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Math Help - Angles between vectors (parallel, orthogonal, ...)

  1. #1
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    Angles between vectors (parallel, orthogonal, ...)

    p = (2 , k)
    q = (3 , 5)


    Find such k that:
    a) p and q are orthogonal
    b) the angle between p and q is \frac {\pi}{4}

    I was using:
    <br />
\cos \phi = \frac {p * q}{\|p\|*\|q\|}<br />
    but somehow I can't solve it right.

    Thanks!
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  2. #2
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    Quote Originally Posted by saeros View Post
    p = (2 , k)
    q = (3 , 5)


    Find such k that:
    a) p and q are orthogonal
    b) the angle between p and q is \frac {\pi}{4}

    I was using:
    <br />
\cos \phi = \frac {p * q}{\|p\|*\|q\|}<br />
    but somehow I can't solve it right.

    Thanks!
    If the vectors are orthogonal, then \mathbf{p}\cdot \mathbf{q} = 0.

    If the vectors are parallel, then \mathbf{p}\cdot \mathbf{q} = |\mathbf{p}||\mathbf{q}|.
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  3. #3
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    I understand, but then I get to a equation I obviously can't solve !

    It's like that:
    <br />
5k - \sqrt{34\cdot(4+k^2)} + 6 = 0<br />

    I guess I'm not wrong at this point.. but I have no clue how to get k out of it. Any clues please?
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  4. #4
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    Hello saeros
    Quote Originally Posted by saeros View Post
    I understand, but then I get to a equation I obviously can't solve !

    It's like that:
    <br />
5k - \sqrt{34\cdot(4+k^2)} + 6 = 0<br />

    I guess I'm not wrong at this point.. but I have no clue how to get k out of it. Any clues please?
    I assume it's part (b) that you're stuck with. So here it is:
    \textbf{p}.\textbf{q}=|\textbf{p}||\textbf{q}|\cos  \phi
    \Rightarrow 5k+6 = \sqrt{34(4+k^2)}.\frac{1}{\sqrt2}, when \phi = 45^o
    Square both sides:
    \Rightarrow 25k^2+60k+36 = 17(4+k^2)
    =68+17k^2
    \Rightarrow 8k^2 + 60k -32 = 0

    \Rightarrow 2k^2+15k-8=0

    \Rightarrow (2k-1)(k+8)=0

    \Rightarrow k = \tfrac12,\;-8
    When you substitute back in the original equation, the negative root is invalid. So the answer is:
    k = \tfrac12
    Grandad
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