# Area of a triangle by vectors - getting different answers

1. Jan 24, 2010

### thomas49th

1. The problem statement, all variables and given/known data
A = (7,6,8)
B = (2,3,-1)
C = (2, -7, 9)

Find the area of the triangle these 3 points make

I'm getting slightly different results for different combinations I use

if I take

CA = a - c = (5 , 13 , -1)
CB = b - c = (0, 10, -10)

Taking another combination
AB = b - a = (-5, -3, -9)
AC = c - a = (-5, -13, 1)

Taking the modulus of the cross product and halving yields similar, but different answers. I've check and redone my work

Why?

Thanks
Thomas

Last edited: Jan 24, 2010
2. Jan 24, 2010

### rasmhop

You computed AB wrong, it should be (-5,-3,-9) (added changed 3 to -3 in second coordinate). Then you get the same area.

3. Jan 24, 2010

### thomas49th

sorry that was a typo, not actually what I did (i've edited it now)

When I did it the first combination I get 5 sqrt(246)
But when I did the second combination I got 5sqrt(233)

Do you concur?

Thanks
Thomas

4. Jan 24, 2010

### rasmhop

No.

I get the cross product as: (-120,50,50) with both your combinations which gives us an area of 5sqrt(194)

5. Jan 24, 2010

### raytrace

If you type in for CB = <0, +10, +10> and do the cross product of CA x CB you will end up with [140, -50, 50] which results in your 5$$\sqrt{246}$$.