Vector Algebra
Projection vector and area of parallelogram
nta_pyq_2025_apr
Grade 12
Question:
Let $\vec{c}$ be the projection vector of $\vec{b}=\lambda\hat{i}+4\hat{k}$, $\lambda>0$, on the vector $\vec{a}=\hat{i}+2\hat{j}+2\hat{k}$. If $|\vec{a}+\vec{c}|=7$, then the area of the parallelogram formed by the vectors $\vec{b}$ and $\vec{c}$ is ________.
Step-by-Step Solution
Key Concept: Compute $\vec{c}=\dfrac{\vec{b}\cdot\vec{a}}{|\vec{a}|^2}\vec{a}$, use $|\vec{a}+\vec{c}|=7$ to find $\lambda$, then compute $|\vec{b}\times\vec{c}|$.
$\vec{c}=\dfrac{\vec{b}\cdot\vec{a}}{|\vec{a}|^2}\vec{a}=\dfrac{\lambda+0+8}{9}(\hat{i}+2\hat{j}+2\hat{k})$.
$\vec{a}+\vec{c}=\left(\dfrac{\lambda+17}{9}\right)\hat{i}+\dfrac{2(\lambda+17)}{9}\hat{j}+\dfrac{2(\lambda+17)}{9}\hat{k}$.
$|\vec{a}+\vec{c}|=\dfrac{\lambda+17}{9}\sqrt{1+4+4}=\dfrac{\lambda+17}{3}=7 \Rightarrow \lambda=4$.
$\vec{c}=\dfrac{12}{9}(\hat{i}+2\hat{j}+2\hat{k})=\dfrac{4}{3}\hat{i}+\dfrac{8}{3}\hat{j}+\dfrac{8}{3}\hat{k}$; $\vec{b}=4\hat{i}+0\hat{j}+4\hat{k}$.
$\vec{b}\times\vec{c}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\4&0&4\\\tfrac{4}{3}&\tfrac{8}{3}&\tfrac{8}{3}\end{vmatrix}=-\dfrac{32}{3}\hat{i}-\dfrac{16}{3}\hat{j}+\dfrac{32}{3}\hat{k}$.
Area $=|\vec{b}\times\vec{c}|=\dfrac{1}{3}\sqrt{32^2+16^2+32^2}=\dfrac{1}{3}\sqrt{2304}=\dfrac{48}{3}=16$.
Correct Answer: 16