Compute area of triangle with cross product

[jwpaine]

Calc III homework. This is a bit elementary, but I can't seem to get the right answer (my online homework will tell me if what I've entered is correct, or not.

"Sketch the triangle with verticies O,

 P = (2,2,0) and Q = (0,2,2)

Since the magnitude of normal is the area of the parallelogram, composed of the vectors P and Q, than I'd assume that (1/2) the magnitude of the normal vector u = P x Q would be the area of the triangle.

For P = (2,2,0) and Q = (0,2,2) I get a cross product P x Q = 4i + 0j + 4k - 0i - 4j - 0k = 4i + 4k - 4j
So wouldn't the area of the triangle be (1/2)sqrt( 3*4^2) = 3.4641?



EDIT: Sorry, apparently that answer is correct. I must have made a typo.

 

[fasteddie65]

Calc III homework. This is a bit elementary, but I can't seem to get the right answer (my online homework will tell me if what I've entered is correct, or not.

"Sketch the triangle with verticies O,

 P = (2,2,0) and Q = (0,2,2)

Since the magnitude of normal is the area of the parallelogram, composed of the vectors P and Q, than I'd assume that (1/2) the magnitude of the normal vector u = P x Q would be the area of the triangle.

For P = (2,2,0) and Q = (0,2,2) I get a cross product P x Q = 4i + 0j + 4k - 0i - 4j - 0k = 4i + 4k - 4j
So wouldn't the area of the triangle be (1/2)sqrt( 3*4^2) = 3.4641?



EDIT: Sorry, apparently that answer is correct. I must have made a typo.

The area of a triangle determined by O, P, and Q is given by (1/2) |OP x OQ|.
OP = (2, 2, 0) and OQ = (0, 2, 2). Find the cross product. Find the magnitude of the cross product. Then take 1/2 of the answer.

 

[jwpaine]

The area of a triangle determined by O, P, and Q is given by (1/2) |OP x OQ|.
OP = (2, 2, 0) and OQ = (0, 2, 2). Find the cross product. Find the magnitude of the cross product. Then take 1/2 of the answer.

Um, that's exactly what I said.

Cheers - John