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    so, this question SHOULD be easy - but i just cant do it - gaaah!!

    l1: r = i - j + s(i + 2j + 3k)
    l2: r = 2i + j + k + t(2i - j + k)

    the points A and B lie on l1 and l2 respectively.
    the line AB is perpendicular to l1 and l2.

    find the coordinates of A and B.

    so errr... the direction vector of the line perpendicular to AB is (i + j - k)
    and i know there's probably some kind of simultaneous equations i'm meant to be solving, but however i do it, i get about 6 variables, and it just seems too complicated compared to the rest of the exercise.

    help much appreciated!!!
    thanks.
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    (Original post by rachel123456)
    so, this question SHOULD be easy - but i just cant do it - gaaah!!

    l1: r = i - j + s(i + 2j + 3k)
    l2: r = 2i + j + k + t(2i - j + k)

    the points A and B lie on l1 and l2 respectively.
    the line AB is perpendicular to l1 and l2.(

    find the coordinates of A and B.

    so errr... the direction vector of the line perpendicular to AB is (i + j - k)
    and i know there's probably some kind of simultaneous equations i'm meant to be solving, but however i do it, i get about 6 variables, and it just seems too complicated compared to the rest of the exercise.

    help much appreciated!!!
    thanks.
    I think what you do is rewrite for l_1
    x_1+y_1+z_1\equiv\

(1+s)i+(2s-1)j+3sk
    comparing we find the values of
    x_1=1+s,y_1=2s-1,3s=z_1
    so A has the above position vector \begin{pmatrix} 1+s \\ 2s-1 \\ 3s \end{pmatrix}..
    similarly find B from l_2
    and also we know (B-A)=direction vector(you have already found out) and solve it from there ?
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    lol - i really should have been able to do that myself!

    you are a saint - thank you! (it worked wonderfully)

    xxx
 
 
 
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