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)

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