If a=5 i-j-3 k & b=i-3 j+5 k the angle between a+b and a-b is:

$\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}$

$\vec{b}=\hat{i}-3 \hat{i}+5 \hat{k}$

$\Rightarrow \vec{a}+\vec{b}=6 \hat{i}-4 \hat{j}+2 \hat{k}$

$\Rightarrow \vec{a}-\vec{b}=4 \hat{i}+2 \hat{j}-8 \hat{k}$

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 $(\vec{a}-\vec{b})$

$\Rightarrow \cos \theta=0$

$\theta=\cos ^{-1}(0)$

$\theta=90°$