Question

The vector having magnitude equal to $5$ and perpendicular to the vector $\overrightarrow{B}=2\hat{i}-3\hat{j}+2\hat{k}$ and making equal angle with the positive $X$-axis and the positive $Y$-axis is

• A. $\frac{10\hat{i}+10\hat{j}+10\hat{k}}{3}$
• B. $\frac{10\hat{i}+10\hat{j}+5\hat{k}}{3}$
• C. $5\hat{k}$
• D. $\frac{5}{2}\hat{i}+\frac{5}{2}\hat{j}$

Answer 1

The correct option is B $\frac{10\hat{i}+10\hat{j}+5\hat{k}}{3}$

Let the vector be
$\overrightarrow{A}=a\hat{i}+b\hat{j}+c\hat{k}$
$a^2+b^2+c^2=25\cdots \left(1\right)$
$\overrightarrow{A}.\overrightarrow{B}=0$
$\Rightarrow (2\hat{i}-3\hat{j}+2\hat{k}).(a\hat{i}+b\hat{j}+c\hat{k})=0$
$\Rightarrow 2a-3b+2c=0\cdots \left(2\right)$

Also, $\cos {\theta }_1=\cos {\theta }_2$ where ${\theta }_1\text{and}{\theta }_2$ are the angles made by $\overrightarrow{A}$ with $X$-axis and $Y$-axis respectively.
$\Rightarrow \frac{(a\hat{i}+b\hat{j}+c\hat{k}).\left(x\hat{i}\right)}{5|x|}=\frac{(a\hat{i}+b\hat{j}+c\hat{k}).\left(y\hat{j}\right)}{5|y|}$
$\Rightarrow a=b$

Putting this value in eqn $\left(2\right),$ we get
$c=\frac{1}{2}a$
Putting this value in eqn $\left(1\right),$ we get
$a^2+a^2+\frac{1}{4}{a}^2=25$
$\Rightarrow \frac{9}{4}{a}^2=25$
$\Rightarrow a=\pm \frac{10}{3}$

Hence, the required vector is
$\overrightarrow{A}=\pm \left(\frac{10\hat{i}+10\hat{j}+5\hat{k}}{3}\right)$