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. $\hat{i} -\hat{j} +2\hat{k}$
• D. $\hat{i} +2\hat{j} +2\hat{k}$

Answer 1

Answer: (B)

$2\hat{i} -\hat{j} +2\hat{k}$

Answer 2

Given, direction cosines are $\frac{2}{3},-\frac{a}{3}, \frac{2}{3}$.
Then, direction ratios are $2,-a, 2$.
According to the question,
$3=\sqrt{2^{2}+(-a)^{2}+2^{2}}$
$9=8+a^{2}$
$a^{2}=1 \Rightarrow a=\pm 1$
$\Rightarrow a=1 \,\,\,[\because a>0]$
So, the required vector is $2 \hat{ i }-\hat{ j }+2 \hat{ k }$.