If $\vec a, \vec b$ and $\vec c$ are three mutually perpendicular vectors, then the projection of the vector $l\frac{\vec a}{|\vec a|}+ m\frac{\vec b}{|\vec b|} + n\frac{(\vec a×\vec b)}{|\vec a×\vec b|}$ along the angle bisector of the vectors $\vec a$ and $\vec b$ is
Answer & explanation
Correct answer: option 4
A vector parallel to the bisector of the angle between the vectors $\vec a$ and $\vec b$ is $\frac{\vec a}{|\vec a|}+\frac{\vec b}{|\vec b|}=\hat a+\hat b$
∴ Unit vector along the bisector
$=\frac{\hat a+\hat b}{|\hat a+\hat b|}$
$=\frac{1}{\sqrt{2}}(\hat a+\hat b)$ $\begin{bmatrix}∵|\hat a+\hat b|^2=|\hat a|^2+|\hat b|^2+2\hat a.\hat b\\⇒|\hat a+\hat b|^2=1+1+0=2\end{bmatrix}$
∴ Required projection
$=\left\{l\frac{\vec a}{|\vec a|}+ m\frac{\vec b}{|\vec b|} + n\frac{(\vec a×\vec b)}{|\vec a×\vec b|}\right\}.\frac{1}{\sqrt{2}}(\hat a+\hat b)$
$=\frac{1}{\sqrt{2}}(l+m)$ $\left[∵|\hat a|=|\hat b|=1\,and\,\hat a.\hat b=\hat a.(\vec a×\vec b)=\hat b.(\vec a×\vec b)=0\right]$