# Area of A triangle

• Sep 6th 2008, 11:58 AM
Allie
Area of A triangle
Thank you so much to everyone that is helping me
Find the area of the triangle with vertices http://mathcs.chapman.edu/webwork2_f...984cec8771.png, http://mathcs.chapman.edu/webwork2_f...9e21732301.png, and http://mathcs.chapman.edu/webwork2_f...e424343f71.png
so for this one i found 2 vectors
4i-j+0k
-6i+10j+0k

took the cross product of the two to get 40k - (6k) which equals 34k, so
where am i going wrong?
• Sep 6th 2008, 12:20 PM
Soroban
Hello, Allie!

You did nothing wrong . . . leep going!

Quote:

Find the area of the triangle with vertices (0,-3), (4,-4) and (-2,6)

So for this one i found 2 vectors: . $4\vec i - \vec j + 0\vec k\:\text{ and }\;-6\vec i + 10\vec j + 0\vec k$ . . . . Right!

took the cross product of the two to get: . $40\vec k - 6\vec k \:=\:34\vec k$ . . . . Yes!

Where am i going wrong?

Recall that the magnitude of the cross product gives the area of the parallelogram.

Area of parallelogram: . $|34\vec k| \:=\:34$

Therefore, area of triangle: . $\frac{1}{2}\times 34 \;=\;17$

• Sep 6th 2008, 01:03 PM
Allie
ohh yess i see now thank you so much