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Math Help - Vectors and area

  1. #1
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    Vectors and area

    Im having trouble with two problems

    Find the unit vectors perpendicular to the plane determined by the three points (-1, 3, 0), (5,1,2), and (4,-3,1).
    i think here you have to divide something by its magnitude but im not sure.

    Find the area of the triangle with (3,2,1), (2,4,6), and (-1,2,5) as vertices.
    i can't even begin here!

    any suggestions appreciated
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  2. #2
    Member eXist's Avatar
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    For the first problem, you need to find a normal vector that defines the plane. To do this find two vectors in the plane using those points and cross them:

    A = (5, 1, 2) - (-1, 3, 0) = <6, -2, 0>
    B = (4, -3, 1) - (-1, 3, 0) =  <5, -6, 0>

     A X B = <0, 0, -26> = C

    Now this is the normal vector that defines the plane, we need to get the unit vector:

    Unit Vector: \frac{C}{|C|} = \frac{<0, 0, -26>}{26} = <0, 0, -1> also <0, 0, 1>

    We must include both the negative and positive of this unit vector because it is perpendicular either way. I hope that's pretty clear, if not, I'll explain more just ask.

    For the second problem, take the absolute value of the determinant divided by 2:

    \left(\begin{array}{cccc}3&2&1\\2&4&6\\-1&2&5\end{array}\right) = 0

    I believe these points to be co-linear since I calculated the determinant to be 0, but I'm going to check again. So for now, I believe the area of the triangle to be 0.
    Last edited by eXist; September 1st 2009 at 08:23 PM.
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  3. #3
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    Quote Originally Posted by eXist View Post
    For the first problem, you need to find a normal vector that defines the plane. To do this find two vectors in the plane using those points and cross them:

    A = (5, 1, 2) - (-1, 3, 0) = <6, -2, 0>
    B = (4, -3, 1) - (-1, 3, 0) =  <5, -6, 0>

     A X B = <0, 0, -26> = C

    Now this is the normal vector that defines the plane, we need to get the unit vector:

    Unit Vector: \frac{C}{|C|} = \frac{<0, 0, -26>}{26} = <0, 0, -1> also <0, 0, 1>

    We must include both the negative and positive of this unit vector because it is perpendicular either way. I hope that's pretty clear, if not, I'll explain more just ask.

    For the second problem, take the absolute value of the determinant divided by 2:

    \left(\begin{array}{cccc}3&2&1\\2&4&6\\-1&2&5\end{array}\right) = 0

    I believe these points to be co-linear since I calculated the determinant to be 0, but I'm going to check again. So for now, I believe the area of the triangle to be 0.
    No, the points are not collinear but they are co-planar, of course- these are 3 points, not vectors, and 3 points always lie in one plane. The "triple" product \vec{u}\cdot(\vec{v}\times\vec{w}), of three vectors in R^3 gives the volume of the parallelopiped having those three vectors as edges.
    The formula you need is the two dimensional equivalent: the area of the parallelogram having \vec{u} and \vec{v} as sides is the length of the cross product of the sides. Given the points (3,2,1), (2,4,6), and (-1,2,5), form the vectors (2-3)\vec{i}+ (4-2)\vec{j}+ (6- 1)\vec{k} and (-1-3)\vec{i}+ (2-2)\vec{j}+ (5-1)\vec{k}. The length of their cross product is the area of a parallelogram having those vectors as sides and 1/2 that is the area of the triangle.
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