Question:
If the shortest distance between the lines \(\vec{r_1} = \alpha\hat{i} + 2\hat{j} + 2\hat{k} + \lambda(\hat{i} - 2\hat{j} + 2\hat{k})\), \(\lambda \in \mathbb{R}\), \(\alpha \gt 0\) and \(\vec{r_2} = -4\hat{i} - \hat{k} + \mu(3\hat{i} - 2\hat{j} - 2\hat{k})\), \(\mu \in \mathbb{R}\) is 9, then \(\alpha\) is equal to ____________.
If the shortest distance between the lines \(\vec{r_1} = \alpha\hat{i} + 2\hat{j} + 2\hat{k} + \lambda(\hat{i} - 2\hat{j} + 2\hat{k})\), \(\lambda \in \mathbb{R}\), \(\alpha \gt 0\) and \(\vec{r_2} = -4\hat{i} - \hat{k} + \mu(3\hat{i} - 2\hat{j} - 2\hat{k})\), \(\mu \in \mathbb{R}\) is 9, then \(\alpha\) is equal to ____________.
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Use \(d = \frac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{d_1}\times\vec{d_2})|}{|\vec{d_1}\times\vec{d_2}|}\) and keep the root with \(\alpha \gt 0\).
Updated On: Nov 4, 2026
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Correct Answer: 6
Solution and Explanation
\(\vec{d_1} = (1, -2, 2)\), \(\vec{d_2} = (3, -2, -2)\).
\(\vec{d_1}\times\vec{d_2} = ((-2)(-2) - 2(-2),\ 2\cdot 3 - 1\cdot(-2),\ 1\cdot(-2) - (-2)\cdot 3) = (8, 8, 4)\), magnitude 12.
\(\vec{a_2} - \vec{a_1} = (-4-\alpha,\ -2,\ -3)\).
\((\vec{a_2}-\vec{a_1})\cdot(8,8,4) = -8(\alpha+4) - 16 - 12 = -(8\alpha + 60)\).
Distance \(= \frac{|8\alpha + 60|}{12} = 9\) gives \(8\alpha + 60 = \pm 108\), so \(\alpha = 6\) or \(\alpha = -21\). Since \(\alpha \gt 0\), \(\alpha = 6\).
Answer: 6
\(\vec{d_1}\times\vec{d_2} = ((-2)(-2) - 2(-2),\ 2\cdot 3 - 1\cdot(-2),\ 1\cdot(-2) - (-2)\cdot 3) = (8, 8, 4)\), magnitude 12.
\(\vec{a_2} - \vec{a_1} = (-4-\alpha,\ -2,\ -3)\).
\((\vec{a_2}-\vec{a_1})\cdot(8,8,4) = -8(\alpha+4) - 16 - 12 = -(8\alpha + 60)\).
Distance \(= \frac{|8\alpha + 60|}{12} = 9\) gives \(8\alpha + 60 = \pm 108\), so \(\alpha = 6\) or \(\alpha = -21\). Since \(\alpha \gt 0\), \(\alpha = 6\).
Answer: 6
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