Thread: Area of Parallelogram of Forces Problem Help

Can someone check my answer for this problem please.

"Find the area of the "parallelogram of forces" with vertices p1=(-2,1,1), p2=(-1,3,2), p3=(-2,3,-4), and p4=(-3,5,-4)."

I got the square root of 29 as my answer. Can someone double check this answer for me please.

Originally Posted by collegemath

Can someone check my answer for this problem please.

"Find the area of the "parallelogram of forces" with vertices p1=(-2,1,1), p2=(-1,3,2), p3=(-2,3,4), and p4=(-3,5,4)."

I got the square root of 29 as my answer. Can someone double check this answer for me please.

Take the norm of the cross product of two adjacent perpendicular lines.

If you set it up, I will check it out.

what are the 2 adjecent perpendicular lines?
are they p1p2 and p1p3?

Originally Posted by collegemath

what are the 2 adjecent perpendicular lines?
are they p1p2 and p1p3?

Draw the picture.

Hello, collegemath!

There must be a typo.

The four vertices do not form a parallelogram.

sorry there is a typo the z 0f p3 and p4 is -4 not 4.
so the correct problem is "Find the area of the "parallelogram of forces" with vertices p1=(-2,1,1), p2=(-1,3,2), p3=(-2,3,-4), and p4=(-3,5,-4)."