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3D Geometry question

2024 · 9 Apr · Shift 1 · Q44

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JEE Main
Mathematics
3D Geometry
2024 · 9 Apr · Shift 1 · Q44

3D Geometry question

2024 · 9 Apr · Shift 1 · Q44

JEE MainMathematics3D GeometryMCQ+4 / −1

The shortest distance between the lines

\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}

and

\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}

is:

A
$\frac{185}{\sqrt{563}}$
B
$\frac{187}{\sqrt{563}}$
C
$\frac{178}{\sqrt{563}}$
D
$\frac{179}{\sqrt{563}}$

View written solutionFree

Correct answer: B

Write the lines in vector form

The given lines are

\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}

and

\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}.

So we can identify:

A point on line $L_1$ : $A(3,-7,1)$
Direction vector of $L_1$ : $\vec d_1=(4,-11,5)$
A point on line $L_2$ : $B(5,9,-2)$
Direction vector of $L_2$ : $\vec d_2=(3,-6,1)$

Formula for shortest distance between two skew lines

For lines through points $A$ and $B$ with direction vectors $\vec d_1$ and $\vec d_2$ , the shortest distance is

D=\frac{|(\overrightarrow{AB})\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.

Compute $\vec d_1\times \vec d_2$

\vec d_1\times \vec d_2= \begin{vmatrix} \hat i & \hat j & \hat k\\ 4 & -11 & 5\\ 3 & -6 & 1 \end{vmatrix}

=\hat i\big((-11)(1)-5(-6)\big)-\hat j\big(4(1)-5(3)\big)+\hat k\big(4(-6)-(-11)(3)\big)

=\hat i(-11+30)-\hat j(4-15)+\hat k(-24+33)

=(19,11,9).

Hence,

|\vec d_1\times \vec d_2|=\sqrt{19^2+11^2+9^2}=\sqrt{361+121+81}=\sqrt{563}.

Compute $\overrightarrow{AB}$

\overrightarrow{AB}=B-A=(5-3,\,9-(-7),\,-2-1)=(2,16,-3).

Compute the scalar triple product

\overrightarrow{AB}\cdot(\vec d_1\times \vec d_2) =(2,16,-3)\cdot(19,11,9)

=2\cdot 19+16\cdot 11-3\cdot 9 =38+176-27=187.

Find the shortest distance

D=\frac{|187|}{\sqrt{563}}=\frac{187}{\sqrt{563}}.

Match with options

This corresponds to:

Option B

Comparison with stored correct answer

Stored correct answer is B, which matches our derived answer.

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