The Normal Vector

  1. 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.
     
  2. jcsd
  3. D H

    Staff: Mentor

    You are looking for the area of a triangle, not a cube.

    How does the cross product relate to the area of a parallelogram?
     
  4. Might be I was unclear (perhaps of my poor English), but I found the answer, so this might not qualify for the homework help part of the forum, I don't know. The question is mostly why do the normal vector (cross product as many like to refer to it) have the properties of the area of a cube stretched by sides AB and AD (or in this case AC). In my perception it just a line which stretches outwards from the plane, 90 degrees in both 2D axises. I can't seem to comprehend why this line is so important, but somehow it is.
     
  5. D H

    Staff: Mentor

    I realize we have a language issue here. This is a cube:
    [​IMG]


    A parallelogram is a plane figure:

    [​IMG]

    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:

    [tex]A=ab\sin\theta[/tex]

    This is also the magnitude of the cross product of the displacement vectors corresponding to those adjacent sides.

    [​IMG]

    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.
     
  6. There it is! Thank you. I see now how cube was a wrong choice of words. What I meant to say was a square (3D vs. 2D), but as you say the correct would be a parallelogram. I think just the feeling of knowing somewhat what the normal vector is provides a lot when it comes to the presentation. I want to thank you again for the help.
     
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