Practicing Success
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 |
$\frac{l^2+m^2}{\sqrt{l^2+m^2+n^2}}$ $\sqrt{l^2+m^2+n^2}$ $\frac{\sqrt{l^2+m^2}}{\sqrt{l^2+m^2+n^2}}$ $\frac{l+m}{\sqrt{2}}$ |
$\frac{l+m}{\sqrt{2}}$ |
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 $=\left\{l\frac{\vec a}{|\vec a|}+ m\frac{\vec b}{|\vec b|} + n\frac{(\vec a×\vec b)}{|\vec a×\vec b|}\right\}.\frac{1}{\sqrt{2}}(\hat a+\hat b)$ $=\frac{1}{\sqrt{2}}(l+m)$ $\left[∵|\hat a|=|\hat b|=1\,and\,\hat a.\hat b=\hat a.(\vec a×\vec b)=\hat b.(\vec a×\vec b)=0\right]$ |