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Parametric equation of a line and a plane

  1. Jul 30, 2007 #1
    I CAN'T SEEM TO GET THE ANSWER THAT IS CONSISTENT WITH MY UNDERSTANDING OF THE USE OF DOT AND CROSS PRODUCTS AND THE USE OF THE PARAMETRIC EQUATION OF THE LINE.

    LOOK AT THIS PLEASE:

    the parametric equation of the line is:
    x = 2 + 3t
    y = -4t
    z = 5 + t

    the plane is 4x + 5y - 2z = 18

    The simple part of course is to figure out the direction vector.

    V = <3, -4, 1>

    and if we use this basic vector fact, r = ro + vt where vt = a (vector)

    r0 = <2, 0, 5>

    this was the easy part.

    I found that the plane and the line intersect when the parameter = -2, which gives us a point, p.

    p(-4, 8, 3)

    And, we all know that the NORMAL VECTOR of the given plane is
    N = <4, 5, -2>

    MY QUESTION FOR YOU....I NEED HELP!!!
    SHOULDN'T THE CROSS PRODUCT BETWEEN THE VECTORS R0 AND THE DIRECTION VECTOR, V, GIVE US THE NORMAL VECTOR????? OR AT LEAST A NORMAL VECTOR WITH COMPONENTS THAT ARE A MULTIPLE OF <4, 5, -2>?????

    WHAT ABOUT THE VECTOR THAT FORMS FROM THE POINT OF INTERSECTION OF PLANE AND LINE??? VECTOR P = <-4, 8, 3>
    IF I CROSS VECTOR P AND VECTOR R0, I OUGHT TO GET THE NORMAL VECTOR....BUT I DON'T!!!!!

    WHAT PART OF MY UNDERSTANDING IS INCORRECT?????????
     
  2. jcsd
  3. Jul 31, 2007 #2
    You already have an understanding of direction vector for line.
    What you want now is a direction set for your plane.

    Here's one way to compute them.
    r1=(2,0,-5) is on the plane, and so is r2=(0,2,-4), and so is your p=(-4,8,3).
    Claim: {r1-p, r2-p} constitutes a direction set for your plane (i.e., a linearly independent
    doubleton set of direction vectors).

    If you take the cross product of r1-p and r2-p, you should (will)
    get a vector that's a scalar multiple of N.

    How to "see" it? Visualize where the direction vectors reside.
    If you cross them, what do you get?
     
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