Question

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

• A. $\frac{i}{2}+\frac{j}{2}+\frac{k}{\sqrt{2}}$
• B. $\frac{i}{2}+\frac{j}{2}-\frac{k}{\sqrt{2}}$
• C. $-\frac{i}{2}-\frac{j}{2}+\frac{k}{\sqrt{2}}$
• D. None of these

Answer 1

The correct option is C $-\frac{i}{2}-\frac{j}{2}+\frac{k}{\sqrt{2}}$

Let $\hat{a}=li+mj+nk$ where $l^2+m^2+n^2=1$
$a$ makes an angle $\frac{\pi }{4}$ with z-axis
$\therefore n=\frac{1}{\sqrt{2}}$

$\Rightarrow l^2+m^2=\frac{1}{2}....\left(i\right)$
then, $\hat{a}=li+mj+\frac{k}{\sqrt{2}}$
Now, $\hat{a}+i+j=(l+1)i+(m+1)j+\frac{k}{\sqrt{2}}...\left(ii\right)$
Its magnitude is $1$, hence
${(l+1)}^2+{(m+1)}^2=\frac{1}{2}...\left(ii\right)$
From Eqs (i) and (ii) we get
$2(l+m+1)=0$
$\Rightarrow {(l+m)}^2={(-1)}^2$
$l^2+m^2+2lm=1$
$\Rightarrow 2lm=\frac{1}{2}\Rightarrow l=m=-\frac{1}{2}$
Hence, $\hat{a}=-\frac{i}{2}-\frac{j}{2}+\frac{k}{\sqrt{2}}$