The shortest distance between the lines \fracx-32=\fracy+15-7=\fracz-95 and \fracx+12=\fracy-11=\fracz-9-3 is

3D Geometry PYQ — JEE Main 2024 · 6 Apr · Shift 1 · Q39 with Solution

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Correct answer: D

The given lines are

\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}

and

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

So we can identify:

For line $L_1$ , a point is $A=(3,-15,9)$ and direction vector is $\vec d_1=(2,-7,5).$
For line $L_2$ , a point is $B=(-1,1,9)$ and direction vector is $\vec d_2=(2,1,-3).$

The shortest distance between lines $L_1: \vec r=\vec a+\lambda \vec d_1, \qquad L_2: \vec r=\vec b+\mu \vec d_2$

\text{S.D.}=\frac{|(\vec b-\vec a)\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.

Here,

\vec b-\vec a = B-A = (-1-3,\,1-(-15),\,9-9)=(-4,16,0).

\vec d_1\times \vec d_2= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix}

= \hat i\big((-7)(-3)-5(1)\big) - \hat j\big(2(-3)-5(2)\big) + \hat k\big(2(1)-(-7)(2)\big)

= \hat i(21-5)-\hat j(-6-10)+\hat k(2+14)

=16\hat i+16\hat j+16\hat k

So,

\vec d_1\times \vec d_2=(16,16,16)=16(1,1,1).

Its magnitude is

|\vec d_1\times \vec d_2|=\sqrt{16^2+16^2+16^2}=16\sqrt{3}.

(\vec b-\vec a)\cdot(\vec d_1\times \vec d_2) =(-4,16,0)\cdot(16,16,16)

= (-4)(16)+(16)(16)+0(16) = -64+256=192.

Thus,

\text{S.D.}=\frac{|192|}{16\sqrt{3}}=\frac{12}{\sqrt{3}}=4\sqrt{3}.

\boxed{4\sqrt{3}}

So the correct option is D.