Find the length and direction (when defined) of u × v.
-$\mathbf { u } = - \frac { 1 } { 2 } \mathbf { i } + \frac { 3 } { 2 } \mathbf { j } + \mathbf { k } , \mathbf { v } = \mathbf { i } + \mathbf { j } + 2 \mathbf { k }$
A) $2 \sqrt { 2 } ; - \frac { \sqrt { 2 } } { 2 } \mathbf { i } + \frac { \sqrt { 2 } } { 2 } j - \frac { \sqrt { 2 } } { 2 } \mathbf { k }$
B) $8 ; \frac { 1 } { 4 } \mathbf { i } - \frac { 1 } { 4 } \mathrm { j } + \frac { 1 } { 2 } \mathbf { k }$
C) $2 \sqrt { 3 } ; \frac { \sqrt { 3 } } { 3 } \mathbf { i } + \frac { \sqrt { 3 } } { 3 } \mathbf { j } - \frac { \sqrt { 3 } } { 3 } \mathbf { k }$
D) $8 ; \frac { 1 } { 2 } \mathbf { i } - \frac { 1 } { 4 } \mathbf { j } - \frac { 1 } { 4 } \mathbf { k }$

Question

Quizplus · Accepted Answer

We can find the cross product of u and v using the formula $$\mathbf{u}	imes\mathbf{v}= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -\frac12 & \frac32 & 1 \ 1 & 1 & 2 \end{vmatrix}$$ Expanding the determinant, we get: $$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -\frac12 & \frac32 & 1 \ 1 & 1 & 2 \end{vmatrix} = \mathbf{i}\begin{vmatrix} \frac32 & 1 \ 1 & 2 \end{vmatrix} - \mathbf{j}\begin{vmatrix} -\frac12 & 1 \ 1 & 2 \end{vmatrix} + \mathbf{k}\begin{vmatrix} -\frac12 & \frac32 \ 1 & 1 \end{vmatrix} = -\frac52\mathbf{i} -\frac52\mathbf{j} -2\mathbf{k}$$ To find the length of the cross product, we take the magnitude $$|\mathbf{u}	imes\mathbf{v}|=\sqrt{\left(-\frac52ight)^2 + \left(-\frac52ight)^2 + (-2)^2} = 2\sqrt3$$ Finally, to find the direction, we normalize the cross product by dividing by its magnitude: $$\frac{\mathbf{u}	imes \mathbf{v}}{|\mathbf{u}	imes \mathbf{v}|} = \frac{-\frac52\mathbf{i} -\frac52\mathbf{j} -2\mathbf{k}}{2\sqrt3} = \frac{\sqrt3}{3}\mathbf{i}+\frac{\sqrt3}{3}\mathbf{j}-\frac{\sqrt3}{3}\mathbf{k}$$ Therefore, the answer is $\boxed{	extbf{(C)}\ 2\sqrt{3};\frac{\sqrt{3}}{3}\mathbf{i}+\frac{\sqrt{3}}{3}\mathbf{j}-\frac{\sqrt{3}}{3}\mathbf{k}}$.

Find the Length and Direction (When Defined) of U ×