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Math Help - vectors

  1. #1
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    vectors

     \bold{r}(t) = t^{2} \bold{i} - 4t \bold{j} -t^{2} \bold{k}

     \bold{v}(t) = 2t \bold{i} -4 \bold{j} -2t \bold{k}

     r(t) = \sqrt{2t^{4} + 16t^{2}}

     v(t) = \sqrt{8t^{2} + 16} .

    Find the cosine of the angle between  \bold{r} and  \bold{v} . For what values of  t is  \bold{r} perpendicular to  \bold{v} ? Parallel?

    So  \cos \theta = \frac{ \bold{r}(t) \cdot \bold{v}(t)}{r(t) \cdot v(t)}

     \cos \theta = \frac{4t^{3} + 16t}{2t \sqrt{4t^{4} + 40t^{2} + 64}}

    So  t = 0 or  t = \pm 2i for  \bold{r} \perp \bold{v} .

    But for  \bold{r} \parallel \bold{v} I get  t = 0 . But this is undefined. So there are no values of  t for which  \bold{r} and  \bold{v} are parallel?
    Last edited by heathrowjohnny; February 2nd 2008 at 10:40 PM.
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  2. #2
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    Quote Originally Posted by heathrowjohnny View Post
     \bold{r}(t) = t^{2} \bold{i} - 4t \bold{j} -t^{2} \bold{k}

     \bold{v}(t) = 2t \bold{i} -4 \bold{j} -2t \bold{k}

     r(t) = \sqrt{2t^{4} + 16t^{2}}

     v(t) = \sqrt{8t^{2} + 16} .

    Find the cosine of the angle between  \bold{r} and  \bold{v} . For what values of  t is  \bold{r} perpendicular to  \bold{v} ? Parallel?

    So  \cos \theta = \frac{ \bold{r}(t) \cdot \bold{v}(t)}{r(t) \cdot v(t)}

     \cos \theta = \frac{4t^{3} + 16t}{2t \sqrt{4t^{4} + 40t^{2} + 64}}

    So  t = 0 or  t = \pm 2i for  \bold{r} \perp \bold{v} .

    But for  \bold{r} \parallel \bold{v} I get  t = 0 . But this is undefined. So there are no values of  t for which  \bold{r} and  \bold{v} are parallel?
    To me all your calculations seem to be OK. You only forgot to cancel the t when calculating the \cos(\theta).

    If t = 0 then \vec r = 0 and \vec v = (0, -4, 0). So \vec r is the nullvector which is parallel and perpendicular to any other vector.
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  3. #3
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    Quote Originally Posted by heathrowjohnny View Post
     \bold{r}(t) = t^{2} \bold{i} - 4t \bold{j} -t^{2} \bold{k}

     \bold{v}(t) = 2t \bold{i} -4 \bold{j} -2t \bold{k}

     r(t) = \sqrt{2t^{4} + 16t^{2}}

     v(t) = \sqrt{8t^{2} + 16} .

    Find the cosine of the angle between  \bold{r} and  \bold{v} . For what values of  t is  \bold{r} perpendicular to  \bold{v} ? Parallel?

    So  \cos \theta = \frac{ \bold{r}(t) \cdot \bold{v}(t)}{r(t) \cdot v(t)}

     \cos \theta = \frac{4t^{3} + 16t}{2t \sqrt{4t^{4} + 40t^{2} + 64}}

    So  t = 0 or  t = \pm 2i for  \bold{r} \perp \bold{v} .

    But for  \bold{r} \parallel \bold{v} I get  t = 0 . But this is undefined. So there are no values of  t for which  \bold{r} and  \bold{v} are parallel?
    The vectors are perpendicular when t = 0.

    They will be parallel when  \bold{r}(t) = \lambda \bold{v}(t), where \lambda is a non-zero scalar.

    So you need the following three equations to be satisfied:

    t^2 = \lambda 2t ... (1).

    -4t = -4 \lambda ... (2).

    -t^2 = -\lambda 2t ... (3) (and this is equivalent to (1)).

    From (2): t = \lambda. Sub this into (1):

    \lambda^2 = 2\lambda^2 \Rightarrow \lambda = 0 which is no good since \lambda is a non-zero scalar.

    So there is NO value of t such that the vectors are parallel,

    Edit: unless the trivial solution of both vectors being the zero vector is granted.
    Last edited by mr fantastic; February 2nd 2008 at 11:08 PM. Reason: Saw earboth's reply.
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  4. #4
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    But doesn't this imply that  \bold{r} is the null vector as earboth said?
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  5. #5
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    Quote Originally Posted by heathrowjohnny View Post
    But doesn't this imply that  \bold{r} is the null vector as earboth said?
    Yes. Unfortunately.

    I don't like it. But that's the case.
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