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    How do I prove that a point lies on a line?
    I don't know what to do...I'm not talking about a particular exercise but in general as I wanna know how....
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    (Original post by toxi)
    How do I prove that a point lies on a line?
    I don't know what to do...I'm not talking about a particular exercise but in general as I wanna know how....
    I'm pretty sure you have to equate the coefficients of your i's j's and k's, then you should get λ=...
    If the point lies on the line the value of λ should be the same in all 3 cases
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    (Original post by toxi)
    How do I prove that a point lies on a line?
    I don't know what to do...I'm not talking about a particular exercise but in general as I wanna know how....
    ok so you have a line r=ai+bj+cj+Δ(di+ej+fk).
    Now if a point P with position vector : xi+yj+zk lies on the line,then (a+Δd) would be equal to x,So find Δ.Now do the same for the j and k coefficients and u will find that the value of Δ would be the same for all of them.
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    (Original post by IntegralAnomaly)
    ok so you have a line r=ai+bj+cj+Δ(di+ej+fk).
    Now if a point P with position vector : xi+yj+zk lies on the line,then (a+Δd) would be equal to x,So find Δ.Now do the same for the j and k coefficients and u will find that the value of Δ would be the same for all of them.

    ok i think ill re write this in another equation

    assuming
    r = 5i + 6j +2k + k(i + 2j + 3k)
    and you are asked to prove that line 6i + 8j + 5k is on this equation
    (bear in mind that im taking this numbers out of random)
    kk..
    now let k as 1
    (this depends on the question...cause k can be anything...so you can use 2 or 3 or whatever depending on what line touches...im using 1 cause i need the sum of 5 + k = 6...thus k has to be one)

    now if i let k as 1
    r = (5+1)i + (6+2)j + (3+2)k
    r = 6i + 8j + 5k (proven)
    therefore,the line is in the equation
 
 
 
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