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Math Help - Vector problem

  1. #1
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    Vector problem

    HI
    Can someone tell me how would i complete the following question:

    If  a=2i-3j+k, b=2i-4j+5k and c=-i-4j+2k find the values of p and q such that a+pb+qc is parallel to the x-axis.

    This is what i just started:
    2i-3j+k+p(2i-4j+5k)+q(-i-4j+2k)<br />
    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    HI
    Can someone tell me how would i complete the following question:

    If  a=2i-3j+k, b=2i-4j+5k and c=-i-4j+2k find the values of p and q such that a+pb+qc is parallel to the x-axis.

    This is what i just started:
    2i-3j+k+p(2i-4j+5k)+q(-i-4j+2k)<br />
    P.S
    If the resultant vector is parallel to the x axis, then its \mathbf{j} and \mathbf{k} components are 0.
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  3. #3
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    the answers says p=\frac{1}{6} and q=\frac{-11}{12}
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  4. #4
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    Do what I said.

    Collect the \mathbf{i},\mathbf{j} and \mathbf{k} terms.

    Set the \mathbf{j} and \mathbf{k} terms to 0.

    Solve the resulting equations.
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  5. #5
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    ok i'll tried that
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  6. #6
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    I still cannot get it can someone tell me the working to this.
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  7. #7
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    continuing from where u left off

    <br />
2\mathbf{i}-3\mathbf{j}+\mathbf{k}+p(2\mathbf{i}-4\mathbf{j}+5\mathbf{k})+q(-\mathbf{i}-4\mathbf{j}+2\mathbf{k})=(2+2p-2q)\mathbf{i}-(3+4p+4q)\mathbf{j}+(1+5p+2q)\mathbf{k}<br />

    as stated in the question, the vector a+pb+qc is parallel to the x-axis

    therefore, the \mathbf{j} and \mathbf{k} component are null
    using that result;

    3+4p+4q = 0 --> (1)
    1+5p+2q = 0 --> (2)

    (2) multiply by 2

    2+10p+4q=0 --> (2a)

    (2a)-(1);

    -1+6p = 0
    p = \frac{1}{6}

    substitute p = \frac{1}{6} into (1)
    q = \frac{-11}{12}
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