Find the line of intersection between two planes.

Adding both equations we get $$3y+3z=6$$ or $$y=2-z$$ substituting $z=t$ we obtain $$y=2-t$$ and $$x=2-t+2t=1$$ gives us $$x=-1-t$$ So our line has the equation $$[x,y,z]=[-1,2,0]+t[-1,-1,1]$$ where $t$ is a real number.


As an alternative, the normal vectors to the planes are:

  • $n_1=(1,1,2)$
  • $n_2=(-1,2,1)$

then a direction vector for the line is

$$n_1 \wedge n_2 = (-3,-3,3)$$

Tags:

Geometry

Analytic Geometry

Algebra Precalculus

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