Compute the direction cosines of a given vector

Let $\mathbf{i}, \mathbf{j}, \mathbf{k}$ denote the unit coordinate vectors $(1,0,0), \ (0,1,0), \ (0,0,1)$ in $\mathbb{R}^3$ and let $A = (6,3,-2)$ . Further, let $a,b,c$ denote the angles between $A$ and the unit coordinate vectors. Compute $\cos a, \ \cos b, \cos c$ .
Find all vectors in $\mathbb{R}^3$ that are parallel to $A$ and have unit length.

We compute
$\begin{align*} \cos a &= \frac{A \cdot \mathbf{i}}{\lVert A \rVert \lVert \mathbf{i} \rVert} = \frac{6}{\sqrt{49}} = \frac{6}{7}. \\[9pt] \cos b &= \frac{A \cdot \mathbf{j}}{\lVert A \rVert \lVert \mathbf{j} \rVert} = \frac{3}{\sqrt{49}} = \frac{3}{7}. \\[9pt] \cos c &= \frac{A \cdot \mathbf{k}}{\lVert A \rVert \lVert \mathbf{k} \rVert} = \frac{-2}{\sqrt{49}} = -\frac{2}{7}. \end{align*}$
If $B$ is a vector in parallel to $A$ then we know there is a nonzero $t \in \mathbb{R}$ such that
$B = tA = (6t, 3t, -2t).$

Since $B$ has unit length we then have

$\lVert B \rVert = 1 \quad \implies \quad \sqrt{36t^2 + 9t^2 + 4t^2} = 1 \quad \implies \quad \pm 7t = 1 \quad \implies \quad t = \pm \frac{1}{7}.$

Therefore,

$B = \left( \frac{6}{7}, \frac{3}{7}, -\frac{2}{7} \right) \qquad \text{or} \qquad B = \left( -\frac{6}{7}, -\frac{3}{7}, \frac{2}{7} \right).$