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Math Help - Scalar product of vectors question

  1. #1
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    Scalar product of vectors question

    Ive studied vectors but I cant do this question in a book.

    PQRS is a rhombus of side 4 units. K, L, M and N are the mid-points of PQ, QR, RS and SP, respectively. SN is a representative of vector u and SM a representative of vector v. Show that SK.SL = 5u.v + 16.

    The rhombus has P at the top left corner and S directly below. Q is slightly lower than P and R is directly below Q.
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  2. #2
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    Quote Originally Posted by Stuck Man View Post
    Ive studied vectors but I cant do this question in a book.

    PQRS is a rhombus of side 4 units. K, L, M and N are the mid-points of PQ, QR, RS and SP, respectively. SN is a representative of vector u and SM a representative of vector v. Show that SK.SL = 5u.v + 16.

    The rhombus has P at the top left corner and S directly below. Q is slightly lower than P and R is directly below Q.
    If you draw a picture, you should be able to see that \vec{SK} = 2\mathbf{u} + \mathbf{v} and \vec{SL} = \mathbf{u} + 2\mathbf{v}. Also, \mathbf{u.u} = \mathbf{v.v} = 2\times2 = 4. Now you just have to work out (2\mathbf{u} + \mathbf{v})\mathbf{.}(\mathbf{u} + 2\mathbf{v}).
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    I had got the first part. I didn't realise that the magnitude of u is 2 and the same for v. I suppose the answer can also be 36. Thanks.
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    The scalar product of two vectors is the two magnitudes multiplied and multiplied by the cos of the angle (<=180). Why does the question not expect the result to be mutiplied by the cos of the unknown angle? So the answer should be 36 X cos of angle.
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  5. #5
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    Quote Originally Posted by Stuck Man View Post
    I had got the first part. I didn't realise that the magnitude of u is 2 and the same for v. I suppose the answer can also be 36. Thanks.
    Quote Originally Posted by Stuck Man View Post
    The scalar product of two vectors is the two magnitudes multiplied and multiplied by the cos of the angle (<=180). Why does the question not expect the result to be mutiplied by the cos of the unknown angle? So the answer should be 36 X cos of angle.
    I think you're confusing \mathbf{u.u} with \mathbf{u.v}. When you take the dot product of a vector with itself, the angle is 0, so you just get the square of the length of the vector. But when the vectors are different, as in \mathbf{u.v}, that no longer applies. That is why the answer to the question leaves \mathbf{u.v} as it is, without attempting to evaluate it. Unless you know the angle of the rhombus, you can't evaluate \mathbf{u.v}.
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