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Vector Applications

  1. Oct 3, 2005 #1
    Given the four points A(0,1,3), B(1,-3,-2), C(4,2,-1) and D(3,6,4) use vector methods to:

    (a) show these points are coplanar: I just did a determinant of AB BC and CD
    and got an answer of 0 so it is complanar

    Im dont know how to do the next 3:

    (b)find the equation of the plane;
    (c) find the parametric equation of the line perpendicular to the plane that passes through the point X=(-3, 0, -2);

    (d) Find the shortest distance of the point X = (-3,0, -2) to the plane

    Thanks for your time.
  2. jcsd
  3. Oct 3, 2005 #2


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    Homework Helper

    You can do a by calculating one determinant as well:

    [tex]\left| {\begin{array}{*{20}c}
    0 & 1 & 3 & 1 \\
    1 & { - 3} & { - 2} & 1 \\
    4 & 2 & { - 1} & 1 \\
    3 & 6 & 4 & 1 \\

    \end{array} } \right|[/tex]

    b can be done using a determinant too, expanding to the first row of the following determinant. Of course, you only need to use 3 of the 4 points.

    [tex]\left| {\begin{array}{*{20}c}
    x & y & z & 1 \\
    1 & { - 3} & { - 2} & 1 \\
    4 & 2 & { - 1} & 1 \\
    3 & 6 & 4 & 1 \\

    \end{array} } \right|=0[/tex]

    Vectorially, the equation of the plane is given by [itex]P = P_1 + \lambda \left( {P_2 - P_1 } \right) + \mu \left( {P_3 - P_1 } \right)[/itex] where the two directions are given by the difference of two points.

    c) If you have the plane in cartesian equation, the coëfficiënts of x, y and z form a normal vector (so perpendicular to the plane). Then set up the equation of a line with that normal vector as direction and through your point.

    d) Either use a formula, or if you have to compute it yourself, use c! Set up the line through the point, perpendicular to the plane and intersect it with the plane. Then use distance between two points. Note; "distance" is always the "shortest distance".
  4. Oct 3, 2005 #3


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    Staff Emeritus
    Science Advisor

    b) Or use the fact that the normal vector, that you calclulated in (a) is the (A,B, C) in A(x-x0)+ B(y-y0)+ C(z-z0)= 0
    equation for a plane

    c) and, indeed, that same vector is the (A,B,C) in the parametric equations
    x= At+ x0, y= Bt+ y0, z= Ct+ z0 for a normal line!

    d) As TD said, the distance from a point to a plane is measured along the shortest line from the plane to the plane- which is, of course, perpendicular to the plane. Since you calculated the parametric equations of that line in (c), find the point where that line intersects the plane: replace x, y, z in the equation of the plane that you found in (b) with the parametric equations you found in (c) so that you have a single equation in the parameter t. Solve for that parameter, then find the corresponding x, y, z coordinates. Finally, of course find the distance from that point to (-3, 0, -2).

    As TD also mentioned, there is a comparatively simple formula for that that is probably in your text book.
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