Question

A unit vector $\overrightarrow{a}$ makes an angle of $45°$ with positive $z-$axis and if $\overrightarrow{a}+\hat{i}+\hat{j}$ is a unit vector, then $\overrightarrow{a}=$

• A. $(-\frac{1}{2},-\frac{1}{2},\frac{1}{\sqrt{2}})$
• B. $(\frac{1}{2},-\frac{1}{2},\frac{1}{\sqrt{2}})$
• C. $(\frac{1}{2},\frac{1}{2},\frac{1}{\sqrt{2}})$
• D. $(-\frac{1}{2},-\frac{1}{2},-\frac{1}{\sqrt{2}})$

Answer 1

The correct option is A $(-\frac{1}{2},-\frac{1}{2},\frac{1}{\sqrt{2}})$

Let $\overrightarrow{a}=x\hat{i}+y\hat{j}+\cos 45°\hat{k}$
$\Rightarrow \overrightarrow{a}=x\hat{i}+y\hat{j}+\frac{\hat{k}}{\sqrt{2}}$
Since, $\overrightarrow{a}$ is a unit vector.
$\therefore x^2+y^2+\frac{1}{2}=1$
$\Rightarrow x^2+y^2=\frac{1}{2}\cdots \left(1\right)$
Now, $\overrightarrow{a}+\hat{i}+\hat{j}=(x+1)\hat{i}+(y+1)\hat{j}+\frac{\hat{k}}{\sqrt{2}}$

Since, it is a unit vector.
$\therefore (x+1{)}^2+(y+1{)}^2+\frac{1}{2}=1\Rightarrow (x+1{)}^2+(y+1{)}^2=\frac{1}{2}\Rightarrow x^2+y^2+2(x+y)=-\frac{3}{2}\Rightarrow x+y=-1\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$
$x=-\frac{1}{2};y=-\frac{1}{2}$

$\overrightarrow{a}=-\frac{1}{2}\hat{i}-\frac{1}{2}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}$