# Find the line of intersection between two planes.

Consider the planes:

$$\begin{eqnarray}x + y + 2z = 1\\ −x + 2y + z = 5\end{eqnarray}$$

1. Find the angle between the two planes.
2. Find the line of intersection of the two planes.

I was able to answer part (1), and I found that the angle was $$60$$ degrees ($$\pi/3$$ radians). The part that I'm having trouble with is finding the line of intersection of the two planes. I have no idea where to start, any help would be much appreciated.

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.

• Thanks for the help. The only part I don't understand is how you came up with the second last line. I'm a bit confused. – N_Mathematics_B Sep 25 '19 at 8:35
• @N_Mathematics_B The usual method for this question: let one of the variables be a "parameter" (that you can rename $t$), then solve the system for the other two variables: you get a parametric equation of the line. Here $z$ is the parameter, so the parametric equation is $(x=f(t), y=g(t), z=t)$, with linear functions $f$ and $g$. – Jean-Claude Arbaut Sep 25 '19 at 8:38
• I understand now, thanks for the help. – N_Mathematics_B Sep 25 '19 at 8:39

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

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

• +1 Interestingly, normal vectors also give very easily an answer to the first question. – Jean-Claude Arbaut Sep 25 '19 at 8:36
• @Jean-ClaudeArbaut Yes, that's a very good point! – user Sep 25 '19 at 8:50