# Homework Help: Cross Product Magnitude for Triangle Area

1. Sep 1, 2012

### dr721

1. The problem statement, all variables and given/known data

So I have several problems on my homework which deal with the application of cross product and its magnitude to find areas. I know how to do cross product, and I know how to find the magnitude, however, no matter how many times I try to calculate it, I get the wrong answer.

Take for example:

Find the area of the triangle with the three vertices [0, 0, -5], [-2, 1, -4], and [-3, -1, -6].

2. The attempt at a solution

This is where I'm not sure if I'm doing something incorrectly. I have generally been trying to create the two vectors for the cross product by subtracting one vertex from the other two.

So,

[-2, 1, -4] - [0, 0, -5] = [-2, 1, 1]

and

[-3, -1, -6] - [0, 0, -5] = [-3, -1, -1]

Then, computing the cross product to be [0, 5, 5]

So the magnitude comes out to be $\sqrt{50}$

And then the area of the triangle would be $\frac{1}{2}$$\sqrt{50}$

Is this right? If not, where did I go wrong? Is there some proper approach to deciding which vertex to use as the starting point?

2. Sep 1, 2012

### Dick

I don't see anything wrong with what you are doing.

3. Sep 2, 2012

### HallsofIvy

50= (2)(52) so it might be better to write it as
$$\frac{\sqrt{50}}{2}= \frac{5\sqrt{2}}{2}$$