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Math Help - Vectors, problem with variables

  1. #1
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    Vectors, problem with variables

    We have three vectors, a = (1,2,3) b = (2,-1,x) c = (y,z,1). We need to find x, y and z, if the vectors are right-angled.

    I am trying to solve it with the dot product. I know that the dot product of two vectors is equal to zero, if they are right-angled.
    But i always get x = 0, y = 0, z = 0.

    and sorry for no arrows, i have to learn Latex, i know.
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  2. #2
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    Quote Originally Posted by Nforce View Post
    We have three vectors, a = (1,2,3) b = (2,-1,x) c = (y,z,1). We need to find x, y and z, if the vectors are right-angled.

    I am trying to solve it with the dot product. I know that the dot product of two vectors is equal to zero, if they are right-angled.
    But i always get x = 0, y = 0, z = 0.

    and sorry for no arrows, i have to learn Latex, i know.
    Yes, x = 0. Please show all your working for how you got your value of y and z.
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  3. #3
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    Well when i get x = 0. I repeat the process and try to get x and z. I get an equation with two variables. When i solve it, i get y = 0 and z = 0 again. It's a little strange but it could be. So i check it for the dot product of vectors a and c and it's not equal to zero, i get 3 = 0.
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  4. #4
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    Quote Originally Posted by Nforce View Post
    We have three vectors, a = (1,2,3) b = (2,-1,x) c = (y,z,1). We need to find x, y and z, if the vectors are right-angled.
    There is no such thing as a "right angled vector". Do you mean that a and b are at right angles to each other and that b and c are at right angles to each other? (From that it follows that a and c are at right angles to each other.)

    As mr fantastic said, x= 0. Please show how you calculated y and z. I suspect that you dropped a number in your equations.

    I am trying to solve it with the dot product. I know that the dot product of two vectors is equal to zero, if they are right-angled.
    But i always get x = 0, y = 0, z = 0.

    and sorry for no arrows, i have to learn Latex, i know.
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  5. #5
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    Yes i meant at right angles.

    so...

    (2,-1,0)*(y,z,1) = 0
    2y -z = 0
    2y = z

    2y -(+2y) = 0
    2y - 2y = 0
    y(2-2) = 0

    ?
    0 = 0
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  6. #6
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    Quote Originally Posted by Nforce View Post
    Yes i meant at right angles.

    so...

    (2,-1,0)*(y,z,1) = 0
    2y -z = 0
    2y = z
    Okay, that's bc= 0.


    2y -(+2y) = 0
    2y - 2y = 0
    y(2-2) = 0

    ?
    0 = 0
    But where did you get "2y- (2y)= 0" from? Surely not using "bc= 0" again? That equation already gave you z= 2y and because of that setting z= 2y in it will give a solution for all y. That's why it reduces to "0= 0" which is always true. The requirement you haven't used yet is that ac= 0 which is (1, 2, 3)(y, z, 1)= y+ 2z+ 3= 0.
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  7. #7
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    the solution for ac = 0 is

    y = -2z - 3

    if i understand you properly, now i have to solve this two equations?:

    y = -2z - 3
    z = 2y

    sorry, my bad... i double check it. And it's right. Thanks for helping me out.

    Problem solved.
    Last edited by Nforce; October 17th 2010 at 08:54 AM. Reason: sorry my bad
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