Question

If the direction cosines of a vector of magnitude $3$ are $\frac{2}{3},\frac{-a}{3},\frac{2}{3},a>0$, then the vector is

• A. $2\hat{i}+\hat{j}+2\hat{k}$
• B. $2\hat{i}-\hat{j}+2\hat{k}$
• C. $\hat{i}-2\hat{j}+2\hat{k}$
• D. $\hat{i}+2\hat{j}+2\hat{k}$
• E. $\hat{i}+2\hat{j}-2\hat{k}$

Answer 1

The correct option is B $2\hat{i}-\hat{j}+2\hat{k}$

Given, direction cosines are $\frac{2}{3},-\frac{a}{3},\frac{2}{3}$.
Then, direction ratios are $2,-a,2$.
According to the question, we have
$3=\sqrt{2^2+(-a{)}^2+2^2}$
$=98+a^2$
Thus $a^2=1$

$\Rightarrow a=\pm 1$
$\Rightarrow a=1$ $[\because a>0]$
So, the required vector is $2\hat{i}-\hat{j}+2\hat{k}$.