If the shortest distance between the lines
$$\overrightarrow r = \left( { - \widehat i + 3\widehat k} \right) + \lambda \left( {\widehat i - a\widehat j} \right)$$
and $$\overrightarrow r = \left( { - \widehat j + 2\widehat k} \right) + \mu \left( {\widehat i - \widehat j + \widehat k} \right)$$ is $\sqrt {{2 \over 3}}$, then the integral value of a is equal to ___________.
Answer (integer)
2
Solution
$\vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -a & 0 \\ 1 & -1 & 1\end{array}\right|=-a \hat{i}-\hat{j}+(a-1) \hat{k}$ <br/><br/>$\vec{a}_{1}-\vec{a}_{2}=-\hat{i}+\hat{j}+\hat{k}$
<br/><br/>
Shortest distance $=\left|\frac{\left(\vec{a}_{1}-\vec{a}_{2}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right|$
<br/><br/>
$$
\begin{aligned}
&\Rightarrow \quad \sqrt{\frac{2}{3}}=\frac{2(a-1)}{\sqrt{a^{2}+1+(a-1)^{2}}} \\\\
&\Rightarrow 6\left(a^{2}-2 a+1\right)=2 a^{2}-2 a+2 \\\\
&\Rightarrow \quad(a-2)(2 a-1)=0 \Rightarrow a=2 \text { because } a \in z .
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
This question is part of PrepWiser's free JEE Main question bank. 182 more solved questions on Three Dimensional Geometry are available — start with the harder ones if your accuracy is >70%.