Question

A unit vector which is equally inclined to the vectors $\overrightarrow{i}$, $\frac{-2\overrightarrow{i}+\overrightarrow{j}+2\overrightarrow{k}}{3}$ and $\frac{-4\overrightarrow{j}-3\overrightarrow{k}}{5}$ is

• A. $\frac{1}{\sqrt{51}}(-\overrightarrow{i}+5\overrightarrow{j}-5\overrightarrow{k})$
• B. $\frac{1}{\sqrt{51}}(\overrightarrow{i}+5\overrightarrow{j}-5\overrightarrow{k})$
• C. $\frac{1}{\sqrt{51}}(\overrightarrow{i}+5\overrightarrow{j}+5\overrightarrow{k})$
• D. $\frac{1}{\sqrt{51}}(\overrightarrow{i}-5\overrightarrow{j}+5\overrightarrow{k})$

Answer 1

Answer: (A), (D)

$\frac{1}{\sqrt{51}}(-\overrightarrow{i}+5\overrightarrow{j}-5\overrightarrow{k})$
$\frac{1}{\sqrt{51}}(\overrightarrow{i}-5\overrightarrow{j}+5\overrightarrow{k})$

Let the required vector be $a\hat{i}+b\hat{j}+c\hat{k}$
According to the three conditions

$a=\frac{-2a+b+2c}{3}=\frac{-4b-3c}{5}$

$\Rightarrow b+2c=5a$ and $5a+4b+3c=0$

$\Rightarrow b+2c=-4b-3c$ and $b=-c$

And thus, $5a=c$

This is satisfied by options $A$ and $D$ and thus we get the correct answer.