equation of plane passing through one line

Question

Q: Find the equation of the two tangent plane to the sphere $x^{2}+y^{2}+z^{2}=9$ which passes through the line $x+y=6,x-2z=3$

in solution $ : $ the equation of plane passing through the given line is $x+y-6+k(x-2z-3)=0$

I can't understand how the equation of the plane is derived here.

I know how to derive the equation of a plane having two lines, but I don't understand how an equation of a plane can be found from one line. How can a normal vector be found in this?

Please tell me how it is derived; is it even correct?

Answer 1

Well, that is almost correct. Given any two numbers $\alpha,\beta\in\Bbb R$ (such that they aren't both equal to $0$), then the plane$$\alpha(x+y-6)+\beta(x-2z-3)=0$$clearly contains the line defined by $x+y=6$ and $x-2z=3$. Furthermore, every plane containing the line is of this form. Of course, if $\alpha\ne0$, then$$\alpha(x+y-6)+\beta(x-2z-3)=0\iff x+y-6+\frac\beta\alpha(x-2z-3)=0,$$and if you take $k=\frac\beta\alpha$, then you get the equation that you mentioned.

Note that$$x+y-6+k(x-2z-3)=0\label{a}\tag1$$is not one plane. It's a family of planes, depending upon a parameter $k$. However (and this is why I wrote “almost correct”), the plane $x-2z=3$ also contains the given line but it is not of the form \eqref{a}.

Answer 2

For any $k$, $x+y-6+k(x-2z-3)=0$ is the equation of a plane, and if a point lies on the line then $x+y-6=0$ and $x-2z-3=0$, so the point lies on the plane. Strictly speaking, you need $a(x+y-6)+b(x-2z-3)=0$, to allow the plane $x-2z-3=0$, with $a=0$.

Also, the direction vector of the line is perpendicular to the normals to the two planes which define it and any plane including that line will also have a normal which is perpendicular to the direction vector of the line, so a linear combination of the normals to the two planes which define it.

Normal to $x+y-6=0$ is $\left( \begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right)$, normal to $x-2z-3=0$ is $\left( \begin{array}{c} 1 \\ 0 \\ -2 \end{array}\right)$, so normal to any plane including the line of intersection of these two is $$a\left( \begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right)+b\left( \begin{array}{c} 1 \\ 0 \\ -2 \end{array}\right).$$