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Equations of Planes

  1. Oct 22, 2012 #1
    Find an equation for the plane through
    (−5, 1, 2) and perpendicular to n = 3 i − 5 j + 2 k.

    They say that :
    The plane has an equation
    3(x + 5) − 5(y − 1) + 2(z − 2) = 0

    What confuses me is why they switched the add/subtract signs and the order of everything when in my notes they tell us the following:

    vector P'knot'·P = (x − x'knot', y − y'knot', z − z'knot').
    Thus, n ·P'knot'·P = 0 is equivalent to a(x − x'knot') + b(y − y'knot') + c(z − z'knot') = 0

    Thanks in advance
  2. jcsd
  3. Oct 22, 2012 #2


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    Science Advisor
    Homework Helper

    Hi Square! :smile:
    Never mind the formula (which btw i can't read :redface:) …

    isn't it obvious that the LHS will be zero if x = -5, y = 1, z = 2, and not for the "minus" values? :wink:
  4. Oct 22, 2012 #3


    Staff: Mentor

    P0 sounds like P "knot" but it's really P "nought". A knot is formed by joining two pieces of rope; nought means "nothing" or "zero".

    You can make subscripts and exponents by clicking Go Advanced, which opens the advanced menu across the top of the input window. Use the X2 button to make exponents, and use the X2 to make subscripts.
  5. Oct 22, 2012 #4
    Yes tim, yes mark and thanks mark. :)

    Ok I got it though I misread the question. Lesson learnt - don't do homework at stupid hours.
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