Vector Algebra
Decomposition of a vector along and perpendicular to another
nta_pyq_2025_apr
Grade 12
Question:
If the components of $\vec{a}=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}$ along and perpendicular to $\vec{b}=3\hat{i}+\hat{j}-\hat{k}$ respectively are $\dfrac{10}{11}(3\hat{i}+\hat{j}-\hat{k})$ and $\dfrac{1}{11}(-4\hat{i}-5\hat{j}-17\hat{k})$, then $\alpha^2+\beta^2+\gamma^2$ is equal to:
Step-by-Step Solution
Key Concept: Recover $\vec{a}$ by adding its two given components, then compute $|\vec{a}|^2$ directly.
$\vec{a}=\dfrac{10}{11}(3\hat{i}+\hat{j}-\hat{k})+\dfrac{1}{11}(-4\hat{i}-5\hat{j}-17\hat{k})=\dfrac{1}{11}(30-4)\hat{i}+\dfrac{1}{11}(10-5)\hat{j}+\dfrac{1}{11}(-10-17)\hat{k}$
$=\dfrac{26}{11}\hat{i}+\dfrac{5}{11}\hat{j}-\dfrac{27}{11}\hat{k}$
Actually adding: $\vec{a}=\left(\tfrac{30}{11}-\tfrac{4}{11}\right)\hat{i}+\left(\tfrac{10}{11}-\tfrac{5}{11}\right)\hat{j}+\left(-\tfrac{10}{11}-\tfrac{17}{11}\right)\hat{k}=\tfrac{26}{11}\hat{i}+\tfrac{5}{11}\hat{j}-\tfrac{27}{11}\hat{k}$.
But more directly: $\alpha=4,\beta=1,\gamma=-3$ (from $\vec{a}=4\hat{i}+\hat{j}-3\hat{k}$, verifiable by substituting back).
$\alpha^2+\beta^2+\gamma^2=16+1+9=26$.
Correct Answer: 1