Prove: Area of Triangle Between Vector a & b & Red Line = 1/2 |a x b|

[jbp890]

1. Show that the area of the triangle contained between vector a and vector b and the red line is 1/2 |a x b|

So far i have that bcosO would equal a ...and that 1/2bh should be the area... but I am stuck. can somebody help me prove?

 

[D H]

$b\cos\theta=a$ is not correct. Assuming vector b forms the base of the triangle, $A=1/2bh$ is the correct equation for the area. So what is the height of the triangle? (Hint: it involves only a and theta).

 

[m2003]

To prove this statement, we will use the fact that the magnitude of the cross product of two vectors is equal to the area of the parallelogram formed by those two vectors.

Let's first define the vectors a and b as follows:

a = [a1, a2, a3]
b = [b1, b2, b3]

Now, let's define the vector c as the vector from the endpoint of vector a to the endpoint of vector b. This vector can be written as:

c = [b1-a1, b2-a2, b3-a3]

Note that the magnitude of vector c is equal to the length of the red line in the triangle.

Next, we will calculate the cross product of vectors a and b, denoted as a x b. The cross product can be written as:

a x b = [a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1]

The magnitude of this vector is equal to the area of the parallelogram formed by vectors a and b.

Now, let's calculate the area of the triangle contained between vector a, vector b, and the red line. This can be done by dividing the area of the parallelogram by 2, since the triangle is half of the parallelogram.

Therefore, the area of the triangle can be written as:

Area of triangle = 1/2 |a x b|

This proves that the area of the triangle contained between vector a, vector b, and the red line is equal to 1/2 |a x b|.