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

2 Answers
Apr 21, 2018

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

Apr 21, 2018

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