Three Dimensional Geometry
NCERT Class 12
CBSE
Grade 12
Question:
The shortest distance between lines $\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k})$ is:
(a) $\dfrac{3\sqrt{2}}{2}$ units
(b) $\dfrac{1}{\sqrt{2}}$ units
(c) $3\sqrt{2}$ units
(d) $\sqrt{2}$ units
Step-by-Step Solution
$d = \dfrac{9}{3\sqrt{2}} = \dfrac{3\sqrt{2}}{2}$ units. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Evaluating shortest distance: 1.0 Mark
Correct Answer: $\dfrac{3\sqrt{2}}{2}$ units
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