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Vector algebra (i guess?)

  1. Sep 2, 2005 #1
    hello, i'm taking electrodynamics, and to begin we're doing vector algebra/calculus.

    here's the problem i'm faced with. there's a triangle defined by the points.


    the task at hand is finding the unit vector orthogonal to the defined plane. the method that occured to me was to find two of the sides of the triange. from there, i could divide the resulting vector by the magnitude (abcos(theta)), giving me n-hat.

    in any case, it's been along time since trig, and i'm not sure how i should get the angle theta between the two selected sides. maybe i'm not doing it right, i don't know. any advise would be appreciated!

  2. jcsd
  3. Sep 2, 2005 #2


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    I can't understand what you're suggesting, but here's a hint for a way to do it.

    HINT: Cross product.
  4. Sep 2, 2005 #3
    yes, call the x-y line vector a, x-z vector b

    a X b = |ab|cos theta n(hat)

    n(hat) would equal [aXb]/|ab|cos theta


    the question remains on how to get the angle. i can't seem to recall and a friend is borrowing my trig/algebra book.
  5. Sep 2, 2005 #4
    Try taking the determinant.

    [tex]a\times b=\left|\begin{array}{ccc}i & j & k \\ 1 & -2 & 0 \\ 0 & 2 -& 3\end{array}\right|[/tex]

    This will give you one vector orthogonal to the triangle.
  6. Sep 2, 2005 #5


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    Almost. The magnitude of a X b is |ab|sin(theta), not cos(theta). But use apmcavoy's way. Get the vector by the determinant "method", and simply calculate its norm with use of "3-D pythagoras".
  7. Sep 2, 2005 #6
    For me the easiest way is to find the equation of the plane in the form of

    ax + by + cz + d = 0.

    It follows that the unit vector n-hat ix (a,b,c)
  8. Sep 3, 2005 #7
    thanks, i'm retarded. i had the cross product defined, i just didn't think to make a unit vector out of it. it was pretty trivial dividing my result by 7. thanks again.

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