If direction cosines of a vector of magnitude 3 are \(\frac{2}{3},-\frac{1}{3}, \frac{2}{3}\) and \(a>0\), then vector is

Given that, direction cosines are \( \frac{2}{3}, \frac{-a}{3}, \frac{2}{3} \)
We know that,\( l^{2}+m^{2}+n^{2}=1 \)
So, \( \left(\frac{2}{3}\right)^{2}+\left(\frac{-a}{3}\right)^{2}+\left(\frac{2}{3}\right)^{2}=1 \)
\( \Rightarrow \frac{4}{9}+\frac{a^{2}}{9}+\frac{4}{9}=1 \)
\( \Rightarrow \frac{a^{2}+8}{9}=1 \Rightarrow a^{2}=1 \Rightarrow a=\pm 1 \)
Vector is given by
\[
\vec{V}=|\vec{V}|\left(\begin{array}{l}
\hat{i}+m \hat{j}+n \hat{k}
\end{array}\right)
\]
Given magnitude of the vector is \( 3 \).
So, required vector is
\( 3\left(\frac{2}{3} \hat{i}-\frac{1}{3} \hat{j}+\frac{2}{3} \hat{k}\right)=2 \hat{i}-\hat{j}+2 \hat{k} \)