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anonymous

  • 4 years ago

Find the equation of the plane through the origin and parallel to the plane 2x-y+3z=1.

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  1. anonymous
    • 4 years ago
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    And how would you figure it out if they asked for an equation perpendicular to the plane? Is it the same process?

  2. dumbcow
    • 4 years ago
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    Parallel planes by definition do not intersect. so the slopes do not change, for planes this means the normal vector stays the same the components of the normal vector are the coefficients of x,y,z respectively n = <2,-1,3> Given a point (x1,y1,z1) that plane passes through, the equation is --> 2(x-x1) -(y-y1)+3(z-z1) = 0 since it passes through origin, the parallel plane is: 2x -y +3z = 0 For a perpendicular plane , there are many solutions as long as the 2 normal vectors are perpendicular Let n1 be normal vector of a perpendicular plane n1 = <a,b,c> then n*n1 = 0 2a -b +3c = 0 choose any (a,b,c) that satisfies the equation to obtain coefficients of perpendicular plane Let a=1, b=-1, c=-1 --> (x-x1) -(y-y1)-(z-z1) = 0 since it passes through origin, a perpendicular plane is: x -y -z = 0

  3. anonymous
    • 4 years ago
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    Thank you very much!!! This makes sense now! (:

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