Find the area of the parallelogram that has the vectors as adjacent sides. $\begin{array} { l }
\mathbf { u } = 10 \mathbf { i } + 6 \mathbf { j } + 10 \mathbf { k } \
\mathbf { v } = 5 \mathbf { i } - 10 \mathbf { j } + 15 \mathbf { k }
\end{array}$
A) $\text { Area } = \sqrt { 63000 } \text { square units }$
B) $\text { Area } = \sqrt { 29200 } \text { square units }$
C) $\text { Area } = \sqrt { 27090 } \text { square units }$
D) $\text { Area } = \sqrt { 46230 } \text { square units }$
E) $\text { Area } = \sqrt { 36130 } \text { square units }$

Question

Find the area of the parallelogram that has the vectors as adjacent sides.  $\begin{array} { l } 
\mathbf { u } = 10 \mathbf { i } + 6 \mathbf { j } + 10 \mathbf { k } \
\mathbf { v } = 5 \mathbf { i } - 10 \mathbf { j } + 15 \mathbf { k }
\end{array}$ 
A) $\text { Area } = \sqrt { 63000 } \text { square units }$
B) $\text { Area } = \sqrt { 29200 } \text { square units }$
C) $\text { Area } = \sqrt { 27090 } \text { square units }$
D) $\text { Area } = \sqrt { 46230 } \text { square units }$
E) $\text { Area } = \sqrt { 36130 } \text { square units }$

Quizplus · Accepted Answer

The area of the parallelogram is given by the magnitude of the cross product of the vectors $\mathbf{u}$ and $\mathbf{v}$. The cross product $\mathbf{u} \times \mathbf{v}$ is $(-180, -50, -130)$, and its magnitude is $\sqrt{(-180)^2 + (-50)^2 + (-130)^2} = \sqrt{63000}$.

Find the Area of the Parallelogram That Has the Vectors