Vector Algebra
Cross Product Condition — |d|²
nta_pyq_2023_apr
Grade 12
Question:
$\vec{a}=\hat{i}+4\hat{j}+2\hat{k}$, $\vec{b}=3\hat{i}-2\hat{j}+7\hat{k}$, $\vec{c}=2\hat{i}-\hat{j}+4\hat{k}$. $\vec{d}\times\vec{b}=\vec{c}\times\vec{b}$, $\vec{d}\cdot\vec{a}=24$. Then $|\vec{d}|^2$ is equal to
Step-by-Step Solution
Key Concept: $\vec{d}\times\vec{b}=\vec{c}\times\vec{b}\Rightarrow(\vec{d}-\vec{c})\times\vec{b}=0\Rightarrow\vec{d}=\vec{c}+\lambda\vec{b}$. Use $\vec{d}\cdot\vec{a}=24$.
$|\vec{d}|^2=413$.
Correct Answer: 4