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Math Help - position vector given points, but ratio between the two points?

  1. #1
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    position vector given points, but ratio between the two points?

    Use vectors to find the position vector of point 'P' if 'P' divides AB in the ratio 3:2 given A(-1,6,4) and B(4,1,-1).

    This is what I'm thinking so far..
    Point P is 3/5 the way along from A to B. When you divide something in the ratio 3:2, one part will be 3/5 of the total and the other will be 2/5 of the total.

    coordinates of P are (x, y, z), the position vector of point P will be the vector (x - (-1), y - 6, z - 4). This vector has the same direction as AP and the same magnitude.

    Am I on the right track here fellas?
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  2. #2
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    First find the column vector AB = (5, -5, -5)

    vector AP = 3/5 vector AB
    vector AP = 3/5(5, -5, -5)
    vector AP = (3, -3, -3)

    So move 3, -3 and -3 from A (-1, 6, 4) to find the co-ordinates of P.
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  3. #3
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    Position vectors

    Hello everyone -

    You'll find that the answer is simply

    \tfrac{2}{5}\vec{a} + \tfrac{3}{5}\vec{b}

    where, of course \vec{a} and \vec{b} are the position vectors of A and B. Which simply says that P's position is two-fifths of A's plus three-fifths of B's.

    Can you see why this is? As you were saying, you move to P by moving to A and then three-fifths of the way from A to B. So:

    \vec{p} = \vec{a} + \tfrac{3}{5}\vec{AB}

    = \vec{a} + \tfrac{3}{5}(\vec{b} - \vec{a})

    = \tfrac{2}{5}\vec{a} + \tfrac{3}{5}\vec{b}

    And you can generalise this, and say that if P divides AB in the ratio \lambda:\mu, then

    \vec{p} = \frac{\mu}{\lambda+\mu}\vec{a} + \frac{\lambda}{\lambda+\mu}\vec{b}

    (Notice how the \lambda and \mu have 'swapped' over.)

    I hope you find this useful.

    Grandad
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