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 u and v.

Short answer

The area of the parallelogram determined by the vectors u and v is $\sqrt{213}$.

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 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 $u\times v$

localid="1649437514111" $u\times v=\det \left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 2& 1& -3\\ 4& 0& 1\end{array}\right]$

Now solving the determinant.

localid="1649437505083" role="math" $$\begin{gathered} u\times v=\left(\right(1\left)\right(1)-(-3\left)\right(0\left)\right)i+\left(\right(2\left)\right(1)-(-3\left)\right(4\left)\right)j+\left(\right(2\left)\right(0)-(1\left)\right(4\left)\right)k \\ u\times v=(1+0)i+(2+12)j+(0-4)k \\ u\times v=1i+14j-4k \end{gathered}$$

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

$$\begin{gathered} ||u\times v||=||1i+14j-4k|| \\ ||u\times v||=\sqrt{{\left(1\right)}^2+{\left(14\right)}^2+{(-4)}^2} \\ ||u\times v||=\sqrt{1+196+16} \\ ||u\times v||=\sqrt{213} \end{gathered}$$