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Wrong answer using correct logic

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data

    A plane has the equation aX + bY + cZ + d = 0. A line L goes from (0,0,0) and crosses the plane at some point. L and plane are orthogonal. express the coordinates of the crossing point P by: a, b, c and d.

    2. Relevant equations
    |d|/sqrt(a^2 + b^2 + c^2) is the distance between point P and origo.
    [a,b,c] is the directional vector for the line and a normal vector on the plane.

    3. The attempt at a solution

    So, if I divide [a,b,c] by the length of the directional vector, sqrt(a^2 + b^2 + c^2) and then multiply it by the length of the OP vector, |d|/sqrt(a^2 + b^2 + c^2), I get:

    a|d|/(a^2 + b^2 + c^2), b|d|/(a^2 + b^2 + c^2), c|d|/(a^2 + b^2 + c^2)=P.

    This is the crossing point of the line and the plane (or a point completely opposite to the plane).

    The answer is supposed to be (-ad/(a2 + b2 + c^2),-bd/(a2 + b2 + c2),-cd/(a2 + b2 + c2))=P

    I know that my answer is completely wrong, because |d| will remain always positive, while [a,b,c] could be negative or positive, with no consequence to |d|.

    Can somebody tell me how I can get to the correct answer?
  2. jcsd
  3. Sep 24, 2011 #2


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    Homework Helper

    Hi, Nikitin,

    Your logic is not that correct:smile:. You made a vector of direction the same as the normal of the plane and length equal to the distance between the origin and the plane. This is OK, but is this vector a point of the plane? Try. Substitute the coordinates you got into the equation of the plane. You get |d|+d =0, is it correct?

    Write the parametric equation of the line: It is (at, bt, ct). Substitute for x,y,z in the equation of the plane. You get an equation for t. Solve and get the coordinates of he crossing point using the value of t.

  4. Sep 24, 2011 #3
    yes, I got help previously and was told that one could do it your way.

    Tho I understand the fault in my reasoning.. the normal-vector must be pointing towards the plane.

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