Let us consider three points in the space A120 B002 and C13

Let us consider three points in the space : A(-1,2,0), B(0,0,2) and C(-1,3,-2).

a) Find coordinates of the vector 2AB + BC

b) find the cosine of the angel between the vectors AB and AC

C) Find the area if the triangle ABC

Solution

Vector AB = 1i - 2j + 2k Vector BC = -1i + 3j -4 k Vector AC = j - 2k So 2AB + BC = i - j So coordinates = (1,-1,0) Cosine of Angle between AB and AC = |-2 -4|/(3*sqrt(5)) = 2/sqrt(5) Area of Triangle ABC = .5|AB(cross)AC| = 3/2 square units = 1.5 square units