If a → and b → are unit vectors and a → + 3 b → is perpendicular to 7 a → - 5…

We have, $\left(\overrightarrow{a}+3\overrightarrow{b}\right)\perp \left(7\overrightarrow{a}-5\overrightarrow{b}\right)$

Therefore, $\left(\overrightarrow{a}+3\overrightarrow{b}\right)·\left(7\overrightarrow{a}-5\overrightarrow{b}\right)=0$

$\Rightarrow 7{\left|\overrightarrow{a}\right|}^2-15{\left|\overrightarrow{b}\right|}^2+16\overrightarrow{a}·\overrightarrow{b}=0 \dots \left(1\right)$

and $\left(\overrightarrow{a}-4\overrightarrow{b}\right)·\left(7\overrightarrow{a}-2\overrightarrow{b}\right)=0$

$\Rightarrow 7{\left|\overrightarrow{a}\right|}^2+8{\left|\overrightarrow{b}\right|}^2-30\overrightarrow{a}·\overrightarrow{b}=0 \dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$ we get

$2\left|\overrightarrow{a}\right|\cos \theta =\left|\overrightarrow{b}\right|$

$\therefore \cos \theta =\frac{\left|\overrightarrow{b}\right|}{2\left|\overrightarrow{a}\right|}$

$\Rightarrow \theta =60°.$