐 = 2𝑖̂ − 𝑗̂ + 𝑘̂
The vector perpendicular to the vectors ̅𝑏 and ̅𝑐 is given by
𝑖̂
𝑗̂ 𝑘̂
𝑏̅ × 𝑐̅ = |1 1 1| = 𝑖̂(1 + 1) − 𝑗̂(1 − 2) + 𝑘̂(−1 − 2) = 2𝑖̂ + 𝑗̂ − 3𝑘̂
2 −1 1
Since line is perpendicular to the vector 𝑏̅ and 𝑐̅ it is parallel to
𝑏̅ × 𝑐̅.
Page 4 of 11
The vector equation of the line passing through A(𝑎̅) and parallel to 𝑏̅ × 𝑐̅ is
𝑟̅ = 𝑎̅ + 𝜆 (𝑏̅ × 𝑐̅)
∴ Vector equation of the required line is
𝑟̅ = (𝑖̂ + 2𝑗̂ + 3𝑘̂) + λ (2𝑖̂ + 𝑗̂ − 3𝑘̂ ).
6) Find the vector equation of the line passing through the point
(-1, -1, 2) and parallel to the line 2𝑥 − 2 = 3𝑦 + 1 = 6𝑧 − 2
Sol: Let 𝑎̅ be the position vector of the point A(-1, -1, 2)
∴ 𝑎̅ = −𝑖̂ − 𝑗̂ + 2𝑘̂
Given equation of line is
2𝑥 − 2 = 3𝑦 + 1 = 6𝑧 − 2
1
1
2(𝑥 − 1) = 3 (𝑦 + ) = 6(𝑧 − )
𝑥−1
1
2
=
𝑦+
1
3
1
3
=
3
1
𝑧−
1
6
3
3
∴ Direction ratios are
1 1 1
, , i.e. 3, 2, 1
2 3 6
Let 𝑏̅ be the vector parallel to required line
𝑏̅ = 3𝑖̂ + 2𝑗̂ + 𝑘̂
∴ The vector equation of the line passing through A(𝑎̅) and parallel to 𝑏̅ is
𝑟̅ = 𝑎̅ + 𝜆𝑏̅
𝑟̅ = (−𝑖̂ − 𝑗̂ + 2𝑘̂) + λ (3𝑖̂ + 2𝑗̂ + 𝑘̂ )
7) The Cartesian equation of a line is 3𝑥 − 1 = 6𝑦 + 2 = 1 − 𝑧. Find the direction
ratios and write its equation in vector form.
Sol: The Cartesian equation of a line is
3𝑥 − 1 = 6𝑦 + 2 = 1 − 𝑧
1
1
3 (𝑥 − ) = 6 (𝑦 + ) = −(𝑧 − 1)
x −1/3
1/3
3
=
y+1/3
1/6
=
3
z −1
−1
1
1
3
3
∴ The line pass through the point A( , − , 1) and having
1 1
Direction ratios , , −1 i.e. 2, 1, −6
3 6
Let 𝑎̅ be the position vector of the point A and 𝑏̅ be the vector
parallel to the line.
1
1
𝑎̅ = 𝑖̂ − 𝑗̂ + 𝑘̂ and 𝑏̅ = 2𝑖̂ + 𝑗̂ − 6𝑘̂
3
3
∴ vector equation is
𝑟̅ = 𝑎̅ + 𝜆𝑏̅
1
1
r = ( 𝑖̂ − 𝑗̂ + 𝑘̂) + λ (2𝑖̂ + 𝑗̂ − 6𝑘̂)
3
3
Page 5 of 11
8) Show that the points whose position vectors are 5𝑖̂ + 5𝑘̂ , −4𝑖̂ + 3𝑗̂ − 𝑘̂ and 2𝑖̂ +
𝑗̂ + 3𝑘̂ are collinear.
Sol: Let A, B, C be the points with position vector 𝑎̅, 𝑏̅, 𝑐̅
𝑎̅ = 5𝑖̂ + 5𝑘̂, 𝑏̅ = −4𝑖̂ + 3𝑗̂ − 𝑘̂, 𝑐̅ = 2𝑖̂ + 𝑗̂ + 3𝑘̂
vector equation of line AB is
𝑟̅ = 𝑎̅ + 𝜆(𝑏̅ − 𝑎̅)
𝑏̅ − 𝑎̅ = −4𝑖̂ + 3𝑗̂ − 𝑘̂) – (5𝑖̂ + 5𝑘̂) = −9𝑖̂ + 3𝑗̂ − 6𝑘̂
∴ 𝑟̅ = (5𝑖̂ + 5𝑘̂) + λ (−9𝑖̂ + 3𝑗̂ − 6𝑘̂) …..(1)
Now the points A, B, C are collinear if the point C lies on the line AB i.e. the
position vector 𝑐̅ of 𝐶 must satisfy the equation (i) for some scalar λ.
∴
Replacing 𝑟̅ by 𝑐̅ in equation (1) we get
2𝑖̂ + 𝑗̂ + 3𝑘̂ = (5𝑖̂ + 5𝑘̂) + λ (−9𝑖̂ + 3𝑗̂ − 6𝑘̂)
2𝑖̂ + 𝑗̂ + 3𝑘̂ − (5𝑖̂ + 5𝑘̂) = λ (−9𝑖̂ + 3𝑗̂ − 6𝑘̂)
−3𝑖̂ + 𝑗̂ − 2𝑘̂ = λ (−9𝑖̂ + 3𝑗̂ − 6𝑘̂)
1
(−9𝑖̂ + 3𝑗̂ − 6𝑘̂) = λ (−9𝑖̂ + 3𝑗̂ − 6𝑘̂)
3
∴
1
λ=
3
Thus the points A,B, C are collinear.
1)
2)
3)
4)
Examples for practice
Find the vector equation of a line which passes through the point with position
vector 4𝑖̂ − 𝑗̂ + 2𝑘̂ and is in the direction of −2𝑖̂ + 𝑗̂ + 𝑘̂. Also reduce the
equation to Cartesian form.
Find the vector equation of the line passing through the point
with position vector 2𝑖̂ + 𝑗̂ − 𝑘̂ and parallel to the line joining the points –−𝑖̂ +
𝑗̂ + 4𝑘̂ and ̂𝑖 + 2𝑗̂ + 2𝑘̂. Also find the Cartesian form of this equation.
Find the Cartesian equation of line passing through the points A(4, 2, 1) and
B (2, -1, 3). Also, reduce it to vector form.
𝑥−1
𝑦−3
𝑧
𝑥−2
𝑦+1
𝑧−4
Find the angle between the lines
=
= and
=
=
.
4
1
8
2
2
1
P(𝛼̅))
Distance of a point from a line:
Let P (𝛼̅) be a point and
Let 𝑟̅ = 𝑎̅ + 𝜆𝑏̅ be the vector
equation of a line.
A(𝑎̅)
The line passes through a point A(𝑎̅).
Let M be the foot of the perpendicular drawn from the point
P (𝛼̅) on the line 𝑟̅ = 𝑎̅ + 𝜆𝑏̅.
̅̅̅̅̅ ∥ 𝑏̅
Now 𝐴𝑀
̅̅̅̅̅ = 𝜆𝑏̅
∴ 𝐴𝑀
Page 6 of 11
M
𝑏̅.
and 𝑙(AM) =
̅̅̅̅.𝐴𝑀
̅̅̅̅̅
𝐴𝑃
̅̅̅̅̅
|𝐴𝑀 |
=
̅̅̅̅̅
(𝛼
̅ −𝑎̅).𝐴𝑀
̅̅̅̅̅
|𝐴𝑀|
=
(𝛼
̅ −𝑎̅).𝑏̅
|𝑏̅|
……………..(1)
Now, ∆AMP is right angled triangle.
