Vector Algebra
Cross Product and Lagrange Identity
nta_pyq_2023_apr
Grade 12
Question:
Let $\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}$, $\vec{b}=\hat{i}-2\hat{j}-2\hat{k}$, $\vec{c}=-\hat{i}+4\hat{j}+3\hat{k}$. If $\vec{d}\perp\vec{b},\vec{c}$ and $\vec{a}\cdot\vec{d}=18$, then $|\vec{a}\times\vec{d}|^2$ is equal to
Step-by-Step Solution
Key Concept: $\vec{d}=\lambda(\vec{b}\times\vec{c})$. $\vec{b}\times\vec{c}=2\hat{i}-\hat{j}+2\hat{k}$. Use $\vec{a}\cdot\vec{d}=18$ to find $\lambda$.
$|\vec{a}\times\vec{d}|^2=720$.
Correct Answer: 3