Finding plane at $60^{\circ}$ angle to other plane

Question

I need to find an equation of the plane $\pi_{2}$that makes an angle $60^\circ$ with plane $\pi_{1}: x + y - z + 10 = 0$. Points $M(1,0,0)$ and $N(0,0,1)$ are contained within the plane I am looking for.

I am struggling with this one. What I know:

$\vec{MN} = [-1, 0, 1]$

$\vec{n}_{1} = [1, 1, -1] \quad \land \quad ||\vec{n}_{1}|| = \sqrt{3}$

$\pi_{2}: Ax + By + Cz + D = 0$

$\vec{n}_{2} = [A,B,C] \quad \land \quad ||\vec{n}_{2}|| = \sqrt{A^2 + B^2 + C^2}$

And also:

$$\cos{60^{\circ}}=\frac{\vec{n}_{1} \circ \vec{n}_{2}}{||\vec{n}_{1}||||\vec{n}_{2}||}$$

$$\frac{1}{2} = \frac{A + B - C}{\sqrt{3}\sqrt{A^2 + B^2 + C^2}}$$

$$2A + 2B - 2C = \sqrt{3A^2 + 3B^2 + 3C^2}$$

Lost here. I am not sure how to plug in $\vec{MN}$ here, or just $M, N$ points in general. I suppose what I need to find is vector perpendicular to $\vec{MN}$, but that is too few information for me.

Also:

$$\vec{n}_{2} \perp \vec{MN}$$

$$\vec{n}_{2} \circ \vec{MN} = 0$$

$$-A + C = 0 \Rightarrow A = C$$

So I can simplify:

$$2A + 2B - 2C = \sqrt{3A^2 + 3B^2 + 3C^2} \Longrightarrow 2B = \sqrt{6A^2 + 3B^2}$$

Tips appreciated.

Answer 1

Note that $\vec{MN}\cdot\vec{n_2}=0$ then $A=C$ and $B^2=6A^2$ Now how $(1,0,0)\in \pi_2$ and $(0,0,1)\in \pi_2$ then $A+D=0$ and $C+D=0$, then $D=-A$. If $A=1$, $$A=1\implies \pi_2=:x+\sqrt{6}y+z-1=0$$

Answer 2

A possibly simpler method: $\overrightarrow{MN}=(-1,0,1)^T$. $(1,0,1)^T$ and $(0,1,0)^T$ are linearly-independent vectors orthogonal to this, so there is a one-parameter family of planes through points $M$ and $N$ with equations $(t,1-t,t)\cdot(X-M) = 0$. We want a plane from this family that makes a 60° angle with the given plane, which, as you noted in your question, is the same as the angle between their normals. So, solve $$(t,1-t,t)\cdot(1,1,-1)=\lVert(t,1-t,t)\rVert \lVert(1,1,-1)\rVert \cos(60°)$$ for $t$. (Squaring both sides gets you a straightforward quadratic equation in $t$.)