1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Find Find the point in this plane where it intersects the z axis

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data

    find the point in this plane where it intersects the z axis (x=y=0).

    P = {<x;y;z;> = <1;-1;0;> + a<1;-2;0;> + b<2;0;-1;> , for all a,b}

    (by <l;m;n;> i mean to denote a column)

    2. Relevant equations

    3. The attempt at a solution
    I talked to my TA last friday and he made it seem easy, but I don't remember what he said anymore. I have something that looks like

    I set up a matrix that looks like

    [<1;-1;0;> <1;-2;0;> <2;0;-1;>] [<1;a;b>] = [<0;0;2;>]

    I'm not really sure what to do from here. It says it is asking for a point, does this mean which (a,b) works? What can you help me with?

    Thanks in advance.
  2. jcsd
  3. Sep 26, 2011 #2


    User Avatar
    Homework Helper

    find (a,b) where it works by row-reducing your matrix

    then use (a,b) in
    P= P(a,b)=<x,y,z>=<x(a,b),y(a,b),z(a,b)>

    a & b are parameters in the parameterisation of the plane, each (a,b) corresponds to a point <x,y,z> on the plane P
  4. Sep 26, 2011 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    This is all you should need.
    <x;y;z;> = <1;-1;0;> + a<1;-2;0;> + b<2;0;-1;> ​

    It says that
    x = 1 + 1∙a + 2∙b

    y = -1 - 2∙a + 0∙b

    z = 0 + 0∙a - 1∙b

    Now "Find the point in this plane where it intersects the z axis (x=y=0)." So, find what a & b make this happen?

    Edited per ehild's note.

    Thanks goes to ehild.
    Last edited: Sep 27, 2011
  5. Sep 27, 2011 #4


    User Avatar
    Homework Helper


    There is a typo in your equations (shown in red)

    It should be

    y=-1-2∙a +0∙b

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook