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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.

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1 Answer 1

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$Ax+By+Cz=D$ 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.

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  • $\begingroup$ Yeah, seconds after posting the question I'd realized that I was an idiot, since the vector is the same if scaled by a non-zero coefficient. Thanks anyway. $\endgroup$
    – Koy
    May 16, 2019 at 8:00

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