Question

In Exercises 22–29 compute the indicated quantities when $u=(2,1,-3),v=(4,0,1),\mathrm{and}w=(-2,6,5)$

Find the area of the parallelogram determined by the vectors v and w.

Short answer

The area of the parallelogram determined by the vectors v and w is$\sqrt{1096}$.

Step-by-step solution

Step 1. Given Information 

In Exercises 22–29 compute the indicated quantities when $u=(2,1,-3),v=(4,0,1),\mathrm{and}w=(-2,6,5)$

We have to find the area of the parallelogram determined by the vectors v and w.

Step 2. Before finding area of the parallelogram determined by the vectors v and w we have to finding the cross product of v and w. 

The cross product of $v\times w$

role="math" localid="1649414070213" $$\begin{gathered} v\times w=\det \left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 4& 0& 1\\ -2& 6& 5\end{array}\right] \\ \mathrm{Now}\mathrm{solving}\mathrm{the}\mathrm{determinant}. \\ v\times w=\left(\right(0\left)\right(5)-(1\left)\right(6\left)\right)i+\left(\right(4\left)\right(5)-(1\left)\right(-2\left)\right)j+\left(\right(4\left)\right(6)-(0\left)\right(-2\left)\right)k \\ v\times w=(0-6)i+(20+2)j+(24+0)k \\ v\times w=-6i+22j+24k \end{gathered}$$

Step 3. The area of the parallelogram determined by v and w is v×w

$$\begin{gathered} ||v\times w||=||-6i+22j+24k|| \\ ||v\times w||=\sqrt{{(-6)}^2+{\left(22\right)}^2+{\left(24\right)}^2} \\ ||v\times w||=\sqrt{36+484+576} \\ ||v\times w||=\sqrt{1096} \end{gathered}$$