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 cosine of the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is equal to

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

Answer 1

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

$(3\overrightarrow{a}-5\overrightarrow{b}).(2\overrightarrow{a}+\overrightarrow{b})=0$

$\Rightarrow 6{\left|\overrightarrow{a}\right|}^2-5{\left|\overrightarrow{b}\right|}^2=7\overrightarrow{a}.\overrightarrow{b}$

Also, $(\overrightarrow{a}+4\overrightarrow{b}).(\overrightarrow{b}-\overrightarrow{a})=0$

$\Rightarrow -{\left|\overrightarrow{a}\right|}^2+4{\left|\overrightarrow{b}\right|}^2=3\overrightarrow{a}.\overrightarrow{b}$

$\Rightarrow \frac{6}{7}{\left|\overrightarrow{a}\right|}^2-\frac{5}{7}{\left|\overrightarrow{b}\right|}^2=-\frac{1}{3}{\left|\overrightarrow{a}\right|}^2+\frac{4}{3}{\left|\overrightarrow{b}\right|}^2$

$\Rightarrow 25{\left|\overrightarrow{a}\right|}^2=43{\left|\overrightarrow{b}\right|}^2$

Also $3\overrightarrow{a}.\overrightarrow{b}=-{\left|\overrightarrow{a}\right|}^2+4{\left|\overrightarrow{b}\right|}^2=\frac{57}{25}{\left|\overrightarrow{b}\right|}^2$

$\Rightarrow 3\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta =\frac{57}{25}{\left|\overrightarrow{b}\right|}^2$

$3\sqrt{\frac{43}{25}}{\left|\overrightarrow{b}\right|}^2\cos \theta =\frac{57}{25}{\left|\overrightarrow{b}\right|}^2\Rightarrow \cos \theta =\frac{19}{5\sqrt{43}}$