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Math Help - scalar products

  1. #1
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    scalar products

    The position vectors of three points A, B, and C with respect to a fixed origin O are 2i - 2j + k , 4i + 2j + k and i + j + 3k respectively.
    Find the unit vectors in the directions of CA and CB. <
    Calculate the ange ACB in degrees.

    Thank you in advance!
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  2. #2
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    Quote Originally Posted by Oasis1993 View Post
    The position vectors of three points A, B, and C with respect to a fixed origin O are 2i - 2j + k , 4i + 2j + k and i + j + 3k respectively.
    Find the unit vectors in the directions of CA and CB. <
    Calculate the ange ACB in degrees.

    Thank you in advance!
    CA = CO + OA

     = -OC + OA

     = OA - OC

     = 2\mathbf{i} - 2\mathbf{j} + \mathbf{k} - (4\mathbf{i} + 2\mathbf{j} + \mathbf{k})

     = -2\mathbf{i} - 4\mathbf{j} + 0\mathbf{k}.


    To find the unit vector in this direction, divide it by its length.

    |CA| = \sqrt{(-2)^2 + (-4)^2 + 0^2}

     = \sqrt{4 + 16 + 0}

     = \sqrt{20}

     = 2\sqrt{5}.



    So \frac{OC}{|OC|} = \frac{-2\mathbf{i} - 4\mathbf{j} + 0\mathbf{k}}{2\sqrt{5}}

     = \frac{-\mathbf{i} - 2\mathbf{j} + 0\mathbf{k}}{\sqrt{5}}

     = \frac{-\sqrt{5}\mathbf{i} - 2\sqrt{5}\mathbf{j} + 0\mathbf{k}}{5}

     = -\frac{\sqrt{5}}{5}\mathbf{i} - \frac{2\sqrt{5}}{5}\mathbf{j} + 0\mathbf{k}.



    Can you do the same to find the unit vector in the direction of CB?
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  3. #3
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    Btw once you have both unit vectors, use the fact that

    \mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos{\theta} (where \theta is the angle in between \mathbf{a} and \mathbf{b}) to find \theta.
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  4. #4
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    hello. Thank you for your help but my book has a different answer than yours. For CA it says (i - 3j - 2k)/ sqrt 14
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    Oops. It looks like I actually worked out BA.

    The process is basically the same though...
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