Vector Algebra
Dot Product and Projection
Grade 12
Question:
<p>If the vector \(6\hat{i}-3\hat{j}-6\hat{k}\) is decomposed into vectors parallel and perpendicular to the vector \(\hat{i}+\hat{j}+\hat{k}\), then the two vectors are</p>
<li>\(-(\hat{i}+\hat{j}+\hat{k})\) and \((7\hat{i}-2\hat{j}-7\hat{k})\)</li>
<li>\(-(\hat{i}+\hat{j}+\hat{k})\) and \((7\hat{i}-2\hat{j}-5\hat{k})\)</li>
<li>\((\hat{i}+\hat{j}+\hat{k})\) and \(-(7\hat{i}-2\hat{j}-5\hat{k})\)</li>
<li>\((\hat{i}+\hat{j}+\hat{k})\) and \((5\hat{i}-4\hat{j}-7\hat{k})\)</li>
Step-by-Step Solution
Key Concept: Parallel component = (v \cdot n̂)n̂ where n̂=(i+j+k)/\sqrt{3.} Perpendicular = v - parallel component.
Let $\vec{v}=6\hat{i}-3\hat{j}-6\hat{k}$, $\vec{n}=\hat{i}+\hat{j}+\hat{k}$, $|\vec{n}|^2=3$.
Parallel component: $\vec{v}_\parallel = \dfrac{\vec{v}\cdot\vec{n}}{|\vec{n}|^2}\vec{n}=\dfrac{6-3-6}{3}(\hat{i}+\hat{j}+\hat{k})=\dfrac{-3}{3}(\hat{i}+\hat{j}+\hat{k})=-(\hat{i}+\hat{j}+\hat{k})$.
Perpendicular component: $\vec{v}_\perp = \vec{v}-\vec{v}_\parallel=(6\hat{i}-3\hat{j}-6\hat{k})-(-\hat{i}-\hat{j}-\hat{k})=7\hat{i}-2\hat{j}-5\hat{k}$.
Answer: (B) $-(\hat{i}+\hat{j}+\hat{k})$ and $7\hat{i}-2\hat{j}-5\hat{k}$.
Correct Answer: B