JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Vector Algebra

Two vectors are perpendicular if their dot product is zero.

From the first condition:
$(\vec{a}+3\vec{b})\cdot (7\vec{a}-5\vec{b})=0$
$7|\vec{a}{|}^2-5\vec{a}\cdot \vec{b}+21\vec{a}\cdot \vec{b}-15|\vec{b}{|}^2=0$
$7|\vec{a}{|}^2+16\vec{a}\cdot \vec{b}-15|\vec{b}{|}^2=0$ ... (1)

From the second condition:
$(\vec{a}-4\vec{b})\cdot (7\vec{a}-2\vec{b})=0$
$7|\vec{a}{|}^2-2\vec{a}\cdot \vec{b}-28\vec{a}\cdot \vec{b}+8|\vec{b}{|}^2=0$
$7|\vec{a}{|}^2-30\vec{a}\cdot \vec{b}+8|\vec{b}{|}^2=0$ ... (2)

Subtracting (2) from (1):
$46\vec{a}\cdot \vec{b}-23|\vec{b}{|}^2=0$
$\vec{a}\cdot \vec{b}=\frac{|\vec{b}{|}^2}{2}$

Substituting back into (1):
$7|\vec{a}{|}^2+8|\vec{b}{|}^2-15|\vec{b}{|}^2=0$
$7|\vec{a}{|}^2=7|\vec{b}{|}^2$
$|\vec{a}|=|\vec{b}|$

Now, $\cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{|\vec{b}{|}^2/2}{|\vec{b}{|}^2}=\frac{1}{2}$

Therefore, $\theta =60°$.