$\text { Given } \mathbf { u } = \mathbf { i } + 3 \mathbf { j } \text { and } \mathbf { v } = - 8 \mathbf { i } + \mathbf { j } + \mathbf { k } \text {, find } \mathbf { u } \times \mathbf { v } \text {. }$
A) $6 \mathbf { i } - \mathbf { j } + 26 \mathbf { k }$
B) $3 \mathbf { i } + \mathbf { j } + 24 \mathbf { k }$
C) $3 \mathbf { i } - \mathbf { j } + 25 \mathbf { k }$
D) $4 \mathbf { i } + \mathbf { j } - 25 \mathbf { k }$
E) $3 \mathbf { i } - \mathbf { j } - 25 \mathbf { k }$

Question

$\text { Given } \mathbf { u } = \mathbf { i } + 3 \mathbf { j } \text { and } \mathbf { v } = - 8 \mathbf { i } + \mathbf { j } + \mathbf { k } \text {, find } \mathbf { u } \times \mathbf { v } \text {. }$ 
A) $6 \mathbf { i } - \mathbf { j } + 26 \mathbf { k }$
B) $3 \mathbf { i } + \mathbf { j } + 24 \mathbf { k }$
C) $3 \mathbf { i } - \mathbf { j } + 25 \mathbf { k }$
D) $4 \mathbf { i } + \mathbf { j } - 25 \mathbf { k }$
E) $3 \mathbf { i } - \mathbf { j } - 25 \mathbf { k }$

Quizplus · Accepted Answer

We can use the formula for the cross product of two vectors: 
$$\mathbf{u}	imes \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 3 & 0 \ -8 & 1 & 1 \end{vmatrix}$$

Expanding out the determinant, we get:

\begin{align*}
\mathbf{u}	imes \mathbf{v} &= \left((3)(1)-(0)(1)ight)\mathbf{i} -\left((1)(1)-(0)(-8)ight)\mathbf{j} +\left((1)(3)-(1)(-8)ight)\mathbf{k}\
&= 3\mathbf{i} -\mathbf{j}+25 \mathbf{k}
\end{align*}

Therefore, the answer is $\boxed{	extbf{(C)}}$

A) 6i−j+26k6 \mathbf { i } - \mathbf { j } + 26 \mathbf { k }6i−j+26k

A) $6 \mathbf { i } - \mathbf { j } + 26 \mathbf { k }$