Question

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|2\vec{a}+3\overrightarrow{b}|=|3\vec{a}+\vec{b}|$ and the angle between $\vec{a}$ and $\vec{b}$ is $60^{\circ }.$ If $\frac{1}{8}\vec{a}$ is a unit vector, then $|\vec{b}|$ is equal to:

• A. 4
• B. 6
• C. 5
• D. 8

Answer 1

Answer: (C)

$|3\vec{a}+\vec{b}{|}^2=|2\vec{a}+3\vec{b}{|}^2$
$(3\vec{a}+\vec{b})\cdot (3\vec{a}+\vec{b})$
$=(2\vec{a}+3\vec{b})\cdot (2\vec{a}+3\vec{b})$
$9\vec{a}\cdot \vec{a}+6\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{b}$
$=4\vec{a}\cdot \vec{a}+12\vec{a}\cdot \vec{b}+9\cdot \vec{b}\cdot \vec{b}$
$5|\vec{a}{|}^2-6\vec{a}\cdot \vec{b}=8|\vec{b}{|}^2$
$5(8{)}^2-6.8\cdot |\vec{b}|\cos 60^{\circ }$
$=8|\vec{b}{|}^2\left(\begin{array}{c}\because \frac{1}{8}|\vec{a}|=1\\ \Rightarrow |\vec{a}|=8\end{array}\right)$
$40-3|\vec{b}|=|\vec{b}{|}^2$
$\Rightarrow |\vec{b}{|}^2+3|\vec{b}|-40=0$
$|\vec{b}|=-8,$
$|\vec{b}|=5$