Question

Find the area of the triangle t with vertices o(0,0,0), p(2,1,4), and q(5,6,3). (the area of a triangle is half the area of the corresponding parallelogram.)

Answer 1

Obtain two vectors that lie in the plane containing the triangle.

`\langle2,1,4\rangle-\langle0,0,0\rangle=\langle2,1,4\rangle`

`\langle5,6,3\rangle-\langle0,0,0\rangle=\langle5,6,3\rangle`

Compute their cross product to get the normal vector to the plane.

`\langle2,1,4\rangle*\langle5,6,3\rangle=\langle-21,14,7\rangle`

Now, the angle
`\theta` made by
`\langle2,1,4\rangle` and
`\langle5,6,3\rangle` is such that

`\|\langle-21,14,7\rangle\| = \|\langle2,1,4\rangle\| \|\langle5,6,3\rangle\| \sin(\theta)`

and this expression is exactly the area of the parallelogram spanned by the two vectors. Cut this in half to get the area of the triangle in question.

We have

`\|\langle-21,14,7\rangle\| = √((-21)^2 + 14^2 + 7^2) = 7√(14)`

and so the area of
`T` is 7√14/2.