JEE Main 2023 — Mathematics Vectors & 3D Geometry

Given equations are

$\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2}$3x−2​=2y+1​=2z−6​ is passing through a point $(2,-1,6)$(2,−1,6) and it's direction ratios are $3,2,2$3,2,2.

So,

$${\vec{a}}_{1}=2\hat{i}-\hat{j}+6\hat{k}$$a1​=2i^−j^​+6k^

$${\vec{b}}_{1}=3\hat{i}+2\hat{j}+2\hat{k}$$b1​=3i^+2j^​+2k^

And,

$\frac{x-6}{3}=\frac{y-1}{-2}=\frac{z+8}{0}$3x−6​=−2y−1​=0z+8​ is passing through a point $(6,1,-8)$(6,1,−8) and it's direction ratios are $3,-2,0$3,−2,0.

$${\vec{a}}_{2}=6\hat{i}+\hat{j}-8\hat{k}$$a2​=6i^+j^​−8k^

$${\vec{b}}_{2}=3\hat{i}-2\hat{j}+0\hat{k}$$b2​=3i^−2j^​+0k^

Now,

$${\vec{a}}_{2}-{\vec{a}}_{1}=4\hat{i}+2\hat{j}-14\hat{k}$$a2​−a1​=4i^+2j^​−14k^

$${\vec{b}}_{1}\times {\vec{b}}_{2}=|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 2 \\ 3 & -2 & 0\end{matrix}|$$b1​×b2​=∣i^33​j^​2−2​k^20​∣

$$\Rightarrow {\vec{b}}_{1}\times {\vec{b}}_{2}=4\hat{i}+6\hat{j}-12\hat{k}$$⇒b1​×b2​=4i^+6j^​−12k^

So,

$$\Rightarrow |{\vec{b}}_{1}\times {\vec{b}}_{2}|=\sqrt{16+36+144}=14$$⇒∣b1​×b2​∣=16+36+144​=14

We know that the shortest distance between two skew lines is $\frac{|({\vec{a}}_{2}-{\vec{a}}_{1})\cdot ({\vec{b}}_{1}\times {\vec{b}}_{2})|}{|{\vec{b}}_{1}\times {\vec{b}}_{2}|}$∣b1​×b2​∣∣(a2​−a1​)⋅(b1​×b2​)∣​

$$=|\frac{(4\hat{i}+2\hat{j}-14\hat{k})\cdot (4\hat{i}+6\hat{j}-12\hat{k})}{14}|$$=∣14(4i^+2j^​−14k^)⋅(4i^+6j^​−12k^)​∣

$=\frac{|16+12+168|}{14}=14$=14∣16+12+168∣​=14 units