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Math Help - Vector's

  1. #1
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    Vector's

    I've been given this question and am struggling to complete it.

    Relative to a fixed origin, O, the points A and B have position vectors

    (i + 5j -1k) and (6i + 3j - 6k) respectively

    Find in exact, simplified form,
    (i) the cosine of angle AOB - i have done this and it comes out as root 3 over 3
    (ii) the area of the triangle
    (iii) the shortest distance from A to the line OB

    Any help on this would be very much appreciated. Thanks
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  2. #2
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    Quote Originally Posted by lemmingsrevolt View Post
    I've been given this question and am struggling to complete it.

    Relative to a fixed origin, O, the points A and B have position vectors

    (i + 5j -1k) and (6i + 3j - 6k) respectively

    Find in exact, simplified form,
    (i) the cosine of angle AOB - i have done this and it comes out as root 3 over 3
    (ii) the area of the triangle
    (iii) the shortest distance from A to the line OB

    Any help on this would be very much appreciated. Thanks
    (ii) If you know about the cross product you can calculate \frac{1}{2} | \vec{OA} \times \vec{OB} |.

    Otherwise use the sine rule: \frac{1}{2} | OA | \cdot | OB | \sin (\angle AOB).


    (iii) Calculate the scalar resolute of the vector \vec{OA} perpendicular to the vector \vec{OB}.
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  3. #3
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    Thank you very much that's really helpful.
    My main problem is how to find the position vector of O so i can do these things? would you be able to tell me how to do this?
    thanks a lot
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  4. #4
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    Hi

    O is the origin therefore its position vector is 0i + 0j + 0k

    The cross product of \overrightarrow{OA} and \overrightarrow{OB} is -27i - 27k

    The area of the triangle is \frac12\:\sqrt{27^2+27^2} = \frac{27\:\sqrt{2}}{2}
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