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Maths Higher Tomorrow!

  1. May 20, 2004 #1
    I have the biggest exam of my life tomorrow -_-

    and i need to know how to do these questions :

    The point Q divides the line joining P(-1, -1, 0) to R(5, 2, -3) in the ratio 2:1. Find the co-ordinates of Q


    The answer is (3, 1, -2)

    Help is appreciated.
  2. jcsd
  3. May 20, 2004 #2


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    [tex]\mbox{Let }\vec{Q}\ =\ (a,\ b,\ c)[/tex]
    [tex]\vec{Q}\ -\ \vec{P}\ =\ 2(\vec{R}\ -\ \vec{Q})[/tex]
    [tex](a,\ b,\ c)\ -\ (-1,\ -1,\ 0)\ =\ 2[(5,\ 2,\ -3)\ -\ (a,\ b,\ c)][/tex]
    [tex]a\ +\ 1\ =\ 10\ -\ 2a[/tex]

    From here, you should be able to figure out two more equations, and should find them very easy to solve, giving you a, b, and c, the co-ordinates of Q.
  4. May 24, 2004 #3
    Disagree with the symbol of AKG, remember POIN is a non-direction Quantity, so that u cant use the VECTOR symbol . And your answer is right. When u take Q-P u will get the vector(PQ) , and the same with R-Q :)
  5. May 24, 2004 #4


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    Every point in [itex]\mathbb R^3[/itex] has a vector representation. The point [itex]Q(a, b, c)[/itex] has a representation [itex]\vec Q = (a, b, c) = a\mathbf i + b\mathbf j + c\mathbf k[/itex]. It made more sense to work with the vectors than to work with the points.
  6. May 24, 2004 #5
    I think
    1/If we get Q(a,b,c) >> a,b,c are the co-ordinate of the POIN Q
    2/If we get vector(Q(a,b,c)) >> a,b,c are the co-ordinate of vector(OQ), with O(0,0,0)

    I agree with you 100% in this kind, may be i understand Q = POINT, but u want to mean Q = vector in your solving :)
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