Finding the locus of planes by given conditions

Question

There is point (1,2,3) on a line,then how to find the locus of all planes which are at a distance of 4 units from the point?

I thought the shifting of origin will help. But how to proceed further??

Answer 1

Thinking to the sphere parameterized by $\theta,\varphi$ (https://en.wikipedia.org/wiki/Spherical_coordinate_system),

$$\vec{N}=\begin{pmatrix}\cos\varphi\cos \theta, \cos\varphi\sin \theta, \sin\varphi \end{pmatrix}\tag{1}$$

is a generic vector with norm $1$ ("unit vector").

Therefore a generic equation for a plane $P_{\theta,\varphi}$ is :

$$x\cos\varphi\cos \theta+y\cos\varphi\sin \theta+z\sin\varphi-K=0 \tag{2}$$

where $K$ is a constant to be determined by the constraint that point $P(1,2,3)$ is at distance $4$ from this plane. Happily, there is a formula for the distance $d$ from a point $(x_0,y_0,z_0)$ to a plane $ax+by+cz-k=0$ with unit vector $(a,b,c)$ which is (http://mathworld.wolfram.com/Point-PlaneDistance.html):

$$d=|ax_0+by_0+cz_0-k|$$

This gives here :

$$d=|(1\cos\varphi\cos \theta+2\cos\varphi\sin \theta+3\sin\varphi)-K|=4$$

which provides two values of constant $K$

$$K=1\cos\varphi\cos \theta+2\cos\varphi\sin \theta+3\sin\varphi \pm 4 \tag{3}$$

(in fact $+$ sign is enough : do you see why ?). It remains to plug (3) into (2) to give the general result :

$$(x-1)\cos\varphi\cos \theta+(y-2)\cos\varphi\sin \theta+(z-3)\sin\varphi=4$$

which is a very natural result interpreted as the following dot product :

$$\vec{CM}.\vec{N}=4$$

where $M$ is the generic point in the plane, $C=(1,2,3)$ and $\vec{N}$ defined by (1).