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How can I compare the vector equation of a line to a vector in the form (x,y,z)

  1. Sep 3, 2011 #1
    I am having trouble understanding how the vector equation of a line : (x,y,z) + t(x1,y1,z1) can be used in conjuction with another vector in the form (x,y,z).

    I visualise it like this. Points are : O to P and O to Q. So there are two vectors. The vector P to Q is given by Q - P which is the same as (x1,y1,z1).

    So that means if someone gives me a vector (x2,y2,z2) to determine if it is perpendicular or parallel I can just use (x1,y1,z1) and do the necessary tests dot product = 0 for perp. and cross product = 0 for parallel ?

    Is my understanding correct ?
     
  2. jcsd
  3. Sep 3, 2011 #2

    LCKurtz

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    That isn't an equation since there is no equals sign. The vector equation of a line has the form

    <x,y,z> = <x0,y0,z0> + t<d1,d2,d3>

    where (x0,y0,z0) is a point on the line and <d1,d2,d3> is a direction vector for the line.
    Yes. OP is the position vector <x0,y0,z0> and P is the point (x0,y0,z0). And the coordinates of Q-P are the components of the direction vector <d1,d2,d3>.
    Yes. Although the test for parallel is easier performed by checking that one vector is a multiple of the other; that's easier than the cross product.

    Also note the distinction between the point P = (x0,y0,z0) and the position vector
    OP=<x0,y0,z0> It is best not to confuse points with position vectors even though the components of the position vector are the same as the coordinates of the point.
     
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