Question

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are two non-parallel unit vectors and the vector $\alpha \overrightarrow{a}+\overrightarrow{b}$ bisects the internal angle between $\overrightarrow{a}$ and $\overrightarrow{b},$ then $\alpha$ is

• A. $\frac{1}{2}$
• B. $1$
• C. $2$
• D. $\frac{5}{2}$

Answer 1

The correct option is B $1$

$\overrightarrow{c}=t(\frac{\overrightarrow{a}}{|\overrightarrow{a}|}+\frac{\overrightarrow{b}}{|\overrightarrow{b}|})$
$\alpha \overrightarrow{a}+\overrightarrow{b}=t(\frac{\overrightarrow{a}}{1}+\frac{\overrightarrow{b}}{1})$
Since $|\overrightarrow{a}|=|\overrightarrow{b}|=1$
On comparing we get,
$\alpha =t$ and $t=1$
Hence $\alpha =1$