Vector Algebra
Cross Product Identity — Finding $\vec{c}\cdot(-2\hat{i}+\hat{j}+\hat{k})$
nta_pyq_2024_apr
Grade 12
Question:
Let $\vec{a}=2\hat{i}+5\hat{j}-\hat{k}$, $\vec{b}=2\hat{i}-2\hat{j}+2\hat{k}$ and $\vec{c}$ be three vectors such that $(\vec{c}+\hat{i})\times(\vec{a}+\vec{b}+\hat{i})=\vec{a}\times(\vec{c}+\hat{i})$. If $\vec{a}\cdot\vec{c}=-29$, then $\vec{c}\cdot(-2\hat{i}+\hat{j}+\hat{k})$ is equal to:
Step-by-Step Solution
Key Concept: Let $\vec{v}=\vec{a}+\vec{b}+\hat{i}=5\hat{i}+3\hat{j}+\hat{k}$ and $\vec{p}=\vec{c}+\hat{i}$. Condition: $\vec{p}\times\vec{v}=\vec{a}\times\vec{p}\Rightarrow\vec{p}\times(\vec{v}+\vec{a})=0\Rightarrow\vec{p}=\lambda(\vec{v}+\vec{a})=\lambda(7\hat{i}+8\hat{j})$.
$\lambda=-1/2$. $\vec{c}\cdot(-2\hat{i}+\hat{j}+\hat{k})=5$.
Correct Answer: 4