# Plane with a given point, a parallel line and an angle with another plane

The problem states:

Find the equation of a plane $$\alpha$$ that has a point $$A(1,1,-1)$$, is parallel to the line $$p$$ given as $$p:x+y=0$$, $$2x+y-2=0$$ and forms a $$\frac{\pi}{4}$$ angle with the plane $$\beta: x-4y-z-2=0$$.

The way I've tried to solve it is:

First, since $$A \in \alpha$$, we know that the equation of the plane will look like

$$\alpha: a(x-1)+b(y-1)+c(z+1) = 0$$ where $$\vec{n_{\alpha}} = (a, b, c)$$ is its normal vector. Now we have to find that vector. From the fact that $$\alpha \parallel p$$ we know that $$\vec{n_{\alpha}}$$ and $$\vec{p}$$ must be perpendicular ($$\vec{p}$$ being the direction vector of the line $$p$$). The direction vector of the line we find by taking the cross product of normal vectors of the two planes that the line is given as: $$\vec{p} = \vec{n_1} \times \vec{n_2}=(1,1, 0) \times(2, 1, 0) = (0,0,-1)$$ And now we have: $$\vec{n_{\alpha}} \cdot \vec{p} = 0$$ $$(a, b, c) \cdot(0,0,-1) = 0$$ $$-c = 0 \Rightarrow c = 0$$ So, for now, we only know that $$\vec{n_{\alpha}} = (a, b, 0)$$

From the third condition we have that

$$\cos(\frac{\pi}{4}) = \frac{ \vec{n_{\alpha}} \cdot\vec{n_{\beta}} } { |\vec{n_{\alpha}}| |\vec{n_{\beta}}| }$$ $$\frac{\sqrt{2}}{2} = \frac{(a, b, 0) \cdot(1, -4, -1)}{\sqrt{a^2+b^2}\sqrt{18}}$$

From that we get: $$3\sqrt2\sqrt2\sqrt{a^2+b^2} = 2(a-4b)$$ $$3\sqrt{a^2+b^2} = a-4b$$ Squaring that gives us $$8a^2-7b^2 -8ab=0$$

Dividing by $$a^2$$ and introducing $$t = \frac{b}{a}$$ we get:

$$7t^2+8t-8 = 0$$ where we find

$$\frac{b}{a} = \frac{-4 \pm 6\sqrt2}{7}$$

Since I've only been able to obtain the ratio of $$a$$ and $$b$$ does that mean that there are infinitely many planes that satisfy the given conditions? Or have I done something wrong?

Thanks.

$$Ax+By+Cz=D$$ is the same plane as $$kAx+kBy+kCz=kD$$ if $$k\ne0$$. So the problem is solved if the ratio of $$a$$ and $$b$$ is found.