geometrylinear algebra
I need to find the equation(s) of the plane containing the line
$$
\left\{
\begin{array}{c}
x=1+k \\
y=1 \\
z = -2k
\end{array}
\right.
$$
lying at distance 1 from the point (-1,0,1).
I was trying to use "Distance from point to a plane" formula and compose a system, given that I know collinear points on the plane (from the line), however, I apparently miss something to finalize the solution.
$(A, B)$ is a normal vector for the line, therefore $v = (-B, A)$ is a direction vector and you get all points on the line with
$$ (x, y) = P_0 + t \, v = (X_0, Y_0) + t (-B, A), \quad t \in \mathbb R. $$
Now choose $t$ such that the length of $t(−B,A)$ is equal to the given distance $d$, this gives the two points
$$ (X_2, Y_2) = (X_0, Y_0) \pm \frac d{\sqrt { A^2 + B^2}} (-B, A) \quad . $$
Let us go for the intersection line first. We have the system of equations $$ x + y + z = 1 \\ y + z = 0 $$ which can be simplified to $$ x = 1 \\ y + z = 0 $$ which gives the line $$ (1, y, -y) = (1,0,0) + (0,1,-1) y $$ We can extend that line to a plane by $$ (1,0,0) + (0,1,-1) s + (a, b, c) t $$ where $s, t \in \mathbb{R}$ and $(a,b,c)$ is a vector we need to choose such that the plane contains $A$: $$ (2,1,0) = (1,0,0) + (0,1,-1) s + (a, b, c) t = (1 + at, s + bt, -s + ct) $$ We are now dealing with two unknown parameters and three unknown components, but have only three equations. So we try to require $s=0$ and $t=1$ and look if there is a solution: $$ (2,1,0) = (1,0,0) + (a, b, c) \iff (1,1,0) = (a,b,c) $$ This gives $$ (1,0,0) + (0,1,-1)s + (1,1,0) t = (1+t,s+t,-s) \quad (s, t \in \mathbb{R}) $$ as equation for the plane.
Alternate representation of the solution plane:
We can bring this into a single equation in three coordinates, by finding the normal form $$ n \cdot x = d $$ where $n$ is a unit normal vector of the plane and $d$ is the (signed) distance to the origin.
Maybe there is an easier way to do this, but I do not see it right now.
We can calculate a normal vector from the vector product of the plane spanning vectors: $$ (0,1,-1) \times (1,1,0) = (1, -1, -1) $$ so a unit normal vector is $n = (1,-1,-1)/\sqrt{3}$. The distance of the plane to the origin is $$ d^2 = q = \lVert (1+t, s+t, -s) \rVert^2 = (1+t)^2 + (s+t)^2 + s^2 \\ $$ We look where the gradient vanishes: $$ 0 = \partial q / \partial s = 2(s+t) + 2s = 4s + 2t \\ 0 = \partial q / \partial t = 2(1+t) + 2(s+t) = 2s + 4t + 2 $$ This gives the system $$ 4s + 2t = 0 \\ 2s + 4t + 2 = 0 $$ or $$ 2s + t = 0 \\ 2s + 4t + 2 = 0 $$ or $$ 2s + t = 0 \\ 3t + 2 = 0 $$ so $$ s = 1/3 \\ t = -2/3 $$ so we get $$ q = (1 - 2/3)^2 + (1/3 - 2/3)^2 + (1/3)^2 = 1/9 + 1/9 + 1/9 = 3/9 = 1/3 $$ and $d = 1 / \sqrt{3}$. This gives $$ \frac{1}{\sqrt{3}} (1,-1,-1) \cdot (x,y,z) = \frac{1}{\sqrt{3}} $$ or simply $$ (1,-1,-1) \cdot (x,y,z) = 1 \iff \\ x - y - z = 1 $$
Best Answer
There is a one-parameter family of planes that all include the given line. You can use the distance formula with this to generate a one-variable quadratic equation.
There are several ways to produce this family of planes. For instance, note that their normals must be perpendicular to the line. We can read a direction vector for the line directly from its parameterization: $(1,0,-2)$. By inspection, two linearly independent vectors perpendicular to this are $(0,1,0)$ and $(2,0,1)$, and a point on the line can also be read directly from its parameterization, so, using the point-normal form of plane equation, the equations of two distinct planes that include the line are $y=1$ and $2x+z=2$. Every plane that includes the line has an equation that’s an affine combination of these two equations, yielding $$(1-\lambda)(y-1)+\lambda(2x+z-2) = 2\lambda x+(1-\lambda)y+\lambda z-(\lambda+1)=0.$$ Now use the distance formula with this generic plane equation and solve for $\lambda$.