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Math Help - [SOLVED] Vectors Question

  1. #1
    Member looi76's Avatar
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    [SOLVED] Vectors Question

    Question:
    The position vectors of points A and B are \left(\begin{array}{c}-3\\6\\3\end{array}\right) and \left(\begin{array}{c}-1\\2\\4\end{array}\right) respectively, relative to an origin O.
    (i) Calculate angel AOB.
    (ii) The point C is such that \vec{AC} = 3\vec{AB}. Find the unit vector in the direction of \vec{OC}.


    Attempt:

    (i) \theta = 36.7^o

    (ii) Need help!
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  2. #2
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    Quote Originally Posted by looi76 View Post
    Question:
    The position vectors of points A and B are \left(\begin{array}{c}-3\\6\\3\end{array}\right) and \left(\begin{array}{c}-1\\2\\4\end{array}\right) respectively, relative to an origin O.
    (i) Calculate angel AOB.
    (ii) The point C is such that \vec{AC} = 3\vec{AB}. Find the unit vector in the direction of \vec{OC}.


    Attempt:

    (i) \theta = 36.7^o

    Correct
    Quote Originally Posted by looi76 View Post
    (ii) Need help!
     \vec{AB} = \vec{OB} - \vec{OA}

    Thus:
     \vec{AC} = 3\vec{AB} \Rightarrow \vec{OC} - \vec{OA} = 3(\vec{OB} - \vec{OA}) \Rightarrow \vec{OC}= 3\vec{OB} - 2\vec{OA}
    So complete it now
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  3. #3
    Member looi76's Avatar
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    Quote Originally Posted by Isomorphism View Post

    Correct


     \vec{AB} = \vec{OB} - \vec{OA}

    Thus:
     \vec{AC} = 3\vec{AB} \Rightarrow \vec{OC} - \vec{OA} = 3(\vec{OB} - \vec{OA}) \Rightarrow \vec{OC}= 3\vec{OB} - 2\vec{OA}
    So complete it now
    Thanks Isomorphism
    <br />
\vec{OC} = 3\vec{OB} - 2\vec{OA}

    \vec{OC} = 3\left(\begin{array}{c}-1\\2\\4\end{array}\right) - 2\left(\begin{array}{c}-3\\6\\3\end{array}\right)

    \vec{OC} = \left(\begin{array}{c}-3\\2\\12\end{array}\right) - \left(\begin{array}{c}-6\\12\\6\end{array}\right)

    \vec{OC} = \left(\begin{array}{c}3\\-6\\6\end{array}\right)

    Is my answer correct?
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  4. #4
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    Almost.... You still did not find the unit vector

    But did you understand how I got the expression:
    \vec{AC} = 3\vec{AB} \Rightarrow  \vec{OC} = 3\vec{OB} - 2\vec{OA}

    That is more important than substituting and computing the answer
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  5. #5
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    Quote Originally Posted by Isomorphism View Post
    Almost.... You still did not find the unit vector

    But did you understand how I got the expression:
    \vec{AC} = 3\vec{AB} \Rightarrow  \vec{OC} = 3\vec{OB} - 2\vec{OA}

    That is more important than substituting and computing the answer
    Yes, I have understood, thanks to you . By unit vector you mean i,j,k?
    I mean 3i - 6j + 6k, is this right?
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  6. #6
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    Quote Originally Posted by looi76 View Post
    Yes, I have understood, thanks to you . By unit vector you mean i,j,k?
    I mean 3i - 6j + 6k, is this right?
    No..No..A unit vector is a vector whose magnitude is 1.

    Your second question is:
    (ii) The point C is such that . Find the unit vector in the direction of .
    You have found OC but not the unit vector in the direction of
    To get that vector divide \vec{OC} by magnitude of \vec{OC}
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  7. #7
    Member looi76's Avatar
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    Quote Originally Posted by Isomorphism View Post
    No..No..A unit vector is a vector whose magnitude is 1.

    Your second question is:
    You have found OC but not the unit vector in the direction of
    To get that vector divide \vec{OC} by magnitude of \vec{OC}
    I got the formula from the text book: \hat{v} = \frac{v}{|v|}

    \hat{OC} = \frac{3i - 6j + 6k}{\sqrt{3^2 - 6^2 + 6^2}}

    \hat{OC} = \frac{3i - 6j + 6k}{\sqrt{81}}

    \hat{OC} = \frac{3i - 6j + 6k}{9}

    \hat{OC} = \frac{3}{9}i - \frac{6}{9}j + \frac{6}{9}k

    \hat{OC} = \frac{1}{2}i - \frac{2}{3}j + \frac{2}{3}k

    Is this correct?
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  8. #8
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    Quote Originally Posted by looi76 View Post
    I got the formula from the text book: \hat{v} = \frac{v}{|v|}

    \hat{OC} = \frac{3i - 6j + 6k}{\sqrt{3^2 - 6^2 + 6^2}}

    Is this correct?
    Ya right


    Except for a typo...
    |v| = \sqrt{3^2 + (-6)^2 + 6^2}
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  9. #9
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    Quote Originally Posted by Isomorphism View Post
    Wrong

    |v| = \sqrt{3^2 + (-6)^2 + 6^2}
    \hat{OC} = \frac{3i - 6j + 6k}{\sqrt{3^2 + (-6)^2 + 6^2}}

    \hat{OC} = \frac{3i - 6j + 6k}{\sqrt{81}}

    \hat{OC} = \frac{3i - 6j + 6k}{9}

    \hat{OC} = \frac{3}{9}i - \frac{6}{9}j + \frac{6}{9}k

    \hat{OC} = \frac{1}{2}i - \frac{2}{3}j + \frac{2}{3}k

    Is this correct?
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  10. #10
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    Quote Originally Posted by looi76 View Post
    \hat{OC} = \frac{3i - 6j + 6k}{\sqrt{3^2 + (-6)^2 + 6^2}}

    \hat{OC} = \frac{3i - 6j + 6k}{\sqrt{81}}

    \hat{OC} = \frac{3i - 6j + 6k}{9}

    \hat{OC} = \frac{3}{9}i - \frac{6}{9}j + \frac{6}{9}k

    \hat{OC} = \frac{1}{2}i - \frac{2}{3}j + \frac{2}{3}k

    Is this correct?
    You were right all along... It was just a typing mistake..

    Well done
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