∴
̅
(𝛼
̅̅̅̅)2 – (𝐴𝑀
̅̅̅̅̅)2 = |𝛼̅ − 𝑎̅|2 − [ ̅ −𝑎̅).𝑏]2
̅̅̅̅̅)2 = (𝐴𝑃
(𝑃𝑀
|𝑏̅|
∴
̅ −𝑎
̅ ).𝑏
(𝛼
̅−𝑎
̅ |2 − [ ̅ ]
𝑃𝑀 = √|𝛼
̅ 2
|𝑏|
Hence, distance of a point P (𝛼) from a line is 𝑟̅ = 𝑎̅ + 𝜆𝑏̅.
̅−𝑎
̅ |2 − [
is √|𝛼
̅ 2
̅ −𝑎
̅ ).𝑏
(𝛼
̅|
|𝑏
]
Cartesian form:
Let P (𝑥 ′ , 𝑦 ′ , 𝑧′) be a given point and
P (𝑥 ′ , 𝑦 ′ , 𝑧′)
𝑥−𝑥1
𝑦−𝑦1
𝑧−𝑧1
Let
=
=
be given line.
𝑎
𝑏
𝑐
Let M be the foot of the perpendicular
drawn from P(𝑥 ′ , 𝑦 ′ , 𝑧′)
A
B
𝑥−𝑥1
𝑦−𝑦1
𝑧−𝑧1
On the line
=
=
𝑎
𝑏
𝑐
Let coordinates of M be (𝑥1 + 𝜆𝑎, 𝑦1 + 𝜆𝑏, 𝑧1 + 𝜆𝑐) then direction ratios of
PM are
(𝑥1 + 𝜆𝑎 − 𝑥 ′ , 𝑦1 + 𝜆𝑏 − 𝑦 ′ , 𝑧1 + 𝜆𝑐 − 𝑧′)
The direction ratios of AB are a,b,c.
PM is perpendicular to AB
𝑎(𝑥′−𝑥 )+𝑏(𝑦′−𝑦 )+𝑐(𝑧′−𝑧 )
1
1
1
∴𝜆=
𝑎2 +𝑏2 +𝑐 2
Putting this value of 𝜆 in (𝑥1 + 𝜆𝑎, 𝑦1 + 𝜆𝑏, 𝑧1 + 𝜆𝑐) we obtain coordinate of
M, and using distance formula we get the length PM.
Skew Lines:
If two lines in space interest at a point,
then the shortest distance between them is zero.
Also, if two lines in space are parallel,
then the shortest distance between them
is the perpendicular distance between them.
Further, in space, there are lines
which are neither intersecting nor parallel.
In fact, such pair of lines are
called non-coplanar or skew lines.
Page 7 of 11
Distance between skew lines:
𝑙 1 and 𝑙 2 are two skew lines therefore there is only one line perpendicular to each
of lines 𝑙1 and 𝑙 2 which is known as the line of shortest distance. Here distance PQ is
called to be shortest distance.
Let 𝑙 1 and 𝑙 2 be two lines whose equations are
𝑟̅ = 𝑎̅1 + 𝜆𝑏̅1 and 𝑟̅ = 𝑎̅2 + 𝜆𝑏̅2 respectively.
∴ 𝑙 1 passes through A(𝑎̅1 ) and 𝑙 2 passes through B (𝑎̅2 )).
Also, 𝑙 1 and 𝑙 2 are parallel to 𝑏̅1 and 𝑏̅1 respectively.
̅̅̅̅is perpendicular to both 𝑙 1 and 𝑙 2
𝑃𝑄
̅̅̅̅ is parallel to 𝑏̅1 × 𝑏̅2 .
