Three Dimensional Geometry
Oswaal
CBSE
Grade 12
Question:
Shortest distance between two skew lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is:
(a) $\frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$
(b) $\frac{|(\vec{a}_2 - \vec{a}_1) \times (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$
(c) $|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|$
(d) $0$
Step-by-Step Solution
Key Concept: Standard shortest distance formula.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: $\frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$
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