Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Vectors

Question:

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

Options:

$\frac{l^2+m^2}{\sqrt{l^2+m^2+n^2}}$

$\sqrt{l^2+m^2+n^2}$

$\frac{\sqrt{l^2+m^2}}{\sqrt{l^2+m^2+n^2}}$

$\frac{l+m}{\sqrt{2}}$

Correct Answer:

$\frac{l+m}{\sqrt{2}}$

Explanation:

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

$=\left\{l\frac{\vec a}{|\vec a|}+ m\frac{\vec b}{|\vec b|} + n\frac{(\vec a×\vec b)}{|\vec a×\vec b|}\right\}.\frac{1}{\sqrt{2}}(\hat a+\hat b)$

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