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Plane geometry problem

  1. Sep 2, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the equation of plane which contains the line [itex]l:\left\{\begin{array}{l} x=t+2 \\y=2t-1\\z=3t+3 \end{array}\right.[/itex], and makes the angle of [itex]\frac{2\pi}{3}[/itex] with the plane [itex]\pi:x+3y-z+8=0[/itex].

    3. The attempt at a solution

    My attempt was to find the normal vector of plane which contains the line by the cross product of vector [itex]\vec{a}[/itex] of line and some vector created between points [itex]P(2,-1,3)[/itex] of line and point [itex]Q(x,y,z)[/itex], than by using the formula to find angles, but this leads me to complications.

    I think there's another way to solve this, which I don't know.

    Thank you
     
  2. jcsd
  3. Sep 2, 2011 #2

    HallsofIvy

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    If two planes make a given angle, [itex]\theta[/itex], with each other, then their normal vectors make the same angle. The dot product of two vectors, u and v, is given by [itex]u\cdot v= |u||v|cos(\theta)[/itex]. So if we take <a, b, c> to be a unit vector perpendicular to the desired plane, we must have [itex]<a, b, c>\cdot<1, 3, -1>= \sqrt{1+ 9+ 1} cos(2\pi/3)[/itex] or [itex]a+ 3b- c= -.5\sqrt{11}[/itex]. That, together with [itex]a^2+ b^2+ c^2= 1[/itex] gives two equations to solve for a, b, and c.
     
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