$A$ vector $\vec{r}$ has magnitude $14$ and direction ratios $2, 3, -6$. Find the direction cosines and components of $\vec{r}$,given that $\vec{r}$ makes an acute angle with the $x$-axis.

(A) Let the direction ratios be $a=2k, b=3k, c=-6k$ for some constant $k$.
The magnitude of the vector is given by $|\vec{r}| = \sqrt{a^2 + b^2 + c^2} = 14$.
$\sqrt{(2k)^2 + (3k)^2 + (-6k)^2} = 14$
$\sqrt{4k^2 + 9k^2 + 36k^2} = 14$
$\sqrt{49k^2} = 14$
$7|k| = 14 \Rightarrow |k| = 2$.
Since $\vec{r}$ makes an acute angle with the $x$-axis,the direction cosine $l = \frac{a}{|\vec{r}|} = \frac{2k}{14} = \frac{k}{7}$ must be positive. Thus,$k=2$.
The direction cosines are $l = \frac{2(2)}{14} = \frac{4}{14} = \frac{2}{7}$,$m = \frac{3(2)}{14} = \frac{6}{14} = \frac{3}{7}$,and $n = \frac{-6(2)}{14} = \frac{-12}{14} = -\frac{6}{7}$.
The components of $\vec{r}$ are $a = 2(2) = 4$,$b = 3(2) = 6$,and $c = -6(2) = -12$.
Thus,$\vec{r} = 4\hat{i} + 6\hat{j} - 12\hat{k}$.