Intersection of plane How would I do this question.
Find a plane that contains the point A(3,1,−1) and touches the cylinder with radius 3 whose axis is the line p : x = 0, y = z.
It was on a test I did a few days ago and has been bugging me since as I don't even know where to approach this from. Any help would be much appreciated.
Thanks.
 A: Assume the plane is $\pi:ax+by+cz+d=0$. The vector $(0,1,1)$ is on the cylinder axis so it is perpendicular to the normal of the plane:
$$(0,1,1)\cdot(a,b,c)=0\implies c=-b$$
Also, $A(3,1,-1)$ is on the plane so: $$3a+b-c+d=0 \implies b=-\frac{1}{2}d-\frac{3}{2}a$$
We know $(0,0,0)$ is on the axis of the cylinder so the distance between the plane and the point is 3. plugging this into the formula for distance between a plane and a point we get:
$$\frac{|d|}{\sqrt{a^2+b^2+c^2}}=3 \\ d^2 = 9a^2+9b^2+9c^2\\d^2=18(\frac{1}{2}d+\frac{3}{2}a)^2+9a^2\\99a^2+(54d)a+7d^2=0$$
solving this for $a$ gives us:
$$a_{1,2}=\frac{-54d\pm\sqrt{54^2d^2-4\cdot99\cdot7d^2}}{2\cdot99}=\frac{-54d\pm12d}{198}$$
$$a_1=-\frac{7}{33}d,\, a_2=-\frac{1}{3}d$$
Note: Remember that after squaring an equation we have to substitute the solutions in it and check that they really are solutions. In this case both solutions are correct.
$$$$
Case 1:
Substituting $a_1$ in equations for $b,c$ gives us $b = -\frac{2}{11}d,\, c= \frac{2}{11}d$. Plugging this into $\pi$ gives us: $$\pi:-(\frac{7}{33}d)x-(\frac{2}{11}d)y+(\frac{2}{11}d)z+d=0\\$$ $$\pi:7x+6y-6z-33=0\\$$
Case 2:
Substituting $a_2$ in equations for $b,c$ gives us $b = 0,\, c= 0$. Plugging this into $\pi$ gives us: $$\pi:-(\frac{1}{3}d)x + d=0\\$$ $$\pi:x-3=0\\$$
So the two planes are $\,7x+6y-6z-33=0,\, x-3=0$.
