Three Dimensional Geometry
NCERT Class 12
CBSE
Grade 12
Question:
Find the shortest distance between the lines $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 2\hat{k})$ and $\vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k})$.
Step-by-Step Solution
$\vec{a_2}-\vec{a_1} = 3\hat{i}+3\hat{j}+3\hat{k}, \vec{b_1}\times\vec{b_2} = -9\hat{i}+3\hat{j}+9\hat{k}$. [1.5 Marks]
$d = \dfrac{9}{3\sqrt{19}} = \dfrac{3\sqrt{19}}{19}$ units. [1.5 Marks]
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🎯 Official CBSE Marking Scheme:
Evaluating cross product and scalar numerator: 1.5 Marks
Evaluating shortest distance $= 3\sqrt{19}/19$ units: 1.5 Marks
Correct Answer:
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