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Homework Help: Plane equation given two points and distance to parallel line

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the plane through the points P = (1, 1, −1) and Q = (2, 1, 1) and parallel to a line r
    {(1, 0, 2) + t(1, 0, 2), t ∈ R} with distance 1 to the plane

    2. Relevant equations

    A=(a1,a2,a3),B(b1,b2,b3),A and B are vectors. AxB= det[i j k;a1 a2 a3;b1 b2 b3] ";" means change of line.

    A.B= a1b1+a2b2+a3b3

    vector projector of A in B is proj(B)A = (A . B /||B||² ) B

    norm of vector projector of A in B is ||proj(B)A|| = ||(A . B /||B||² ) B||= (|A.B|/||B||²)||B||= |A.B|/||B||

    3. The attempt at a solution

    I have the vector PQ=Q-P=(1,0,2) and named a point M(a,b,c) which gives me the vector PM=(a-1,b-1,c+1).
    PQxPM will give me a vector perpendicular to the plane (scalar multiple of vector Normal N).
    N= (-2b+2,-c+2a-3,b-1)
    And the distance of the line to the plane is equal to the norm of the vector projection PP1 in the vector Normal when P1 is a point in the line and P in the plane. P1(1,0,2) and P(1,1,-1)

    ||proj(N)PP1||= |PP1 . N|/||N||=1

    And I get

    3b-2a+c= √[(-2b+2)²+(-c+2a-3)²+(b-1)²] an equation with three unknown numbers.
    I don't know how to proceed to find a,b and c.

    Last edited: Nov 11, 2012
  2. jcsd
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