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. $2i+j+2k$
• B. $2i-j+2k$
• C. $i-2j+2k$
• D. $i+2j+2k$

Answer 1

The correct option is B

$2i-j+2k$

Explanation for the correct option.

Given, the direction cosines of a vector of magnitude $3$ are $\frac{2}{3},\frac{-a}{3},\frac{2}{3},a>0$.

Then, direction ratios are $2,-a,2$
As magnitude is given to be $3$ so we have:
$\begin{array}{rcl}& \Rightarrow & 3=\sqrt{2^2+(-a{)}^2+2^2}\\ & \Rightarrow & 9=8+a^2\\ & \Rightarrow & a^2=1\end{array}$

$$\begin{gathered} \Rightarrow \begin{array}{rcl}& & a\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \pm \end{array}\begin{array}{rcl}& & 1\end{array} \\ \Rightarrow a=1[\because a>0] \end{gathered}$$

Substitute $a=1$in $\frac{2}{3},\frac{-a}{3},\frac{2}{3},a>0$ we get the required vector as:
$2\hat{i}-\hat{j}+2\hat{k}$

Hence, option B is correct.