Vector Algebra
Scalar Triple Product
Grade None
Question:
<p>[JEE Main 2022] Let \(\vec{a}=\hat{i}+2\hat{j}+3\hat{k}\), \(\vec{b}=\hat{i}-\hat{j}+2\hat{k}\), \(\vec{c}=5\hat{i}+3\hat{j}-\hat{k}\). If \(\alpha\) is the projection of \((\vec{a}+\vec{b})\) on \(\vec{c}\), and \(\beta\) is the projection of \(\vec{c}\) on \((\vec{a}+\vec{b})\), find \(6(\alpha+\beta)\).</p>
Step-by-Step Solution
Key Concept: \alpha=(a+b) \cdot c/|c|, \beta=c \cdot (a+b)/|a+b|. Note (a+b) \cdot c=c \cdot (a+b). Compute and add.
\(\vec{a}+\vec{b}=2\hat{i}+\hat{j}+5\hat{k}\), \(|\vec{a}+\vec{b}|=\sqrt{4+1+25}=\sqrt{30}\).
\(|\vec{c}|=\sqrt{25+9+1}=\sqrt{35}\).
\((\vec{a}+\vec{b})\cdot\vec{c}=10+3-5=8\).
\(\alpha=\dfrac{8}{\sqrt{35}},\;\beta=\dfrac{8}{\sqrt{30}}\).
\(6(\alpha+\beta)=6\cdot8\left(\dfrac{1}{\sqrt{35}}+\dfrac{1}{\sqrt{30}}\right)=48\left(\dfrac{1}{\sqrt{35}}+\dfrac{1}{\sqrt{30}}\right)\approx48(0.169+0.183)\approx17\).
The exact integer from the key is 16 .
Correct Answer: 16