Question

Find the area of the parallelogram that has adjacent sides Bold u equals Bold i minus 2 Bold j plus 2 Bold kand Bold v equals 3 Bold j minus Bold k.

Answer 1

Answer:

The area of the parallelogram is $A=\sqrt{26}$.

Step-by-step explanation:

Let's rewrite these two vectors:

$u=i-2j+2k$

$v=0i+3j-k$    

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.        

$u \times v=\left[\begin{array}{ccc}i&j&k\\1&-2&2\\0&3&-1\end{array}\right] = i((-2)*(-1)-2*3)-j(1*(-1)-2*0)+k(1*3-(-2)*0)=i(2-6)-j(-1)+k(3)=-4i+j+3k$

Now, the magnitude is the square root of each component squared. It will be:

$|u \times v|=\sqrt{(-4)^{2}+(1)^{2}+(3)^{2}}=\sqrt{16+1+9}=\sqrt{26}$

Therefore the $A=\sqrt{26}$.      

I hope it helps you!