If $\mathbf{a}$ and $\mathbf{b}$ are two unit vectors such that $\mathbf{a}+\mathbf{b}$ is also $\mathbf{a}$…
If $\mathbf{a}$ and $\mathbf{b}$ are two unit vectors such that $\mathbf{a}+\mathbf{b}$ is also $\mathbf{a}$ unit vector, then $|\mathbf{a}-\mathbf{b}|^2=$
1
2
3
0
Solution
If resultant of two unit vectors is unit vector, then angle between them is $\frac{2 \pi}{3}$, so
$
\begin{aligned}
|\mathbf{a}-\mathbf{b}|^2 & =|a|^2+|b|^2-2|\mathbf{a}||\mathbf{b}| \cos \frac{2 \pi}{3} \\
& =1+1-2(1)(1)(-1 / 2)=2+1=3 .
\end{aligned}
$