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

Answer 1

Answer: (B), (C)

$5\hat{i}-\hat{j}-5\hat{k}$
$-5\hat{i}+\hat{j}+5\hat{k}$

Let the required vector be $\overrightarrow{a}=x\hat{i}+y\hat{j}+z\hat{k}$;

Given: $\overrightarrow{b}=\frac{1}{3}(\hat{i}-2\hat{j}+2\hat{k})$ ; $\overrightarrow{c}=\frac{1}{5}(-4\hat{i}-3\hat{k})$ ; and $\overrightarrow{d}=\hat{k}$

Since, given that $\overrightarrow{a}$ makes equal angles with the given vector;

$[\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{a}.\overrightarrow{c}=\overrightarrow{a}.\overrightarrow{d}]$

Solve $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{a}.\overrightarrow{d}$;

$(x\hat{i}+y\hat{j}+z\hat{k}).\frac{1}{3}(\hat{i}-2\hat{j}+2\hat{k})=(x\hat{i}+y\hat{j}+z\hat{k})(0\hat{i}+\hat{j}+0\hat{k})$

$\frac{1}{3}(x-2y+2z)=y$

$x-2y+2z=3y$

$x-5y+2z=\mathrm{0.......}Eq:01$

Solve: $\overrightarrow{a}.\overrightarrow{c}=\overrightarrow{a}.\overrightarrow{d}$;

$(x\hat{i}+y\hat{j}+z\hat{k}).\frac{1}{5}(0\hat{i}-4\hat{i}-3\hat{k})=(x\hat{i}+y\hat{j}+z\hat{k}).(0\hat{i}+\hat{j}+0\hat{k})]$

$\frac{1}{5}(-4x-3z)=y$

$4x+5y+3z=\mathrm{0......}Eq:02$

Solve Eq:01and Eq:02 using method of cross-multiplication;

$x-2y+2z=0$

$4x+5y+3z=0$

$\frac{x}{10+15}=\frac{y}{3-8}=\frac{z}{-20-5}$

$\frac{x}{25}=\frac{y}{-5}=\frac{z}{-25}$

$\frac{x}{5}=\frac{y}{-1}=\frac{z}{-5}$

Required vector: $\overrightarrow{a}=-5\hat{i}+\hat{j}+5\hat{k}$ or $\overrightarrow{a}=5\hat{i}-\hat{j}-5\hat{k}$.

Option (B) and (C) Both are correct.