Let vector a vector b be two vectors such that

Question:

Let $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ be two vectors such that $|2 \vec{a}+3 \vec{b}|=|3 \vec{a}+\vec{b}|$ 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:

1. 4

2. 6

3. 5

4. 8

Correct Option: , 3

Solution:

$|3 \vec{a}+\vec{b}|^{2}=|2 \vec{a}+3 \vec{b}|^{2}$

$(3 \vec{a}+\vec{b}) \cdot(3 \vec{a}+\vec{b})=(2 \vec{a}+3 \vec{b}) \cdot(2 \vec{a}+3 \vec{b})$

$9 \vec{a} \cdot \vec{a}+6 \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{b}=4 \vec{a} \cdot \vec{a}+12 \vec{a} \cdot \vec{b}+9 \cdot \vec{b} \cdot \vec{b}$

$5|\vec{a}|^{2}-6 \vec{a} \cdot \vec{b}=8|\vec{b}|^{2}$

$5(8)^{2}-6.8 \cdot|\vec{b}| \cos 60^{\circ}=8|\vec{b}|^{2}\left(\begin{array}{c}\because \frac{1}{8}|\vec{a}|=1 \\ \Rightarrow|\vec{a}|=8\end{array}\right)$

$40-3|\vec{b}|=|\vec{b}|^{2}$

$\Rightarrow|\vec{b}|^{2}+3|\vec{b}|-40=0$

$|\overrightarrow{\mathrm{b}}|=-8, \quad|\overrightarrow{\mathrm{b}}|=5$

(rejected)