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!

Vector and Plane Problem

  1. Sep 29, 2012 #1
    1. The problem statement, all variables and given/known data

    c = 10i+2j-3k = p + q

    Calculate p and q where the vector p is parallel and q is perpendicular to the following plane:

    2x+3y-z=4

    2. Relevant equations

    a[itex]\bullet[/itex]b= axbx+ayby+azbz

    3. The attempt at a solution

    I know that because p is perpendicular to q the dot product between these two vectors should be zero. Also a vector parallel to the plane above is one which satisfies to equation of the plane and any vector perpendicular to the plane is a scalar multiple of the vector:

    2x+3y-z
     
  2. jcsd
  3. Sep 29, 2012 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Good- though i would not write the vector as "2x+3y-z". You have already used x, y, and z as variables in the equation of the plane so it would better to use, say, i, j, and k as the unit vectors in the x, y, and z directions. That is, 2i+ 3j- k. Since 10i+2j-3k = p + q= p+ 2ai+ 3aj- ak, you can write p= (10- 2a)i+ (2- 3a)j+ (-1+ a)k. And, of course, because they are perpendicular, the dot product of p and q is 0: (10- 2a)(10)+ (2- 3a)(2)+ (-1+ a)(-1)= 0.

    Solve that equation for a.
     
  4. Sep 29, 2012 #3
    Hello,

    Thanks for your help, I am still a little confused with the solution, when I solve for p I get:

    p = (10-2a)i+(2-3a)j+(-3+a)k

    but I am also confused about the dot product, should it not be:

    q = 2ai+3aj-ak

    pdotq = (10-2a)i(2a)i + (2-3a)j(3a)j + (-3+a)k(-a)k

    This is what I get when I solve is out but my final answer still does not make sense. I must be making a small error at some point. It would be really great if you could help me out.
     
  5. Sep 30, 2012 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    A vector lying in the plane does not satisfy the equation of the plane. The equation of the plane means that the difference between two vectors of the plane r and ro is perpendicular to the normal of the plane: (r-ro)n=0 which leads to the equation ax+bx+cz=C, where a,b,c, are the components of the normal and C is the dot product of ro with the normal vector.

    The normal vector of the plane 2x+3y-z=4 is n=(2,3,-1), a vector, perpendicular to the plane. The vector c then is the sum of a vector, parallel to n and an other, perpendicular to n. You certainly know how to get the projection of a vector to a certain direction?

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Vector and Plane Problem
  1. Vectors and Planes (Replies: 3)

  2. Plane and 3d vector (Replies: 1)

Loading...