Question

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors such that $\overrightarrow{a}+3\overrightarrow{b}$ is perpendicular to $7\overrightarrow{a}-5\overrightarrow{b}$ and $\overrightarrow{a}-4\overrightarrow{b}$ is perpendicular to $7\overrightarrow{a}-2\overrightarrow{b}$ then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is D $\frac{\pi }{3}$

If two vectors are perpendicular their dot product is zero.

$\therefore (\overline{a}+3\overline{b)}.(7\overline{a}-5\overline{b})=0$

$7-5\overline{a}.\overline{b}+21\overline{a}.\overline{b}-15=0$

$16\overline{a}.\overline{b}-8=0$

$\overline{a}.\overline{b}=\frac{1}{2}$

$\overline{a}.\overline{b}=ab\cos \theta$

But $a$ and $b$ are unit vectors and value of cosine is

$\frac{1}{2}$ for $\theta =\frac{\pi }{3}$

Hence, the answer is option $D$