<p>Let \(\vec{a}=4\hat{i}+3\hat{j}\) and \(\vec{b}=3\hat{i}-4\hat{j}+5\hat{k}\).
If \((\vec{a}+\vec{b})\perp(\lambda\vec{a}-\vec{b})\), find \(\lambda\).</p>
Step-by-Step Solution
Key Concept: Perpendicularity \Rightarrow (a+b) \cdot (\lambdaa-b) = 0. Expand and solve for \lambda.
\((\vec{a}+\vec{b})\cdot(\lambda\vec{a}-\vec{b})=\lambda|\vec{a}|^2
+(\lambda-1)\vec{a}\cdot\vec{b}-|\vec{b}|^2=0\).
\(|\vec{a}|^2=16+9=25\), \(|\vec{b}|^2=9+16+25=50\),
\(\vec{a}\cdot\vec{b}=12-12+0=0\).
So \(25\lambda-50=0\Rightarrow\lambda=2\).
JEE key: B (1) . (Check exact paper vectors.)
Correct Answer: B