Find the area of the parallelogram that has the vectors as adjacent sides. $\begin{array} { l }
\mathbf { u } = - 3 \mathbf { i } + 6 \mathbf { j } + 9 \mathbf { k } \
\mathbf { v } = 6 \mathbf { i } + 4 \mathbf { j } + 13 \mathbf { k }
\end{array}$
A) $\text { Area } = \sqrt { 8109 } \text { square units }$
B) $\text { Area } = \sqrt { 12717 } \text { square units }$
C) $\text { Area } = \sqrt { 10911 } \text { square units }$
D) $\text { Area } = \sqrt { 10461 } \text { square units }$
E) $\text { Area } = \sqrt { 1905 } \text { square units }$

Question

Find the area of the parallelogram that has the vectors as adjacent sides.  $\begin{array} { l } 
\mathbf { u } = - 3 \mathbf { i } + 6 \mathbf { j } + 9 \mathbf { k } \
\mathbf { v } = 6 \mathbf { i } + 4 \mathbf { j } + 13 \mathbf { k }
\end{array}$
A) $\text { Area } = \sqrt { 8109 } \text { square units }$
B) $\text { Area } = \sqrt { 12717 } \text { square units }$
C) $\text { Area } = \sqrt { 10911 } \text { square units }$
D) $\text { Area } = \sqrt { 10461 } \text { square units }$
E) $\text { Area } = \sqrt { 1905 } \text { square units }$

Quizplus · Accepted Answer

The area of the parallelogram formed by vectors $\mathbf{u}$ and $\mathbf{v}$ is given by the magnitude of their cross product, $\|\mathbf{u} \times \mathbf{v}\|$. The cross product $\mathbf{u} \times \mathbf{v}$ is calculated as follows:$$\mathbf{u} \times \mathbf{v} = \left| \begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \-3 & 6 & 9 \6 & 4 & 13 \\end{array} \right| = \mathbf{i}(6*13 - 9*4) - \mathbf{j}(-3*13 - 9*6) + \mathbf{k}(-3*4 - 6*6)$$$$= \mathbf{i}(78 - 36) - \mathbf{j}(-39 - 54) + \mathbf{k}(-12 - 36)$$$$= \mathbf{i}(42) + \mathbf{j}(93) + \mathbf{k}(-48)$$Then, the magnitude of this vector is:$$\|\mathbf{u} \times \mathbf{v}\| = \sqrt{(42)^2 + (93)^2 + (-48)^2} = \sqrt{1764 + 8649 + 2304} = \sqrt{12717}$$Therefore, the area of the parallelogram is $\sqrt{12717}$ square units.

Find the Area of the Parallelogram That Has the Vectors