∴ 𝑃𝑄
̅ ̅
̅̅̅̅, then 𝑛̂ = ± 𝑏1×𝑏2
Let 𝑛̂ be a unit vector along 𝑃𝑄
|𝑏̅1 ×𝑏̅2 |
̅̅̅̅ = Projection of 𝐴𝐵
̅̅̅̅ on PQ
∴ 𝑃𝑄
̅̅̅̅ = ̅̅̅̅
∴ 𝑃𝑄
𝐴𝐵. 𝑛̂
(𝑎̅2 −𝑎̅1 ).(𝑏1 ×𝑏̅2 )
𝑏̅1 ×𝑏̅2
=
|𝑏̅1 ×𝑏̅2 |
|𝑏̅1 ×𝑏̅2 |
(𝑎̅ −𝑎̅ ).(𝑏 ×𝑏̅ )
PQ = | 2 ̅1 ̅ 1 2 |
|𝑏1 ×𝑏2 |
= (𝑎̅2 − 𝑎̅1 ).
Hence, distance
Cartesian form: The shortest distance between two skew lines
𝑥−𝑥1
𝑦−𝑦1
𝑧−𝑧1
𝑥−𝑥2
𝑦−𝑦2
𝑧−𝑧2
=
=
and
=
=
is
𝑎1
|
𝑏1
𝑐1
𝑥2 − 𝑥1
| 𝑎1
𝑎2
𝑎2
𝑦2 − 𝑦1
𝑏1
𝑏2
𝑏2
𝑧2 − 𝑧1
𝑐1 |
𝑐2
𝑐2
|
|√(𝑏1 𝑐2 − 𝑏2 𝑐1 )2 + (𝑐1 𝑎2 − 𝑐2 𝑎1 )2 +(𝑎1 𝑏2 − 𝑎2 𝑐1 )2 |
Distance between parallel lines:
If two lines 𝑙1 and 𝑙2 are parallel,
then they are coplanar.
Let the lines be given by
𝑟̅ = 𝑎̅1 + 𝜆𝑏̅ and
……………(1)
̅
𝑟̅ = 𝑎̅2 + 𝜇𝑏
…………..(2)
where 𝑎̅1 is the position vector of a
𝑏̅
B(𝑎̅2 )
point A on 𝑙1 and 𝑎̅2 be the position vector
𝑙2
of a point B on 𝑙2
As 𝑙1 , 𝑙2 are coplanar, if the foot of
the perpendicular from B on the line 𝑙1 is M,
Figure 9.7
then the distance between the lines 𝑙1 and 𝑙2 is
|BM|.
𝑙1
Let 𝜃 be the angle between
A(𝑎̅1 )
M
̅̅̅̅ and 𝑏̅
The vectors 𝐴𝐵
̅̅̅̅ | sin 𝜃)𝑛̂
Thus, 𝑏̅ × ̅̅̅̅
𝐴𝐵 = (|𝑏̅| |𝐴𝐵
……………….(3)
Page 8 of 11
Where 𝑛̂ is unit vector perpendicular to
the plane of 𝑏̅ and ̅̅̅̅
𝐴𝐵
𝐵𝑀
𝐵𝑀
̅̅̅̅ | sin 𝜃 = BM
From triangle AMB, sin 𝜃 =
= ̅̅̅̅ ∴ |𝐴𝐵
𝐴𝐵
|𝐴𝐵 |
Now, ̅̅̅̅
𝐴𝐵 = 𝑎̅2 − 𝑎̅1
Therefore, from (3), we get
𝑏̅ × (𝑎̅2 − 𝑎̅1 )
=
|𝑏̅| BM 𝑛̂
i.e.
|𝑏̅ x (𝑎̅2 − 𝑎̅1 )|
=
|𝑏̅| BM
Hence, the distance between the parallel lines is d where
d = BM = |
𝑏̅ × (𝑎̅2 −𝑎̅1 )
|
|𝑏̅ |
= |(𝑎̅2 − 𝑎̅1 ) × 𝑏̂|
Remark: The lines are intersecting, if shortest distance is zero.
