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#