Calculus 3 Vertices

Question

Find the perimeter of the triangle with vertices

​A(−2,3,2)

B(3,-3,4)

C(6,7,3)

Answer 1

A(-2,3,2) to B(3,-3,4) = AB = sqr(5^2+ 6^2 + 2^2) = sqr(25+36+4) =sqr65

B(3,-3,4) to C(6,7,3) = BC = sqr(3^2 +10^2 +1^2) = sqr(110)

A(-2,3,2) to C(6,7,3) = AC = sqr(8^2 +4^2 +1^2) = sqr(64+16+1) = sqr81 = 9

Perimeter = 9 + sqr110 + sqr65

= about 9+ 10.49 + 8.06 =approximately 27.6

distance formula is d = sqr((x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2)

use it 3 times to get AB, AC, and BC. then add them up to get the perimeter