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    Hi i did this question from a past paper and got it right when doing it. I have since thrown the paper away, and now, reviewing the papers the night before the exam, i CANT get the answer. please put me out of my misery;


    the lines have equations r = (5i + j +5k) + s(i - j -2k)

    and r = (i +11j + 2k) + t(-4i - 14j +2k)

    find the equation of the plane containg both lines in the form ax + by + cz = d


    thanks!
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    You want r.n = a.n

    n is found by the cross product of the direction vectors of the lines.

    a is any point on the plane, either of the two position vectors given should do.

    r is of course xi+yj+zk.
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    (Original post by ZJuwelH)
    You want r.n = a.n

    n is found by the cross product of the direction vectors of the lines.

    a is any point on the plane, either of the two position vectors given should do.

    r is of course xi+yj+zk.
    thats what i was doing, but just cant get the rite answer
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    (Original post by Leeroy)
    thats what i was doing, but just cant get the rite answer
    What is the right answer? I get two different ones by using different vectors for a.

    EDIT: a is (5i+j+5k).
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    i get 5x - y + 3z = 39


    edit: that is...

    (-4i -14j + 2k) x (i -j -2k)

    you cross the direction vectors
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    This is how I always remember the cross product:

    Think of it as finding determinants (watch out for the middle one though!)
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    show me how
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    i'm sure you know how to do the cross product,

    just cross the 2 direction vectors to find the normal to the two lines

    the r.n = k

    r is (x y z), dot that with the normal, 'n'. k is a constant

    so when you 'dot' it out, you get 5x -y +3z = k

    put in a know point ie (x y z) to get k
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    (Original post by Leeroy)

    the lines have equations r = (5i + j +5k) + s(i - j -2k)

    and r = (i +11j + 2k) + t(-4i - 14j +2k)

    find the equation of the plane containg both lines in the form ax + by + cz = d

    Okay...

    Cross product the two vectors in the plane to get a normal... Use a determinant to do this like:
    ¦i 1 -4 ¦
    ¦j -1 -14 ¦
    ¦k -2 2 ¦

    expand in terms of the i j and k

    Dot product one of the 'a' portions, i.e the 5i + j +5k with this normal

    to get r . n = d

    where d is the result of the dot product.

    hence you know that (ai + bj + ck) . n = d

    then put in cartesian.
 
 
 
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