If $\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}$ & $\vec{b}=\hat{i}-3 \hat{j}+5 \hat{k}$ the angle between $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ is: |
60° 90° 120° 30° |
90° |
$\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}$ $\vec{b}=\hat{i}-3 \hat{i}+5 \hat{k}$ $\Rightarrow \vec{a}+\vec{b}=6 \hat{i}-4 \hat{j}+2 \hat{k}$ $\Rightarrow \vec{a}-\vec{b}=4 \hat{i}+2 \hat{j}-8 \hat{k}$ so $(\vec{a}+\vec{b}) . (\vec{a}-\vec{b})=|\vec{a}+\vec{b}||\vec{a}-\vec{b}| \cos \theta$ $=(6 \hat{i}-4 \hat{j}+2 \hat{k})(4 \hat{i}+2 \hat{j}-8 \hat{k}) = \sqrt{6^2+4^2+2^2} \sqrt{4^2+2^2+(-8)^2} \cos \theta$ $=(24-8-16)= \sqrt{36+16+4} \sqrt{16+4+64} \cos \theta$ $\Rightarrow \frac{0}{\sqrt{36+16+4}} \sqrt{16+4+64} = \cos \theta$ $\cos \theta$ → angle between $(\vec{a}+\vec{b})$ and $(\vec{a}-\vec{b})$ $\Rightarrow \cos \theta=0$ $\theta=\cos ^{-1}(0)$ $\theta=90°$ |