CupcakeLover72311 CupcakeLover72311
  • 11-02-2020
  • Mathematics
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Find the angle of intersection of the plane −3x−5y+3z=3−3x−5y+3z=3 with the plane −5x+3y−5z=−4−5x+3y−5z=−4. Answer in radians: and in degrees:

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LammettHash
LammettHash LammettHash
  • 25-02-2020

[tex]-3x-5y+3z=3[/tex] has normal vector (-3, -5, 3)

[tex]-5x+3y-5z=-4[/tex] has normal vector (-5, 3, -5)

Compute the angle between these normal vectors:

[tex](-3,-5,3)\cdot(-5,3,-5)=\|(-3,-5,3)\|\|(-5,3,-5)\|\cos\theta\implies\theta\approx1.873\,\mathrm{rad}\approx107.326^\circ[/tex]

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