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Math Help - Find the projection of a vector

  1. #1
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    Find the projection of a vector

    Find the projection of the vector <br />
5 \hat{i} + 2 \hat{j} + 3 \hat{k} along 2\hat{i} + \hat{j} + \hat{k}.<br />
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by zorro View Post
    Find the projection of the vector <br />
5 \hat{i} + 2 \hat{j} + 3 \hat{k} along 2\hat{i} + \hat{j} + \hat{k}.<br />
    Hi zorro.

    Let \mathbf u=5\mathbf i+2\mathbf j+3\mathbf k and \mathbf v=2\mathbf i+\mathbf j+\mathbf k.

    You want to find a vector parallel to \mathbf v of magnitude |\mathbf u|\cos\theta where \theta is the angle between \mathbf u and \mathbf v. This would be the vector \frac{|\mathbf u|\cos\theta}{|\mathbf v|}\mathbf v.

    Since \cos\theta\ =\ \frac{\mathbf{u\cdot v}}{|\mathbf u||\mathbf v|} the required vector is \frac{\mathbf{u\cdot v}}{|\mathbf v|^2}\mathbf v=\frac{5\cdot2+2\cdot1+3\cdot1}{\left(\sqrt{2^2+1  ^2+1^2}\right)^2}\left(2\mathbf i+\mathbf j+\mathbf k\right)=\frac{15}{\sqrt6}\left(2\mathbf i+\mathbf j+\mathbf k\right).
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  3. #3
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    My work?

    Quote Originally Posted by TheAbstractionist View Post
    Hi zorro.

    Let \mathbf u=5\mathbf i+2\mathbf j+3\mathbf k and \mathbf v=2\mathbf i+\mathbf j+\mathbf k.

    You want to find a vector parallel to \mathbf v of magnitude |\mathbf u|\cos\theta where \theta is the angle between \mathbf u and \mathbf v. This would be the vector \frac{|\mathbf u|\cos\theta}{|\mathbf v|}\mathbf v.

    Since \cos\theta\ =\ \frac{\mathbf{u\cdot v}}{|\mathbf u||\mathbf v|} the required vector is \frac{\mathbf{u\cdot v}}{|\mathbf v|^2}\mathbf v=\frac{5\cdot2+2\cdot1+3\cdot1}{\left(\sqrt{2^2+1  ^2+1^2}\right)^2}\left(2\mathbf i+\mathbf j+\mathbf k\right)=\frac{15}{\sqrt6}\left(2\mathbf i+\mathbf j+\mathbf k\right).


    Projection of the vector = 5i + 2j + 3k \times \frac{2i + j + k}{\sqrt{4+1+1}}

    = \frac{15}{\sqrt{6}}..........Is this method correct and is the result also correct ?
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  4. #4
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    Quote Originally Posted by zorro View Post
    Projection of the vector = 5i + 2j + 3k \times \frac{2i + j + k}{\sqrt{4+1+1}}

    = \frac{15}{\sqrt{6}}..........Is this method correct and is the result also correct ?

    You use  \times Do you mean using cross product would also reach the answer ? No ... cross product or vector product gives vector not scalar  \frac{15}{\sqrt{6}} ...
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  5. #5
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    Quote Originally Posted by simplependulum View Post
    You use  \times Do you mean using cross product would also reach the answer ? No ... cross product or vector product gives vector not scalar  \frac{15}{\sqrt{6}} ...
     \times is not a cross product here it just means a multiplication sign .......i am sorry for coding that there
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