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

Updated On: Jun 7, 2024
  • $2\hat{i} +\hat{j} +2\hat{k}$
  • $2\hat{i} -\hat{j} +2\hat{k}$
  • $\hat{i} -\hat{j} +2\hat{k}$
  • $\hat{i} +2\hat{j} +2\hat{k}$
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The Correct Option is B

Solution and Explanation

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