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    Hi every1, was doing maths hw on vectors and came across one question which i found difficult! could someone help please! I will post all parts to the qu - thanks in advance!

    The postition 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.

    a) Write down unit vectors in the direction of the lines CA and CB and calculate the size of angleACB.

    The midpoint of AB is M.
    b) find a vector equation of the straight line passing through C and M
    c) show that AB and CM are perpendicular.
    d)find the position vector of the point N on the line CM such that angleONC =90degrees.

    Any help on any part would be greatly appreciated!!
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    a)

    CA = (2-1)i+(-2-1)j+(1-3)k = i-3j-2k

    unit vector: sqrt(1^2+3^2+2^2)(CA) = rt14(i-3j-2k)

    CB = (4-1)i+(2-1)j+(1-3)k = 3i+j-2k

    you can do the unit vector...

    CA.CB = (1)(3) + (-3)(1) + (-2)(-2) = |CA||CB|cosx

    x = acos(4/(rt14)(rt14)) = acos(2/7) = 73.4
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    Thanks, the answer i got for first part was diff to urs...

    CA = i - 3j - 2k - that i worked out the same.

    lCAl = sqrt(1+9+4)
    =sqrt14
    therefore unit vector = (i-3j-2k)/sqrt14

    is that not correct?
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    oh yeah. i can't add.
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    i got 73.4?

    CA.CB = 3 +(-3)+4
    =4
    cos x = 4/(sqrt14)(sqrt14)
    =4/14
    x = 73.4?
 
 
 
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