A vector $\vec{r}$ has length 21 and direction ratios are proportional to 2, -3, 6. If $\vec{r}$ makes an acute angle with x-axis, then $\vec{r}$ = 

$6\hat{i}-9\hat{j} + 18 \hat{k}$

Recall that if the direction ratios of a vector are proportional to a, b, c, then its direction cosines are 

$± \frac{a}{\sqrt{a^2+b^2+c^2}}, ± \frac{b}{\sqrt{a^2+b^2+c^2}}, ± \frac{c}{\sqrt{a^2+b^2+c^2}}$

Therefore, direction cosines of $\vec{r}$ are

$± \frac{2}{\sqrt{2^2+(-3)^2+6^2}}, ± \frac{b}{\sqrt{2^2+(-3)^2+6^2}}, ± \frac{c}{\sqrt{2^2+(-3)^2+6^2}}$

Since $\vec{r}$ makes an acute angle with x-axis.

$∴ cos \alpha > 0 \, i.e. l > o $

So, direction cosines of $\vec{r}$ are $\frac{2}{7}, -\frac{3}{7}, \frac{6}{7}$

$∴ \vec{r}= 21\left(\frac{2}{7} \hat{i} - \frac{3}{7} \hat{j} +\frac{6}{7}\hat{k}\right) $   $[∵ \vec{r} = |\vec{r}|(l \hat{i} + m \hat{j} + n \hat{k})]$

$⇒ \vec{r}=6\hat{i}-9\hat{j} + 18 \hat{k}$