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Math Help - Perpendicular Vector Lines

  1. #1
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    Perpendicular Vector Lines

    How can I show that these two lines are perpendicular?

    \b{r} = \left(2\b{i}+1\b{j}+1\b{k}\right)+\lambda \left(-2\b{i}+4\b{j}+2\b{k}\right)

    \b{s} = \left(5\b{i}-4\b{j}+2\b{k}\right)+\lambda \left(-1\b{i}+1\b{j}+3\b{k}\right)
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  2. #2
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    I realised that I can show it by direction cosines but is there an alternative method?
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    Senior Member vincisonfire's Avatar
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    Usually you would take the dot product. If it is equal to 0 then the line are perpendicular.
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  4. #4
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    I agree with vincisonfire. Take the dot products of the direction vectors-the vectors multiplied by \lambda.

    (Actually, I see a distinct problem with "showing these two lines are perpendicular"- they are not! Are you sure you haven't lost a sign somewhere?)
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    Quote Originally Posted by HallsofIvy View Post
    I agree with vincisonfire. Take the dot products of the direction vectors-the vectors multiplied by \lambda.

    (Actually, I see a distinct problem with "showing these two lines are perpendicular"- they are not! Are you sure you haven't lost a sign somewhere?)
    Yes, I realised. There was a minus. It should have been:

    \b{s} = \left(5\b{i}-4\b{j}+2\b{k}\right)+\lambda \left(-1\b{i}+1\b{j}-3\b{k}\right)

    Then I considered the direction vectors which work to give angle between them as 90^o.
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  6. #6
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    Quote Originally Posted by Air View Post
    Yes, I realised. There was a minus. It should have been:
    \b{s} = \left(5\b{i}-4\b{j}+2\b{k}\right)+\lambda \left(-1\b{i}+1\b{j}-3\b{k}\right)
    Now the two direction vectors are perpendicular.
    However, in order for the lines to be perpendicular they must intersect.
    Do they intersect?
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  7. #7
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    Quote Originally Posted by Plato View Post
    Now the two direction vectors are perpendicular.
    However, in order for the lines to be perpendicular they must intersect.
    Do they intersect?
    By equating the lines and finding \lambda then substituting that back into one of the equation to get a vector point. Am I correct?
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  8. #8
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    Quote Originally Posted by Air View Post
    By equating the lines and finding \lambda then substituting that back into one of the equation to get a vector point. Am I correct?
    No, you must write the two lines with different parameters.
    r(\lambda ) = \left\{ \begin{gathered}  x = 2 - 2\lambda  \hfill \\  y = 1 + 4\lambda  \hfill \\  z = 1 + 2\lambda  \hfill \\ \end{gathered}  \right.\,\& \,s(\gamma ) = \left\{ \begin{gathered}  x = 5 - \gamma  \hfill \\  y =  - 4 + \gamma  \hfill \\  z = 2 - 3\gamma  \hfill \\ \end{gathered}  \right.

    The set two of them equal. Say \begin{gathered}  2 - 2\lambda  = 5 - \gamma  \hfill \\  1 + 4\lambda  =  - 4 + \gamma  \hfill \\ \end{gathered} .

    Now solve, making sure the solution is consistent in the z position.
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