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Math Help - Vectors question

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    15

    Vectors question

    Relative to a fixed origin, O, the points A and B have position vectors
    (1i, 5j, 1k)
    and
    (6i, 3j, 6k)
    respectively.
    Find, in exact, simplified form,

    (i)
    the cosine of angle AOB, [4]
    (ii)


    the area of triangle OAB, [3]
    (iii) the shortest distance from A to the line OB. [2]

    I successfully found the answer to part (i) to be \frac{\sqrt{3}}{3}.

    I have the mark scheme and the answer to part (ii) is \frac{27}{2}\sqrt{2}
    and the answer to part (iii) is 3\sqrt{2}

    Please explain to me how to do parts (ii) and (iii)
    Thanks
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  2. #2
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4
    Find the length AB, OA and OB
    Then Area = sqrt[s(s-a)(s-b)(s-c)], where s = (a+b+c)/2, a, b and c are the sides of the triangle AOB.

    If the shortest distance from
    A to the line OB. is d, then

    Area AOB = 1/2*d*OB.

    Find OB. You have already calculated area AOB. Find d.
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