Vector Algebra
Cross Product Equation — Solving for Unknown Vector
nta_pyq_2026_jan
Grade 12
Question:
Let $\vec{a}=-\hat{i}+2\hat{j}+2\hat{k}$, $\vec{b}=8\hat{i}+7\hat{j}-3\hat{k}$ and $\vec{c}$ be a vector such that $\vec{a}\times\vec{c}=\vec{b}$ and $\vec{c}\cdot(\hat{i}+\hat{j}+\hat{k})=4$. Then $|\vec{a}+\vec{c}|^2$ is equal to:
Step-by-Step Solution
Key Concept: Let $\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$. From $\vec{a}\times\vec{c}=\vec{b}$: $z-y=4$, $z+2x=7$, $y+2x=3$. Also $x+y+z=4$.
$|\vec{a}+\vec{c}|^2=27$.
Correct Answer: 1