Q. 9 - Calculus

Question

Sketch the parallelogram determined by the two vectors $(1,2)$ and $(3,-1)$. How can you use the cross product to find the area of this parallelogram?

Step-by-Step Solution

Answer

The area of this parallelogram is $7$.

Step 1. Given Information

Sketch the parallelogram determined by the two vectors $(1,2)$ and $(3,-1)$. We have to use the cross product to find the area of this parallelogram.

Step 2. Sketch the parallelogram determined by the two vectors ( 1 , 2 )   and   ( 3 , - 1 ) is

Step 3. Now the vectors are ( 1 , 2 , 0 )   and   ( 3 , - 1 , 0 ) .

Now the cross product is

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

$$\begin{gathered} u\times v=\left(\right(0\left)\right(2)-(-1\left)\right(0\left)\right)i+\left(\right(1\left)\right(0)-(3\left)\right(0\left)\right)j+\left(\right(1\left)\right(-1)-(2\left)\right(3\left)\right)k \\ u\times v=(0+0)i+(0-0)j+(-1-6)k \\ u\times v=0i+0j-7k \end{gathered}$$

Step 4. The area of the parallelogram determined by u and v is

$$\begin{gathered} ||u\times v||=||0i+oj-7k|| \\ ||u\times v||=\sqrt{0^2+0^2+{(-7)}^2} \\ ||u\times v||=\sqrt{49} \\ ||u\times v||=7 \end{gathered}$$