Vector Algebra
Cross product and projection of a vector
nta_pyq_2025_apr
Grade 12
Question:
Let $\vec{a}=3\hat{i}-\hat{j}+2\hat{k}$, $\vec{b}=\vec{a}\times(\hat{i}-2\hat{k})$ and $\vec{c}=\vec{b}\times\hat{k}$. Then the projection of $\vec{c}-2\hat{j}$ on $\vec{a}$ is:
$2\sqrt{14}$
$\sqrt{14}$
$3\sqrt{7}$
$2\sqrt{7}$
Step-by-Step Solution
Key Concept: Compute the chain of cross products step by step, then use the projection formula $\dfrac{\vec{u}\cdot\vec{a}}{|\vec{a}|}$.
$\vec{b}=\vec{a}\times(\hat{i}-2\hat{k})=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&-1&2\\1&0&-2\end{vmatrix}=\hat{i}(2-0)-\hat{j}(-6-2)+\hat{k}(0+1)=2\hat{i}+8\hat{j}+\hat{k}$.
$\vec{c}=\vec{b}\times\hat{k}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&8&1\\0&0&1\end{vmatrix}=\hat{i}(8-0)-\hat{j}(2-0)+\hat{k}(0)=8\hat{i}-2\hat{j}$.
$\vec{c}-2\hat{j}=8\hat{i}-4\hat{j}$.
Projection $=\dfrac{(\vec{c}-2\hat{j})\cdot\vec{a}}{|\vec{a}|}=\dfrac{8\cdot3+(-4)(-1)+0}{\sqrt{9+1+4}}=\dfrac{28}{\sqrt{14}}=2\sqrt{14}$.
Correct Answer: 1