JEE Advanced 2022 Paper 2 · Mathematics · Three Dimensional Geometry · easy

Vec r=(1,2,3)+λ(1,1,1) and L2: vec r=(0,1,-1)+μ(2,1,0). If d denotes the shortest distance between L1 and L2, then 2d^2 equals

Numerical answer

Consider the lines

L_1:\ \vec r=(1,2,3)+\lambda(1,1,1)

and

L_2:\ \vec r=(0,1,-1)+\mu(2,1,0)

. If

d

denotes the shortest distance between

L_1

and

L_2

, then

2d^{2}

equals

Show answer and solution

Answer: 3

The lines are skew, so

d=\dfrac{|(\vec b-\vec a)\cdot(\vec d_1\times\vec d_2)|}{|\vec d_1\times\vec d_2|}

where

\vec a=(1,2,3)

\vec b=(0,1,-1)

\vec d_1=(1,1,1)

\vec d_2=(2,1,0)

. Compute

\vec d_1\times\vec d_2=(1\cdot0-1\cdot1,\ 1\cdot2-1\cdot0,\ 1\cdot1-1\cdot2)=(-1,2,-1)

, whose squared length is

6

. Also

\vec b-\vec a=(-1,-1,-4)

, so the dot product is

1-2+4=3

. Hence

d^2=\dfrac{3^2}{6}=\dfrac32

and

2d^2=3

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