JEE Main 2023 — Mathematics Vectors & 3D Geometry

Given,

Equation of the lines

$x+1=2y=-12z$x+1=2y=−12z and $x=y+2=6z-6$x=y+2=6z−6

$\Rightarrow \frac{x+1}{1}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{-1}{12}}$⇒1x+1​=21​y​=12−1​z​ and $\frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}}$1x​=1y+2​=61​z−1​

We know that, shortest distance between the lines is given by $S.D=\frac{(\vec{b}-\vec{a})\cdot (\vec{p}\times \vec{q})}{|\vec{p}\times \vec{q}|}$S.D=∣p​×q​∣(b−a)⋅(p​×q​)​

Here \vec{p}=\hat{i}+\frac{\hat{j}}{2}+\frac{-\hat{k}}{12}&\vec{q}=\hat{i}+\hat{j}+\frac{\hat{k}}{6} and \vec{a}=\hat{i}&\vec{b}=2\hat{j}-\hat{k}

$$\Rightarrow S.D=(-\hat{i}+2\hat{j}-\hat{k})\cdot \frac{(\vec{p}\times \vec{q})}{|\vec{p}\times \vec{q}|}$$⇒S.D=(−i^+2j^​−k^)⋅∣p​×q​∣(p​×q​)​

Now solving $\vec{p}\times \vec{q}\equiv |\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & \frac{1}{2} & \frac{-1}{12} \\ 1 & 1 & \frac{1}{6}\end{matrix}|$p​×q​≡∣i^11​j^​21​1​k^12−1​61​​∣

$$\Rightarrow \vec{p}\times \vec{q}\equiv \frac{1}{6}\hat{i}-\frac{1}{4}\hat{j}+\frac{1}{2}\hat{k}$$⇒p​×q​≡61​i^−41​j^​+21​k^

$$\Rightarrow \vec{p}\times \vec{q}\equiv 2\hat{i}-3\hat{j}+6\hat{k}$$⇒p​×q​≡2i^−3j^​+6k^

Hence, $S.D=\frac{(-\hat{i}+2\hat{j}-\hat{k})\cdot (2\hat{i}-3\hat{j}+6\hat{k})}{\sqrt{{2}^{2}+{3}^{2}+{6}^{2}}}=|\frac{-14}{7}|=2$S.D=22+32+62​(−i^+2j^​−k^)⋅(2i^−3j^​+6k^)​=∣7−14​∣=2