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    You could try getting three simultaneous equations in lambda and then eliminating lambda?

    EDIT: that might not work, the line is perpendicular to the plane...
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    hm sorry i was dotting it with the point rather than the purpledracula
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    honestly id stay stop wasting your time trying to work it out, this specific question wont come up. go revise! we have tried god knows we havE!
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    (Original post by ZJuwelH)
    You could try getting three simultaneous equations in lambda and then eliminating lambda?

    EDIT: that might not work, the line is perpendicular to the plane...
    in the markscheme they just state the normal as -i + 5j + 3k
    then the rest is fine i just dont know how to find the normal!
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    (Original post by kriztinae)
    in the markscheme they just state the normal as -i + 5j + 3k
    then the rest is fine i just dont know how to find the normal!
    plz check that q again i swear u have misread it, either that or your copy of jun03 is misprinted

    btw a normal is a line perpendicular to another line (p1 knowledge) and was given in the question,
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    (Original post by kriztinae)
    in the markscheme they just state the normal as -i + 5j + 3k
    then the rest is fine i just dont know how to find the normal!
    If you have a plane described as,

    ax + by + cz = d

    then the normal to that plane is given by,

    (a,b,c)
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    (Original post by chandoug)
    plz check that q again i swear u have misread it, either that or your copy of jun03 is misprinted

    btw a normal is a line perpendicular to another line (p1 knowledge) and was given in the question,
    just for you im going to rewrite the question!!

    the plane P1 passes through the point with position vector i + 2j - k, and is perpendicular to the line L with equation r = 3i - 2k + (lamtha)(-i + 2j +3k)

    a) show that the cartesian equation of P1 is x - 5y - 3z = -6

    thats it!
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    just for you im going to rewrite the question!!

    the plane P1 passes through the point with position vector i + 2j - k, and is perpendicular to the line L with equation r = 3i - 2k + (lamtha)(-i + 2j +3k)

    a) show that the cartesian equation of P1 is x - 5y - 3z = -6

    thats it!
    I'm afraid, kriztinae, that's not it!
    Squishy's concerns are correct.
    I have the paper in front of me and the equation of the line is:

    r = 3i - 2k + lamda(-i + 5j + 3k).
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    (Original post by nollaig)
    I'm afraid, kriztinae, that's not it!
    Squishy's concerns are correct.
    I have the paper in front of me and the equation of the line is:

    r = 3i - 2k + lamda(-i + 5j + 3k).
    have we got the same paper??

    i got june 2003?
    i swear thats whats written!!! but yea it makes much more sence your way! argg
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    I've got an actual copy of the exam paper.[Given to me just after last summer's exam - safer to clarify this!!]

    You could, in good faith, have a copy of a paper with a misprint.

    Here's how:

    Many schools use copies of papers distributed electronically. A person, at Edexcel, gallantly retypes the exam papers in Word format and emails them to schools/colleges. It is often more convenient for maths departments to store those papers electronically. He does this typing on a voluntary, after-hours basis and admits that it is easy for mistakes to happen.
 
 
 
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Updated: June 28, 2004
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