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Thx for help tonight yall 3 more questions X_x

  1. Sep 22, 2013 #1
    Thx for help tonight yall :D 3 more questions X_x

    1. The problem statement, all variables and given/known data
    (1 pt)

    First of all i'm a bit confused.. it wants the equation for a plane in all three questions.. the linear equation of a plane?

    (A) If the positive z-axis points upward, an equation for a horizontal plane through the point (-5,4,-4) is

    So it's a horizontal plane which means that the z direction number will be zero..
    a(x-xi) + b(y-yi) + c(z-zi) = 0
    ax - a5 + by - b4 = -
    so i'll make a equal 4 and b = -5?
    4x - 20 -5y +20 = 0
    4x - 5y = 0

    program says this is wrong..

    .

    (B) An equation for the plane perpendicular to the x-axis and passing through the point (-5,4,-4 is:

    attempt at solution:
    a plane that hits the x axis at a 90 degree angel would mean that it always has the same value for x if i'm picturing this correctly? so the direction number for x would be zero i believe.

    a(x-xi) + b(y-yi) + c(z-zi) = 0
    b(y-(-5)) + c(z-(-4)) = 0
    by + 5b + cz + 4z = 0
    so i could make b = 4 and z = -5?
    4y + 20 - 5z - 20 = 0
    4y - 5z = 0...

    program says i'm doing it wrong X_x
    .

    (C) An equation for the plane parallel to the xz-plane and passing through the point (-5,4,-4) is:

    attempt at solution:
    If it's parallel to the xz plane that means the direction on it's y component must be zero..
    a(x-xi) + b(y-yi) + c(z-zi) = 0
    a(x-(-5)) + b(y-4) + c(z-(-4)) = 0
    ax + 5a + by - 4b + cz + 4c = 0
    well b has to be zero so it's parallel to the xz plane...

    so xa + 5a -cz + 4c = 0
    c could be -5 and a could be 4... ?
    4x + 20 +5z -20 = 0
    4x + 5z = 0

    something went wrong here...?
     
  2. jcsd
  3. Sep 22, 2013 #2

    Mark44

    Staff: Mentor

    It's better to post one problem per thread.
    Any vector along the z-axis will be normal to the plane. One that comes to mind is <0, 0, 1>.

    Correct - this plane isn't horizontal.
     
  4. Sep 22, 2013 #3

    Mark44

    Staff: Mentor

    Yes. And what would that value be? This problem is much simpler than you're making it.


    I don't think the concept of a normal to a plane has gelled in your mind just yet. If the plane is perpendicular to the x-axis, give me one normal to this plane.

    Do you understand what the variables in this equation represent?
    A(x - x0) + B(y - y0) + C(z - z0) = 0

    In case you don't, n = <A, B, C> is a normal to the plane, and P0(x0, y0, z0) is a particular point on the plane.

    The equation comes from the fact that if P(x, y, z) is an arbitrary point on the plane, then the vector PP0 = <x - x0, y - y0, z - z0> is perpendicular to n. Therefore, n ##\cdot## PP0 = 0. The equation above comes directly from this dot product.
     
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