Question

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

Answer: (A)

$\left(3\vec{a}-5\vec{b}\right).\left(2\vec{a}+\vec{b}\right)=0$
or $6\left|\vec{a}\right|^{2} -5\left|\vec{b}\right|^{2}=7 \vec{a}.\vec{b}$
Also, $\left(\vec{a}+4\vec{b}\right).\left(\vec{b}-\vec{a}\right)=0$
or $-\left|\vec{a}\right|^{2}+4\left|\vec{b}\right|^{2}=3 \vec{a}.\vec{b}$
or $\frac{6}{7} \left|\vec{a}\right|^{2} -\frac{5}{7}\left|\vec{b}\right|^{2}=-\frac{1}{3}\left|\vec{a}\right|^{2} \frac{4}{3}\left|\vec{b}\right|^{2}$
or $25\left|\vec{a}\right|^{2}=43\left|\vec{b}\right|^{2}$
$\Rightarrow 3.\vec{a}.\vec{b}=-\left|\vec{a}\right|^{2}+4\left|\vec{b}\right|^{2}=\frac{57} {25}\left|\vec{b}\right|^{2}$
or $3\left|\vec{a}\right|\left|\vec{b}\right|cos \theta=\frac{57}{25}\left|\vec{b}\right|^{2}$
or $3\sqrt{\frac{43}{25}}\left|\vec{b}\right|^{2} cos \theta =\frac{57}{25}\left|b\right|^{2} $
or cos $\theta =\frac{19}{5\sqrt{43}}$