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 $\theta =\frac{19}{5\sqrt{43}}$