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Linear algebra, planes, vectors help please

  1. Mar 14, 2014 #1
    1. The problem statement, all variables and given/known data

    Find equation of plan H in R^4 that contains the point P= (2,-1,10,6)
    and is parallel to plain H2: 4a +4b + 5c-6d = 3 then answer the following questions:
    A. find normalized normal of plane H which has an angle theta with the normal n= (4,4,5,-6) of H2 such that cos(theta) >0

    B.Find the distance from (2,2,-1,-2) to the plane H

    2. Relevant equations
    0 = n1(a-p1) + n2(b-p2) + n3(c-p3) + n4(d-p4)



    3. The attempt at a solution

    So for part A:
    I know that if they are parallel then the normal of H2 equals to some constant k times normal of H

    N2 = kN
    and I believe I found equation for H:
    0 = n1(a-2) + n2(b+1) + n3(c-10) + n4(d-6)
    I just do not know where to go from there

    Part B

    d(x,y) ||y-x||
    Does this apply to planes?
     
  2. jcsd
  3. Mar 14, 2014 #2

    Mark44

    Staff: Mentor

    Since you're given plane H2 as 4a + 4b + 5c - 6d = 3, you should be able to write the coordinates of a normal to this plane by nothing more than inspection.

    N = <?, ?, ?, ?>
    Just put in the coordinates of the normal, and you're done.
    There's a formula for the distance between two points in R4.
     
  4. Mar 14, 2014 #3
    Thank you for your reply! Yes by inspection the Normal for H2 is (4,4,5,-6), but does that mean that H and H2 have the same normal since they are parallel?

    Also, what is the equation for distance between two points in R4?
    Thanks again!
     
  5. Mar 15, 2014 #4

    Mark44

    Staff: Mentor

    Yes, sort of. The same vector will be normal to both planes, but the two vectors don't have to be equal - one could be a scalar multiple of the other.
    If A = (a1, a2, a3, a4) and B = (b1, b2, b3, b4) are two points in R4, then ##d(A, B) = \sqrt{ (a_1 - b_1)^2 + (a_2 - b_2)^2 + (a_3 - b_3)^2 + (a_4 - b_4)^2}##
     
  6. Mar 15, 2014 #5
    Okay so H and H2 have same norm of (4,4,5,-6)?
    What then is the "normalized norm n of the plane"?
     
  7. Mar 15, 2014 #6

    Mark44

    Staff: Mentor

    You should not use "norm" to talk about the normal to a plane. The term "norm" is used for something entirely different, and is something like the distance between two things.

    I didn't say that H and H2 have the same normal. For example, I can see that the plane x + 2y - 4z = 0 has <1, 2, -4> as a normal. I can also see by inspection, that the plane 2x + 4y - 8z = 3 has <2, 4, -8> as a normal, but the latter vector is just a scalar multiple of the first. Any two vectors that are normal to these planes have to be scalar multiples of each other.
    Make that "normalized normal." A normalized vector is one whose length is 1.
     
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