Question

How do you find the area of a triangle whose vertices of triangle are (2,0,0) ; (0,3,0) ; (0,0,5)?

Answer 1

`19/2`

Explanation:

The most general way to find the area of a triangle given three or more dimensional coordinates is to use Archimedes' Theorem. A triangle with squared sides `A`, `B`, and `C` has an area `a` that satisfies:

`16a^2 = 4AB - (C-A-B)^2`

Here we have

`A=2^2+3^2=13`
`B= 2^2+5^2=29`
`C =3^2+5^2=34`

`16a^2 = 4(13)(29) - (34 - 13-29)^2 = 1444`

`a = sqrt{ 1444/16} = 19/2`

Answer 2

Given the coordinates of the vertices of a triangle.

Use the points to make two vectors, `veca`, and `vecb`

Compute the cross-product `vecc=veca xx vecb`

`"Area" = 1/2|vecc|`

Explanation:

Compute the vector from `(2,0,0)` to `(0,3,0)`

`veca = (0-2)hati+(3-0)hatj+(0-0)hatk`

`veca = -2hati+3hatj`

Compute the vector from `(2,0,0)` to `(0,0,5)`

`vecb = (0-2)hati+(0-0)hatj+(5-0)hatk`

`vecb = -2hati+5hatk`

This reference tells you how to compute the Cross product.

The cross product `veca xx vecb` is:

`vecc = (-2hati+3hatj) xx (-2hati+5hatk)`

`vecc = 15hati+10hatj+6hatk`

The magnitude of `vecc` is:

`|vecc| = sqrt(15^2+10^2+6^2)`

`|vecc| = 19`

The area of the triangle is half of the magnitude:

`"Area" = 19/2`