(adsbygoogle = window.adsbygoogle || []).push({}); Find: AB X AC (vectors). Prove that the area of the triangle is [tex]\frac{3}{2}[/tex]

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| = [tex]\sqrt{2^{2}+2^{2}+(-1)^{2}}[/tex] = 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.

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# The Normal Vector

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