𝑥2 − 𝑥1 𝑦2 − 𝑦1 𝑧2 − 𝑧1
𝑏1
𝑐1 | = 0
i.e. | 𝑎1
𝑎2
𝑏2
𝑐2
Examples
1) Find the length of the perpendicular from (2, −3, 1) to the line
𝑥+1
𝑦−3
𝑧+2
=
=
.
2
3
−1
Sol: Let M be the foot of perpendicular drawn from the
point P(2, −3, 1) to the given line
The co-ordinate of any point on the line
𝑥+1
𝑦−3
𝑧+2
=
=
are 2λ – 1, 3λ +3, −λ −2
2
3
−1
Let M ≡ (2λ −1, 3λ+3, −λ −2) …………… (i)
Direction ratios PM are
2λ −1−2,
3λ +3+3,
−λ −2−1
i.e. 2λ −3, 3λ +6,
−λ −3
Direction ratios of the given line are 2, 3, −1.
Since PM is perpendicular to the given line.
2(2λ−3) + 3 (3λ+6) – 1 (−λ−3) = 0
4λ−6 + 9λ+18 + λ+3 = 0
14λ+15 = 0
−15
λ=
14
−15
Putting λ =
M≡(
in (i) we get
14
−44 −3 −13
14
,
14
,
14
−44
∴ PM = √(
14
)
−72 2
= √(
2
−3
− 2) + (
14
39 2
2
−13
+ 3) + (
14
−27 2
) + (14) + ( 14 )
14
Page 9 of 11
2
− 1)
(|𝑛̂|=1)
√5184+1521+729
14
√7334
=
PM =
14
units
2) Find the shortest distance between the lines
𝑟̅ = (4𝑖̂ − 𝑗̂) +λ (𝑖̂ + 2𝑗̂ − 3𝑘̂ ) and 𝑟̅ = (𝑖̂ − 𝑗̂ + 2𝑘̂) ) + 𝜇 (𝑖̂ + 4𝑗̂ − 5𝑘̂)).
Sol: We know that the shortest distance between the lines
̅𝑟 = 𝑎̅1 + 𝜆𝑏̅1 and
……………(1)
̅̅̅
𝑟̅ = 𝑎̅2 + 𝜆𝑏2
is
𝑑=|
̅ ×𝑏
̅ ).(𝑎
̅1)
(𝑏
1
2 ̅ 2 −𝑎
|
̅
̅
|𝑏1 ×𝑏2 |
Here ̅̅̅
𝑎1 = 4𝑖̂ − 𝑗̂ , ̅̅̅
𝑎2 = 𝑖̂ − 𝑗̂ + 2𝑘̂
𝑏̅1 = 𝑖̂ + 2𝑗̂ − 3𝑘̂ , ̅̅̅
𝑏2 = 𝑖̂ + 4𝑗̂ − 5𝑘̂
𝑖̂
𝑏̅1 × 𝑏̅2 = |1
1
𝑘̂
−3| = 𝑖̂(−10 + 12) – 𝑗̂ (−5 + 3) + 𝑘̂ (4 − 2)
−5
𝑗̂
2
4
= 2𝑖̂ + 2𝑗̂ + 2𝑘̂
𝑎̅2 − 𝑎̅1 = (𝑖̂ − 𝑗̂ + 2𝑘̂ ) – (4𝑖̂ − 𝑗̂) = −3𝑖̂ + 2𝑘̂
̅1 × 𝑏̅2 ). (𝑎̅2 − 𝑎̅1 ) = (2𝑖̂ + 2𝑗̂ + 2𝑘̂). (−3𝑖̂ + 2𝑘̂)
(𝑏
= 2(−3) + 2(0) + 2(2)
= −6 + 4
= −2
2
|𝑏̅1 × 𝑏̅2 | = √(2) + (2)2 + (2)2
= √4 + 4 + 4 = 2 √3
−2
1
∴ shortest distance = | | =
units.
