Finding the Direction Cosines of a Vector

1. Find the magnitude and the direction cosines of \(\mathbf{v}=-3\mathbf{i }+2\mathbf{j}-6\mathbf{k}\).
2. Write \(\mathbf{v}\) in terms of its magnitude and its direction cosines.

Solution (a) The magnitude of \(\mathbf{v}\) is \[ \begin{equation*} \Vert \mathbf{v}\Vert =\sqrt{(-3)^{2}+2^{2}+(-6)^{2}}=\sqrt{49}=7 \end{equation*} \]

The direction cosines of the vector \(\mathbf{v}\) are \[ \cos \alpha =\dfrac{v_{1}}{\left\Vert \mathbf{v}\right\Vert }=\dfrac{-3}{7} \qquad \cos \beta =\dfrac{v_{2}}{\left\Vert \mathbf{v}\right\Vert }= \dfrac{2}{7} \qquad \cos \gamma =\dfrac{v_{3}}{\left\Vert \mathbf{v}\right\Vert }=\dfrac{-6}{7} \]

(b) Now \(\mathbf{v=}\left\Vert \mathbf{v}\right\Vert \left[ \cos \alpha {\bf i} + \cos \beta \mathbf{j}+\cos \gamma \mathbf{k}\right] =7\!\left( -\dfrac{3}{7}\mathbf{i}+\dfrac{2}{7}\mathbf{j}-\dfrac{6}{7}\mathbf{k }\right)\! .\)