Answer:
The area of the parallelogram is [tex]A=\sqrt{26}[/tex].
Step-by-step explanation:
Let's rewrite these two vectors:
[tex]u=i-2j+2k[/tex]
[tex]v=0i+3j-k[/tex]
Let's recall that the area of the parallelogram is the magnitude of the cross product between these vectors.
We can use the Determinant method to find it.
[tex]u \times v=\left[\begin{array}{ccc}i&j&k\\1&-2&2\\0&3&-1\end{array}\right] = i((-2)*(-1)-2*3)-j(1*(-1)-2*0)+k(1*3-(-2)*0)=i(2-6)-j(-1)+k(3)=-4i+j+3k[/tex]
Now, the magnitude is the square root of each component squared. It will be:
[tex]|u \times v|=\sqrt{(-4)^{2}+(1)^{2}+(3)^{2}}=\sqrt{16+1+9}=\sqrt{26}[/tex]
Therefore the [tex]A=\sqrt{26}[/tex].
I hope it helps you!
