1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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)
    PP1=P1-P=(1,0,2)-(1,1,-1)=(0,-1,3)

    ||proj(N)PP1||= |PP1 . N|/||N||=1
    {|(0,-1,3).(-2b+2,-c+2a-3,b-1)|/√[(-2b+2)²+(-c+2a-3)²+(b-1)²]}=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.

    Thanks
     
    Last edited: Nov 11, 2012
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?



Similar Discussions: Plane equation given two points and distance to parallel line
  1. Point-Plane Distance? (Replies: 0)

  2. Lines and planes (Replies: 0)

Loading...