Question

Find the magnitude of each of the two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, having the same magnitude such that the angle between them is ${60}^{\circ }$ and their scalar product is $\frac{9}{2}$

Answer 1

Given:
The vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ have the same magnitude.
$\therefore |\overrightarrow{a}|=|\overrightarrow{b}|$
Also, $\overrightarrow{a}.\overrightarrow{b}=\frac{9}{2}$
We know that,
$|\overrightarrow{a}|\overrightarrow{b}|\cos \theta =\overrightarrow{a}.\overrightarrow{b}|\overrightarrow{a}||\overrightarrow{a}|\cos {60}^o=\frac{9}{2}[\because |\overrightarrow{a}|=|\overrightarrow{b}|\text{and}\overrightarrow{a}.\overrightarrow{b}=\frac{9}{2}]|\overrightarrow{a}{|}^2.\frac{1}{2}=\frac{9}{2}|\overrightarrow{a}{|}^2=9|\overrightarrow{a}|=\pm 3\text{but we know that magnitude of vector can't be negative}\therefore |\overrightarrow{a}|=3Hence,|\overrightarrow{a}|=|\overrightarrow{b}|=3$