2√3
√3
3) Find the shortest distance between the lines
𝑥+1
𝑦+1
𝑧+1
𝑥−3
𝑦−5
𝑧−7
=
=
and
=
=
.(Oct.13)
7
−6
1
1
−2
1
Sol: shortest distance between the lines
𝑥−𝑥1
𝑦−𝑦1
𝑧−𝑧1
𝑥−𝑥2
𝑦−𝑦2
𝑧−𝑧2
=
=
and
=
=
is
𝑎1
|
𝑏1
𝑐1
𝑎2
𝑥2 − 𝑥1
| 𝑎1
𝑎2
𝑏2
𝑦2 − 𝑦1
𝑏1
𝑏2
𝑐2
𝑧2 − 𝑧1
𝑐1 |
𝑐2
|
|√(𝑏1 𝑐2 − 𝑏2 𝑐1 )2 + (𝑐1 𝑎2 − 𝑐2 𝑎1 )2 +(𝑎1 𝑏2 − 𝑎2 𝑏1 )2 |
Given equations are
𝑥+1
7
=
𝑦+1
−6
=
𝑧+1
1
and
𝑥−3
1
=
𝑦−5
Page 10 of 11
−2
=
𝑧−7
1
∴ 𝑥1 = −1,
𝑦1 = −1, 𝑧1 = −1, 𝑥2 = 3, 𝑦2 = 5, 𝑧2 = 7, 𝑎1 = 7, 𝑏1 = −6,
𝑐1 = 1, 𝑎2 = 1, 𝑏2 = −2, 𝑐2 = 1
𝑥2 − 𝑥1
| 𝑎1
𝑎2
𝑦2 − 𝑦1 𝑧2 − 𝑧1
4 6 8
𝑏1
𝑐1 | = |7 −6 1|
𝑏2
𝑐2
1 −2 1
= 4(−6 + 2) − 6(7 − 1) + 8(−14 + 6)
= −16 − 36 − 64 = −116
(𝑏1 𝑐2 − 𝑏2 𝑐1 )2 + (𝑐1 𝑎2 − 𝑐2 𝑎1 )2 +(𝑎1 𝑏2 − 𝑎2 𝑏1 )2
= (−6 + 2)2 + (1 − 7)2 + (−14 + 6)2 = 16 + 36 + 64 = 116
−116
Shortest distance = |
| = √116 units.
√116
4) If the lines
Sol: The lines
𝑥−1
=
2
𝑥−𝑥1
𝑎1
Given equations are
∴ 𝑥1 = 1,
𝑦+1
=
=
𝑧−1
and
𝑥−3
3
4
𝑦−𝑦1
𝑧−𝑧1
𝑦−𝑘
𝑧
= interest then find the value of k.
2
𝑥−𝑥2
1
𝑦−𝑦2
=
𝑦+1
3
=
𝑧−1
4
and
𝑥−3
1
=
𝑦−𝑘
2
=
𝑧
1
𝑦1 = −1, 𝑧1 = 1, 𝑥2 = 3, 𝑦2 = 𝑘, 𝑧2 = 0, 𝑎1 = 2, 𝑏1 = 3,
𝑐1 = 4, 𝑎2 = 1, 𝑏2 = 2, 𝑐2 = 1
Since the line intersect
2 𝑘 + 1 −1
|2
3
4 |=0
1
2
1
2(3 − 8) − (𝑘 + 1)(2 − 4) − 1(4 − 3) = 0
−10 − (𝑘 + 1)(−2) − 1 = 0
2𝑘 − 9 = 0
𝑘=
𝑧−𝑧
2
=
and
=
=
intersect if
𝑏1
𝑐1
𝑎2
𝑏2
𝑐2
𝑥2 − 𝑥1 𝑦2 − 𝑦1 𝑧2 − 𝑧1
𝑏1
𝑐1 | = 0
| 𝑎1
𝑎2
𝑏2
𝑐2
𝑥−1
2
1
=
9
2
Page 11 of 11