Find are of triangle?

Question

whose vertices A,B,C have coordinates

A=(0,1,2)
B=(1,2,0)
C=(2,0,1)

FInd the area of the triangle

Answer 1

You can use the cross-product to solve for the area of a parallelogram. Just take half of this to find the area (or are) of the two vectors.

I get two vectors from the three points above:

a = (A-C) = < -2, 1, 1 > ,
b = (B-C) = < -1, 2, -1 >

and the cross product of this I believe will be

v = 1 i - 1 j - 5 k,

The area of the triangle will be the magnitude of this vector divided by two.

or half of

sqrt ( 1^2 + 1^2 + 5^2 )

= [ sqrt ( 27 ) ] /2

(Same as the previous guys answer)

Answer 2

AB = sqrt(1^2+1^2 +2^2) = sqrt(6)
BC = sqrt(1^2 +2^2 +1^2 = sqrt(6)
AC = sqrt(2^2+1^2+1^) = sqrt(6)
So it's an equilaterl triangle with side = sqrt(6)
Area = s^2sqrt(3)/4 = sqrt(6)^2sqrt(3)/4 = 3sqrt(3)/2

Answer 3

What is an ARE?!?!?