Vector Algebra
Line of Intersection of Planes — Angle with Given Vector
nta_pyq_2023_apr
Grade 12
Question:
$\vec{a}$ parallel to intersection of planes through $\hat{i}+\hat{j},\hat{i}+\hat{k}$ and $\hat{i}-\hat{j},\hat{j}-\hat{k}$. $\theta$ angle between $\vec{a}$ and $\vec{b}=2\hat{i}-2\hat{j}+\hat{k}$, $\vec{a}\cdot\vec{b}=6$. $(\theta,|\vec{a}\times\vec{b}|)=$
$(\frac{\pi}{3},3\sqrt{6})$
$(\frac{\pi}{4},3\sqrt{6})$
$(\frac{\pi}{3},6)$
$(\frac{\pi}{4},6)$
Step-by-Step Solution
Key Concept: Normals to planes: $\vec{n}_1=(\hat{i}+\hat{j})\times(\hat{i}+\hat{k})$ and $\vec{n}_2=(\hat{i}-\hat{j})\times(\hat{j}-\hat{k})$. $\vec{a}=\vec{n}_1\times\vec{n}_2$.
$\theta=\frac{\pi}{4},|\vec{a}\times\vec{b}|=6$.
Correct Answer: 4