MCQ
If $\vec{a}=(\hat{i}+2 \hat{j}-3 \hat{k})$ and $\vec{b}=(3 \hat{i}-\hat{j}+2 \hat{k})$ then the angle between $(\vec{a}+\vec{b})$ and $(\vec{a}-\vec{b})$ is
  • A
    $\frac{\pi}{2}$
  • B
    $\frac{2 \pi}{3}$
  • C
    $\frac{\pi}{4}$
  • D
    $\frac{\pi}{3}$

Answer

(a) $\frac{\pi}{2}$
Explanation: Given vectors $\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}$
Now, $\vec{a}+\vec{b}=4 \hat{i}+\hat{j}-\hat{k}$ and $\vec{a}-\vec{b}=-2 \hat{i}+3 \hat{j}-5 \hat{k}$
let $\theta$ be the angle between the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$
$\Rightarrow \cos \theta=\frac{-8+3+5}{\sqrt{16+1++} \times \sqrt{4+9+25}}=0=\frac{\pi}{2}$

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