Question

In Exercises 30–35 compute the indicated quantities when $u=(-3,1,-4),v=(2,0,5),\mathrm{and}w=(1,3,13)$

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

Short answer

The area of the parallelogram determined by the vectors u and vis $\sqrt{78}$.

Step-by-step solution

Step 1. Given Information 

The indicated quantities when $u=(-3,1,-4),v=(2,0,5),\mathrm{and}w=(1,3,13)$

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

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

The cross product of $u\times v$

$u\times v=\det \left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ -3& 1& -4\\ 2& 0& 5\end{array}\right]$

Now solving the determinant.

role="math" $$\begin{gathered} u\times v=\left(\right(1\left)\right(5)-(-4\left)\right(0\left)\right)i+\left(\right(-3\left)\right(5)-(-4\left)\right(2\left)\right)j+\left(\right(-3\left)\right(0)-(1\left)\right(2\left)\right)k \\ u\times v=(5-0)i+(-15+8)j+(0-2)k \\ u\times v=5i-7j-2k \end{gathered}$$

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

$$\begin{gathered} ||u\times v||=||5i-7j-2k|| \\ ||u\times v||=\sqrt{{\left(5\right)}^2+{(-7)}^2+{(-2)}^2} \\ ||u\times v||=\sqrt{25+49+4} \\ ||u\times v||=\sqrt{78} \end{gathered}$$