Three Dimensional Geometry
NCERT Class 12
CBSE
Grade 12
Question:
Find the shortest distance between the lines $\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k}$ and $\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}$.
Step-by-Step Solution
Rewrite lines in standard vector form: $\vec{b_1} = -\hat{i}+\hat{j}-2\hat{k}, \vec{b_2} = \hat{i}+2\hat{j}-2\hat{k}$. [1.5 Marks]
$\vec{a_2}-\vec{a_1} = \hat{j}-4\hat{k}, \vec{b_1}\times\vec{b_2} = 2\hat{i}-4\hat{j}-3\hat{k}$. [1.5 Marks]
$|\vec{b_1}\times\vec{b_2}| = \sqrt{29}$. [1.0 Mark]
Shortest distance $d = \dfrac{8}{\sqrt{29}} = \dfrac{8\sqrt{29}}{29}$ units. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Converting non-standard equations to standard vector forms: 1.5 Marks
Evaluating cross product $\vec{b_1} \times \vec{b_2}$: 1.5 Marks
Evaluating cross product magnitude $\sqrt{29}$: 1.0 Mark
Evaluating shortest distance $= 8/\sqrt{29}$ units: 1.0 Mark
Correct Answer:
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