Express the vector as a product of its length and direction.
-$$6 i - \frac { 4 } { 3 } j - 4 k$$
A) $\frac { 22 } { 3 } ( \mathbf { i } - \mathbf { j } - \mathbf { k } )$
B) $\frac { 22 } { 3 } \left( \frac { 9 } { 11 } \mathbf { i } - \frac { 2 } { 11 } \mathbf { j } - \frac { 6 } { 11 } \mathbf { k } \right)$
C) $\frac { 22 } { 3 } \left( 6 \mathbf { i } - \frac { 4 } { 3 } \mathbf { j } - 4 \mathbf { k } \right)$
D) $\frac { 22 } { 3 } \left( \frac { 9 } { 121 } \mathbf { i } - \frac { 2 } { 121 } \mathbf { j } - \frac { 6 } { 121 } \mathbf { k } \right)$

Question

Quizplus · Accepted Answer

To find the length or magnitude of the given vector:
$$|\mathbf{v}|=\sqrt{(6)^2+\left(-\frac43ight)^2+(-4)^2}=\frac{22}{3}.$$
To find the direction, we divide the vector by its length:
\begin{align*}
\frac{1}{|\mathbf{v}|}\cdot(6\mathbf{i}-	frac43\mathbf{j}-4\mathbf{k}) &= \frac{\frac{1}{3}\cdot 22}{22}\left(\frac{9}{11}\mathbf{i}-\frac{2}{11}\mathbf{j}-\frac{6}{11}\mathbf{k}ight) \
&= \frac{22}{3}\left(\frac{9}{11}\mathbf{i}-\frac{2}{11}\mathbf{j}-\frac{6}{11}\mathbf{k}ight).
\end{align*}
Thus, the answer is $\boxed{	ext{(B)}}.$

Express the Vector as a Product of Its Length and Direction