Question:
If the shortest distance between the lines $\bar{r}_1 = \alpha\hat{i} + 2\hat{j} + 2\hat{k} + \lambda(\hat{i} - 2\hat{j} + 2\hat{k}), \lambda \in \mathbb{R}, \alpha > 0$ and $\bar{r}_2 = -4\hat{i} - \hat{k} + \mu(3\hat{i} - 2\hat{j} - 2\hat{k}), \mu \in R$, is $9$, then the value of $\alpha$ is
If the shortest distance between the lines $\bar{r}_1 = \alpha\hat{i} + 2\hat{j} + 2\hat{k} + \lambda(\hat{i} - 2\hat{j} + 2\hat{k}), \lambda \in \mathbb{R}, \alpha > 0$ and $\bar{r}_2 = -4\hat{i} - \hat{k} + \mu(3\hat{i} - 2\hat{j} - 2\hat{k}), \mu \in R$, is $9$, then the value of $\alpha$ is
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Cross product of direction vectors gives the vector perpendicular to both lines, which is used for shortest distance.
Updated On: May 12, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Shortest distance $d = \frac{|(\bar{a}_2 - \bar{a}_1) \cdot (\bar{b}_1 \times \bar{b}_2)|}{|\bar{b}_1 \times \bar{b}_2|}$.
Step 2: Meaning
$\bar{a}_2 - \bar{a}_1 = (-4-\alpha, -2, -3)$.
$\bar{b}_1 \times \bar{b}_2 = (8, 8, 4)$. Magnitude $= \sqrt{64+64+16} = 12$.
Step 3: Analysis
$(\bar{a}_2 - \bar{a}_1) \cdot (\bar{b}_1 \times \bar{b}_2) = 8(-4-\alpha) - 16 - 12 = -32 - 8\alpha - 28 = -60 - 8\alpha$.
$9 = \frac{|-60 - 8\alpha|}{12} \implies 108 = |60 + 8\alpha|$.
Step 4: Conclusion
$8\alpha = 48 \implies \alpha = 6$ (given $\alpha > 0$). Final Answer: (B)
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