If → a and → b are unit vectors and ( → a + 3 → b ) is...

We have, $\left(\overset{\to}{a} + 3\overset{\to}{b}\right) ⊥ \left(7\overset{\to}{a} - 5\overset{\to}{b}\right)</p><p>Therefore, \left(\overset{\to}{a} + 3\overset{\to}{b}\right) \cdot \left(7\overset{\to}{a} - 5\overset{\to}{b}\right) = 0$

$⇒ 7{\left|\overset{\to}{a}\right|}$^{2} - 15{\left|\overset{\to}{b}\right|}^{2} + 16\overset{\to}{a} \cdot \overset{\to}{b} = 0 \ldots \left(1\right)

and \left(\overset{\to}{a} - 4\overset{\to}{b}\right)$\cdot \left(7\overset{\to}{a} - 2\overset{\to}{b}\right)$ = 0$</p><p>$⇒ 7{\left|\overset{\to}{a}\right|}^{2} + 8{\left|\overset{\to}{b}\right|}^{2} - 30\overset{\to}{a} \cdot \overset{\to}{b} = 0 \ldots \left(2\right)

From \left(1\right)$and \left(2\right)$,$we get</p><p>$2\left|\overset{\to}{a}\right|$cos\theta = \left|\overset{\to}{b}\right|$

$∴ cos\theta = \dfrac{\left|\overset{\to}{b}\right|}{2\left|\overset{\to}{a}\right|}$

$⇒ \theta = 60 ° .</p>$