### Learn math, science, English SAT & ACT from

high-quaility study
videos by expert teachers

##### Thank you for watching the preview.

To unlock all 5,300 videos, start your free trial.

# Area With the Cross Product - Problem 2

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Let’s use the cross product of vectors to find the area of a triangle. I have triangle ABC to find by points (-3 0), (6 -5)and (-1 6). Now any two vectors they pick on this triangle will also form they will define or determine a parallelogram. And so I can find the area of that using the cross product of those two vectors. And of course, half of that area will be the area of my triangle. So what I’m really going to find using the cross product is the area of a whole parallelogram.

Let me take these two vectors AC and AB. And the cross product of those will give me the area of this parallelogram. Know that these points are two dimensional points. So when I come up with components for vector AB and AC. I normally get two components, but you need three dimensional vectors to take the cross products. So I’m going to have to add a third component.

Now vectors AB would be 6 minus -3, 9. –5 minus 0, -5. But I have to add a z component here. I need to make this three dimensional vectors. You can just imagine that the xy. This is all the on the xy plane and so in 3 dimensional space, the z value of every one of these points is 0.

So the vector from a to c would be -1 minus -3, –1 plus 3 is 2, 6 minus 0, 6 and then of course 0. So let’s take the cross product of these two. Now remember, the cross product doesn’t directly give me the area of the parallelogram. I have to take the magnitude of the cross product. The cross product is going to give me a vector but one step at a time. The cross product is ijk, then I want the components of the two vectors. So I want <9 -5 0> and <2 6 0>. Now if you look at this, when I expand across the top row, I’ll have i times this determinant; -5 0, 6 0 that’s going to be 0. And I’ll have minus j times this determinant 9 2 0, that determinant is also going to be 0.

Anytime a determinant has a row or column of 0’s it's going to be 0. However k has this minor and that’s not going to be 0. So the answer is going to be 0i minus 0j plus k times the determinant 9 -5 2 and 6. So that’s 54 minus -10, 64. Let me write the 64 first. This is equals 64k.

I need the magnitude of that vector, to give me the area of the parallelogram. And what I really want here, is the area of this triangle, which is going to be half of that. So the area of triangle ABC is ½ the magnitude of this vector. ½ the magnitude of 64k. I don’t need to use the square root to find the magnitude of 64k, it's just 64. It's 64 times a vector with length 1. So its length will be 64 and ½ of 64 is 32. So that’s the area of this triangle, 32.

Now when you are finding the area of a triangle you can always use this method. You don’t have to trace out the parallelogram first. You can always just take 2 sides, and form vectors from them. Take the cross product of those vectors, the magnitude of the cross product, and multiply by a ½. Because it’s going to be a triangle it will have ½ the area of the associated parallelogram.

Please enter your name.

Are you sure you want to delete this comment?

###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

##### Concept (1)

##### Sample Problems (3)

Need help with a problem?

Watch expert teachers solve similar problems.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete