JEE MAIN - Mathematics (2022 - 24th June Morning Shift - No. 22)
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 ___________.
$$\overrightarrow r = \left( { - \widehat i + 3\widehat k} \right) + \lambda \left( {\widehat i - a\widehat j} \right)$$
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
2
Explanation
$\vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -a & 0 \\ 1 & -1 & 1\end{array}\right|=-a \hat{i}-\hat{j}+(a-1) \hat{k}$
$\vec{a}_{1}-\vec{a}_{2}=-\hat{i}+\hat{j}+\hat{k}$
Shortest distance $=\left|\frac{\left(\vec{a}_{1}-\vec{a}_{2}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right|$
$$ \begin{aligned} &\Rightarrow \quad \sqrt{\frac{2}{3}}=\frac{2(a-1)}{\sqrt{a^{2}+1+(a-1)^{2}}} \\\\ &\Rightarrow 6\left(a^{2}-2 a+1\right)=2 a^{2}-2 a+2 \\\\ &\Rightarrow \quad(a-2)(2 a-1)=0 \Rightarrow a=2 \text { because } a \in z . \end{aligned} $$
$\vec{a}_{1}-\vec{a}_{2}=-\hat{i}+\hat{j}+\hat{k}$
Shortest distance $=\left|\frac{\left(\vec{a}_{1}-\vec{a}_{2}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right|$
$$ \begin{aligned} &\Rightarrow \quad \sqrt{\frac{2}{3}}=\frac{2(a-1)}{\sqrt{a^{2}+1+(a-1)^{2}}} \\\\ &\Rightarrow 6\left(a^{2}-2 a+1\right)=2 a^{2}-2 a+2 \\\\ &\Rightarrow \quad(a-2)(2 a-1)=0 \Rightarrow a=2 \text { because } a \in z . \end{aligned} $$
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