# Area of Parallelogram of Forces Problem Help

• Feb 6th 2011, 04:42 PM
collegemath
Area of Parallelogram of Forces Problem Help
"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.
• Feb 6th 2011, 04:57 PM
dwsmith
Quote:

Originally Posted by collegemath
"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.
• Feb 6th 2011, 05:57 PM
collegemath
what are the 2 adjecent perpendicular lines?
are they p1p2 and p1p3?
• Feb 6th 2011, 05:58 PM
dwsmith
Quote:

Originally Posted by collegemath
what are the 2 adjecent perpendicular lines?
are they p1p2 and p1p3?

Draw the picture.
• Feb 6th 2011, 07:03 PM
Soroban
Hello, collegemath!

There must be a typo.

The four vertices do not form a parallelogram.

• Feb 6th 2011, 07:07 PM
collegemath
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)."