Question

A unit vector a makes an angle $\frac{\pi}{4}$ with $Z$-axis, if $\hat{ a }+\hat{ i }+\hat{ j }$ is a unit vector, then $a$ is equal to

• A. $\frac{\hat{ i }}{2}+\frac{\hat{ j }}{2}+\frac{\hat{ k }}{2}$
• B. $\frac{\hat{ i }}{2}+\frac{\hat{ j }}{2}-\frac{\hat{ k }}{\sqrt{2}}$
• C. $-\frac{\hat{ i }}{2}-\frac{\hat{ j }}{2}+\frac{\hat{ k }}{\sqrt{2}}$
• D. $\frac{\hat{ i }}{2}-\frac{\hat{ j }}{2}-\frac{\hat{ k }}{\sqrt{2}}$

Answer 1

Answer: (C)

Let $a =\hat{l i}+m \hat{ j }+n \hat{ k }$ makes an angle $\frac{\pi}{4}$ with Z-axis
Also, $l^2+m^2+n^2=1$
Here, $ n=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}, l^2+m^2=\frac{1}{2} ....$(i)
$ \therefore a \hat{a}=1 \hat{i}+m \hat{j}+\frac{\hat{ k }}{\sqrt{2}} $
$ \Rightarrow a+\hat{i}+\hat{j}=(l+1) \hat{i}+(m+1) \hat{j}+\frac{\hat{k}}{\sqrt{2}} $
$\Rightarrow la+\hat{i}+\hat{j}l=\sqrt{(l+1)^2+(m+1)^2+\left(\frac{1}{\sqrt{2}}\right)^2} $
$ \Rightarrow 1=l^2+m^2+2+2 l+2 m+\frac{1}{2} $
$\Rightarrow l+m=-1 $ [From Eq.(i)]
$\Rightarrow l^2+m^2+2 l m=1 $
$ \Rightarrow 2 l m=\frac{1}{2} \Rightarrow l=m=-\frac{1}{2}$
$\left(\because l=m=\frac{1}{2} \right.$ is not satisfied the given equation)
$\therefore a=-\frac{\hat{i}}{2}-\frac{\hat{j}}{2}+\frac{\hat{k}}{\sqrt{2}}$