Question

A vector which makes equal angles with the vectors $\frac{1}{3}(\hat{i}-2\hat{j}+2\hat{k}),\frac{1}{5}(-4\hat{i}-3\hat{k}),\hat{j}.$ is

• A. $5\hat{i}+\hat{j}+5\hat{k}$
• B. $-5\hat{i}+\hat{j}+5\hat{k}$
• C. $5\hat{i}-\hat{j}-\hat{k}$
• D. $5\hat{i}+\hat{j}-5\hat{k}$

Answer 1

The correct option is B $-5\hat{i}+\hat{j}+5\hat{k}$

Let the required vector be $\overrightarrow{a}=x\hat{i}+y\hat{j}+z\hat{k}$. It makes equal angles with the vectors.
$\overrightarrow{b}=\frac{1}{3}(\hat{i}-2\hat{j}-2\hat{k}),\overrightarrow{c}=\frac{1}{5}(-4\hat{i}-3\hat{k}),\overrightarrow{d}=\hat{j}$
$\therefore \overrightarrow{a}.\overrightarrow{b}=\overrightarrow{a}.\overrightarrow{c}=\overrightarrow{a}.\overrightarrow{d}$ [$\because \overrightarrow{b},\overrightarrow{c},\overrightarrow{d}$ are unit vectors]
$\therefore \frac{1}{3}(x-2y+2z)=\frac{1}{5}(-4x-3z)=y$
$\Rightarrow x-2y+2=3y$ and $-4x-5y-3z=0$
$\Rightarrow x-5y+2z=0$ and $4x+5y+3z=0$
Solving these equations, we get x= -z and x =-5y.
$\therefore \frac{x}{-5}=\frac{y}{1}=\frac{z}{5}or\frac{x}{5}=\frac{y}{-1}=\frac{z}{-5}$
$\Rightarrow \overrightarrow{a}=-5\hat{i}+\hat{j}+5\hat{k}$ or $\overrightarrow{a}=5\hat{i}-\hat{j}-5\hat{k}$