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The dot product

  1. Nov 4, 2004 #1
    Let
    n = (a,b,c)
    v = (x,y,z)

    Whatever the dot product of these vectors equal to, lets call d, the vector n is perpendicular to v. Again, I cannot stay calm and ask WHY?

    If we call n = (a,b); v= (x,y) ==> ax+by= d and the slope of this line is -a/b whereas the slope of the vector n is b/a. Yes, they are perpendicular since -a/b * b/a = -1 (tanx * tan(90+x) = -1)
    I can visualize and experiment it in 2-D but in 3-D I can't.

    Also, how do we define a line that passes through the origin and, for instance (1,1,1). I have trouble transforming my logic from x-y plane to space.
     
  2. jcsd
  3. Nov 4, 2004 #2
    Where did you get this faulty statement from? n and v are perpendicular if and only if their dot product is 0 (i.e. d = 0).
     
  4. Nov 4, 2004 #3

    HallsofIvy

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    Science Advisor

    The line that passes throught (0,0,0) and (1,1,1) is given by:
    x= t, y= t, z= t.

    You know that the equations will be linear and taking t= 0 gives (0,0,0) while taking t= 1 gives (1,1,1).

    In general: any straight line, in 3 dimensions, can be written x= at+ b, y= ct+ d, z= et+ f for some choice of a, b, c, d, e, f. If you are given two points, (x00,z0) and (x1,y1,z1), choose either one to be t= 0, the other t= 1 and plug the values into the equations. That gives 6 equations for the 6 letters.

    If you understand that one, try this: what's the equation of the line passing through (0,0,0,0) and (1, 1, 1, 1) in 4 dimensions?
     
  5. Nov 4, 2004 #4
    What I meant was, the vector (a,b,c) is always perpendicular to the plane ax+by+cz = d for any value of d. WHY?
     
  6. Nov 4, 2004 #5

    matt grime

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    What defines a plane? It is a point in the plane, p and the normal, n. Why? A point, x, is in the plane if and only if the displacement vector from p to x is zero, ie

    n.(p-x)=0, or n.x=n.p

    we set n.p = d.
     
  7. Nov 4, 2004 #6
    no.

    First the dot product gives a scaler, not a vector
    second, the two vectors are only perpendicular if their dot is 0


    let u = n
    let k = v - (<v,u>/||u||^2)*u

    Where <> is the dot product, and ||u|| is u’s norm

    u and k are perpendicular
     
    Last edited: Nov 4, 2004
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