# Prove area of triangle is given by cross products of the vertex vectors....

• steele1
In summary: So when you expand (BxC)+(CxA)+(AxB), you get 2(AxB + BxC + CxA). And since we are taking the absolute value, it doesn't matter the order in which we add the vectors. So we can rearrange it to get 2((AxB)+(CxA)+(CxB)). And dividing by 2 gives us the final result of 1/2|(BxC)+(CxA)+(AxB)|. In summary, the area of a triangle formed by three vectors A, B, and C pointing from the origin O to the three corners can be found by taking the magnitude of the cross product of those vectors and dividing it by 2, giving us 1/2
steele1

## Homework Statement

The three vectors A, B, and C point from the origin O to the three corners of a triangle. Show that the area of the triangle is given by 1/2|(BxC)+(CxA)+(AxB)|.

## The Attempt at a Solution

I know that the magnitude of the cross product of any two vectors gives the area of the parallelogram formed by those two vectors, and taking half of that would give the area of the triangle. I can see how 1/2|BxC|+1/2|CxA|+1/2|AxB| could give the total area of the triangle formed by the three points, because that would give the sum of the areas of the triangles formed by all 3 possible pairs of vectors which would equal the area of the triangle formed by all 3 vectors. I don't believe that this is the same thing as 1/2|(BxC)+(CxA)+(AxB)|. Not sure where to go from here.

steele1 said:

## Homework Statement

The three vectors A, B, and C point from the origin O to the three corners of a triangle. Show that the area of the triangle is given by 1/2|(BxC)+(CxA)+(AxB)|.

## The Attempt at a Solution

I know that the magnitude of the cross product of any two vectors gives the area of the parallelogram formed by those two vectors, and taking half of that would give the area of the triangle. I can see how 1/2|BxC|+1/2|CxA|+1/2|AxB| could give the total area of the triangle formed by the three points, because that would give the sum of the areas of the triangles formed by all 3 possible pairs of vectors which would equal the area of the triangle formed by all 3 vectors. I don't believe that this is the same thing as 1/2|(BxC)+(CxA)+(AxB)|. Not sure where to go from here.
I think you need to draw a picture. If you have a triangle whose sides are the vectors $V$ and $W$, then the area is $\frac{1}{2} V \times W$. But you aren't given $V$ and $W$. Look at this picture:

You are not given $V$ and $W$, you are given $A$, $B$ and $C$. Can you compute $V$ and $W$ from $A, B, C$?

stevendaryl said:
I think you need to draw a picture. If you have a triangle whose sides are the vectors $V$ and $W$, then the area is $\frac{1}{2} V \times W$. But you aren't given $V$ and $W$. Look at this picture:

View attachment 105540

You are not given $V$ and $W$, you are given $A$, $B$ and $C$. Can you compute $V$ and $W$ from $A, B, C$?
I'm honestly not sure how to go about doing that.

steele1 said:
I'm honestly not sure how to go about doing that.

Vector addition! In my diagram, if you start at the origin (the black dot) and follow vector $B$, and then follow vector $V$, then you end up at the same place as if you followed vector $A$. This means:

$B+V = A$

You can use that to get $V$ in terms of $A$ and $B$. You can similarly compute $W$ in terms of $B$ and $C$. Got it?

stevendaryl said:
Vector addition! In my diagram, if you start at the origin (the black dot) and follow vector $B$, and then follow vector $V$, then you end up at the same place as if you followed vector $A$. This means:

$B+V = A$

You can use that to get $V$ in terms of $A$ and $B$. You can similarly compute $W$ in terms of $B$ and $C$. Got it?
haha yup, wow I feel dumb. thought way too into that. So you have V=A-B and W=C-B.

stevendaryl
steele1 said:
haha yup, wow I feel dumb. thought way too into that. So you have V=A-B and W=C-B.
so 1/2|VxW| would give the area, VxW=(A-B)x(C-B), could you foil and get AxC-AxB-BxC+BxB, and BxB=0, and -AxB=BxA, and -BxC=CxB. So I end up with 1/2|(AxC)+(BxA)+(CxB)| which is all backwards. @stevendaryl

steele1 said:
so 1/2|VxW| would give the area, VxW=(A-B)x(C-B), could you foil and get AxC-AxB-BxC+BxB, and BxB=0, and -AxB=BxA, and -BxC=CxB. So I end up with 1/2|(AxC)+(BxA)+(CxB)| which is all backwards. @stevendaryl

Don't forget that AxB = -(BxA), etc.

## 1. How is the area of a triangle related to cross products of the vertex vectors?

The area of a triangle can be calculated by taking half of the magnitude of the cross product of two of its side vectors. This is known as the "cross product method" for finding the area of a triangle.

## 2. Why is the cross product used to calculate the area of a triangle?

The cross product is used because it gives us a vector that is perpendicular to both of the original vectors, and the magnitude of this vector is directly related to the area of the parallelogram formed by the two original vectors. Taking half of this magnitude gives us the area of the triangle.

## 3. Can the cross product method be used for any type of triangle?

Yes, the cross product method can be used for any type of triangle, including acute, right, and obtuse triangles. However, the order of the vertex vectors used in the cross product may need to be adjusted depending on the orientation of the triangle.

## 4. How is the cross product calculated?

The cross product of two vectors, A and B, is calculated by taking the determinant of a 3x3 matrix formed by the components of the vectors. The resulting vector is perpendicular to both A and B and has a magnitude equal to the area of the parallelogram formed by A and B.

## 5. Are there any other methods for finding the area of a triangle?

Yes, there are other methods for finding the area of a triangle, such as using the base and height of the triangle, or using trigonometric functions. However, the cross product method is often preferred as it is more general and can be extended to calculate the area of other shapes in higher dimensions.

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