Three Dimensional Geometry
NCERT Class 12
CBSE
Grade 12
Question:
Find the distance between the parallel lines $\vec{r} = (\hat{i} + \hat{j}) + \lambda(2\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = (2\hat{i} + \hat{j} - \hat{k}) + \mu(2\hat{i} - \hat{j} + \hat{k})$.
Step-by-Step Solution
$|\vec{b}\times(\vec{a_2}-\vec{a_1})| = \sqrt{30}, |\vec{b}| = \sqrt{6}$. [1.5 Marks]
$d = \sqrt{30/6} = \sqrt{5}$ units. [1.5 Marks]
---
🎯 Official CBSE Marking Scheme:
Evaluating cross product magnitude: 1.5 Marks
Evaluating distance between parallel lines $= \sqrt{5}$ units: 1.5 Marks
Correct Answer:
Confused by the solution? Ask the AI to explain a specific step, tell you where you went wrong, or break down the key trap in this question.