Question

If the shortest distance between the lines $r_1=\alpha \hat{i}+2\hat{j}+2\hat{k}+\lambda \left(\hat{i}-2\hat{j}+2\hat{k}\right)$, $\lambda \in \mathrm{ℝ}$, $\alpha >0$ and $r_2=-4\hat{i}-\hat{k}+\mu \left(3\hat{i}-2\hat{j}-2\hat{k}\right)$, $\mu \in \mathrm{ℝ}$ is $9$, then the value of $\alpha$ is

• A. $2$
• B. $4$
• C. $6$
• D. $\sqrt{6}$

Answer 1

The correct option is C

$6$

The explanation for the correct option

The given equations of the straight lines are

$r_1=\alpha \hat{i}+2\hat{j}+2\hat{k}+\lambda \left(\hat{i}-2\hat{j}+2\hat{k}\right)$, $\lambda \in \mathrm{ℝ}$, $\alpha >0$.

$r_2=-4\hat{i}-\hat{k}+\mu \left(3\hat{i}-2\hat{j}-2\hat{k}\right)$, $\mu \in \mathrm{ℝ}$.

The shortest distance between the lines $\overrightarrow{l_1}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\overrightarrow{l_2}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$ can be given by $\left|\frac{\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)·\left(\overrightarrow{b_2}\times \overrightarrow{b_1}\right)}{\left|\overrightarrow{b_2}\times \overrightarrow{b_1}\right|}\right|$

Thus, $\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)=\left(-4\hat{i}-\hat{k}\right)-\left(\alpha \hat{i}+2\hat{j}+2\hat{k}\right)$

$\Rightarrow \left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)=\left(-4-\alpha \right)\hat{i}-2\hat{j}-3\hat{k}$.

Thus, $\left(\overrightarrow{b_2}\times \overrightarrow{b_1}\right)=\left(3\hat{i}-2\hat{j}-2\hat{k}\right)\times \left(\hat{i}-2\hat{j}+2\hat{k}\right)$

$$\begin{gathered} \Rightarrow \left(\overrightarrow{b_2}\times \overrightarrow{b_1}\right)=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -2& -2\\ 1& -2& 2\end{array}\right| \\ \Rightarrow \left(\overrightarrow{b_2}\times \overrightarrow{b_1}\right)=\left(-4-4\right)\hat{i}-\left(6+2\right)\hat{j}+\left(-6+2\right)\hat{k} \\ \Rightarrow \left(\overrightarrow{b_2}\times \overrightarrow{b_1}\right)=-8\hat{i}-8\hat{j}-4\hat{k} \end{gathered}$$

So, the shortest distance between two given lines is $\left|\frac{\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)·\left(\overrightarrow{b_2}\times \overrightarrow{b_1}\right)}{\left|\overrightarrow{b_2}\times \overrightarrow{b_1}\right|}\right|=9$

$$\begin{gathered} \Rightarrow \left|\frac{\left[\left(-\alpha -4\right)\hat{i}-2\hat{j}-3\hat{k}\right]·\left(-8\hat{i}-8\hat{j}-4\hat{k}\right)}{\left|\left(-8\hat{i}-8\hat{j}-4\hat{k}\right)\right|}\right|=9 \\ \Rightarrow \left|\frac{8\alpha +32+16+12}{\sqrt{64+64+16}}\right|=9 \\ \Rightarrow \left|\frac{8\alpha +60}{\sqrt{144}}\right|=9 \\ \Rightarrow \frac{\left|8\alpha +60\right|}{12}=9 \\ \Rightarrow \left|8\alpha +60\right|=108 \\ \Rightarrow 8\alpha +60=\pm 108 \\ \Rightarrow \alpha =6,-21 \end{gathered}$$

It is given that $\alpha >0$.

Therefore, $\alpha =6$.

Hence, (C) is the correct option.