Three Dimensional Geometry Question - CBSE

Question

Find the shortest distance between the parallel lines $\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$ and $\vec{r} = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(2\hat{i} + 3\hat{j} + 6\hat{k})$. Also write the vector equation of the line passing through $(1, 2, 3)$ parallel to these lines.

Accepted Answer

$\vec{b} imes (\vec{a_2}-\vec{a_1}) = -9\hat{i} + 14\hat{j} - 4\hat{k} \Rightarrow |\vec{b} imes (\vec{a_2}-\vec{a_1})| = \sqrt{293}$. [2.5 Marks]
Shortest distance $d = \dfrac{\sqrt{293}}{7}$ units. [1.5 Marks]
Parallel line through $(1,2,3)$: $\vec{r} = (\hat{i}+2\hat{j}+3\hat{k}) + t(2\hat{i}+3\hat{j}+6\hat{k})$. [1.0 Mark]

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🎯 Official CBSE Marking Scheme:
Calculating cross product for parallel lines: 2.5 Marks
Evaluating distance $= \sqrt{293}/7$ units: 1.5 Marks
Evaluating parallel line vector equation: 1.0 Mark

Question:

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