# The Normal Vector

 P: 4 Find: AB X AC (vectors). Prove that the area of the triangle is $$\frac{3}{2}$$ Further information We're given the points A(1,0,0) , B(0,2,2), C(1,1,2) So considering how this is the same question there is no doubt that we have to use the normal vector to find the answer. The attempt at a solution AB X AC = [2,2,-1] |AB X AC| = $$\sqrt{2^{2}+2^{2}+(-1)^{2}}$$ = 3 Dividing this by 2 gives us the answer we're looking for. However, why are the length of the normal vector equal to the area of a cube? Are there any other special characteristics of the normal vector I should know before my oral exam in two days? The theme is geometry considered around third-dimension vectors and planes. All help is good help as I can't really catch the essence of the abstract normal vector.
 Mentor P: 15,170 The Normal Vector I realize we have a language issue here. This is a cube: A parallelogram is a plane figure: The area of a parallelogram is the product of the lengths of any two adjacent sides and the sine of the angle between those sides: $$A=ab\sin\theta$$ This is also the magnitude of the cross product of the displacement vectors corresponding to those adjacent sides. That the triangle has half the area of the parallelogram should be easy to visualize. A diagonal of a parallelogram splits the parallelogram into two identical triangles.