The shortest distance between the lines x - 1 \over 2 = y + 8 \over -7 = z - 4 \over 5 and x - 1 \over 2 = y - 2 \over 1 = z - 6 \over - 3 is :

3D Geometry PYQ — JEE Main 2023 · 29 Jan · Shift 2 · Q37 with Solution

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Correct answer: C

The given lines are

$\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}.$

So their parametric/vector forms are:

For $L_1$ : $x=1+2t,\quad y=-8-7t,\quad z=4+5t$ Hence a point on $L_1$ is $A(1,-8,4)$ and direction vector is $\vec d_1=(2,-7,5).$
For $L_2$ : $x=1+2s,\quad y=2+s,\quad z=6-3s$ Hence a point on $L_2$ is $B(1,2,6)$ and direction vector is $\vec d_2=(2,1,-3).$

For lines with points $A,B$ and direction vectors $\vec d_1,\vec d_2$ ,

$\text{Shortest distance} = \frac{|(\overrightarrow{AB})\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.$

Here, $\overrightarrow{AB}=B-A=(1-1,\,2-(-8),\,6-4)=(0,10,2).$

\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=(16,16,16).$$ Therefore, $$|\vec d_1\times \vec d_2|=\sqrt{16^2+16^2+16^2}=16\sqrt3.$$ --- 4. **Compute the scalar triple product** $$(\overrightarrow{AB})\cdot(\vec d_1\times \vec d_2)=(0,10,2)\cdot(16,16,16)$$ $$=0\cdot16+10\cdot16+2\cdot16=12\cdot16=192.$$ So, $$\text{Shortest distance}=\frac{192}{16\sqrt3}=\frac{12}{\sqrt3}=4\sqrt3.$$ --- 5. **Match with options** $$4\sqrt3$$ corresponds to **Option C**. --- 6. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** They agree.