Vector Algebra
Finding |a×c|² from Multiple Dot/Cross Conditions
nta_pyq_2023_apr
Grade 12
Question:
$\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$, $\vec{b}=\hat{i}+\hat{j}-\hat{k}$. $\vec{c}$: $\vec{a}\cdot\vec{c}=11$, $\vec{b}\cdot(\vec{a}\times\vec{c})=27$, $\vec{b}\cdot\vec{c}=-\sqrt{3}|\vec{b}|$. Then $|\vec{a}\times\vec{c}|^2$ is equal to
Step-by-Step Solution
Key Concept: From $\vec{b}\cdot\vec{c}=-\sqrt{3}\cdot\sqrt{3}=-3$. From $\vec{b}\cdot(\vec{a}\times\vec{c})=[\vec{b}\vec{a}\vec{c}]=27$. Set up $\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$ and solve system.
$|\vec{a}\times\vec{c}|^2=285$.
Correct Answer: 285