Question

Vectors $3\overrightarrow{a}-5\overrightarrow{b}$ and $2\overrightarrow{a}+\overrightarrow{b}$ are mutually perpendicular. If $\overrightarrow{a}+4\overrightarrow{b}$ and $\overrightarrow{b}-\overrightarrow{a}$ are also mutually perpendicular, then the cosine of the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is

• A. $\frac{19}{5\sqrt{43}}$
• B. $\frac{19}{3\sqrt{43}}$
• C. $\frac{19}{2\sqrt{45}}$
• D. $\frac{19}{6\sqrt{43}}$

Answer 1

The correct option is A $\frac{19}{5\sqrt{43}}$

We have

$(3\overrightarrow{a}-5\overrightarrow{b})\cdot (2\overrightarrow{a}+\overrightarrow{b})=0$
or $6|\overrightarrow{a}{|}^2-5|\overrightarrow{b}{|}^2=7\overrightarrow{a}\cdot \overrightarrow{b}$
Also, $(\overrightarrow{a}+4\overrightarrow{b})\cdot (\overrightarrow{b}-\overrightarrow{a})=0$
or $-|\overrightarrow{a}{|}^2+4|\overrightarrow{b}{|}^2=3\overrightarrow{a}\cdot \overrightarrow{b}$
or $\frac{6}{7}|\overrightarrow{a}{|}^2-\frac{5}{7}|\overrightarrow{b}{|}^2=-\frac{1}{3}|\overrightarrow{a}{|}^2+\frac{4}{3}|\overrightarrow{b}{|}^2$
or $25|\overrightarrow{a}{|}^2=43|\overrightarrow{b}{|}^2$
$\Rightarrow 3\overrightarrow{a}\cdot \overrightarrow{b}=-|\overrightarrow{a}{|}^2+4|\overrightarrow{b}{|}^2=\frac{57}{25}|\overrightarrow{b}{|}^2$
or $3|\overrightarrow{a}||\overrightarrow{b}|\cos \theta =\frac{57}{25}|\overrightarrow{b}{|}^2$
or $3\sqrt{\frac{43}{25}}|\overrightarrow{b}{|}^2\cos \theta =\frac{57}{25}|\overrightarrow{b}{|}^2$
or $\cos \theta =\frac{19}{5\sqrt{43}}$
Hence, option '$A$' is correct.