If $\vec r = (x,y,z)$ and $\vec r_1 = (x_1,y_1,z_1)$, describe the set of all points $(x,y,z)$ such that $\left|\vec r-\vec r_1\right| = 1$

Question

If $\vec r = (x,y,z)$ and $\vec r_1 = (x_1,y_1,z_1)$, describe the set of all points $(x,y,z)$ such that $\left|\vec r-\vec r_1\right| = 1$.

Can some one prove the solution for me?

I don't understand why the solution is a a sphere with a radius of one, centered at $(x_1,y_1,z_1)$.

Answer 1

$\Vert r - r_1 \Vert$ = $\Big\Vert [x - x_1; \quad y-y_1; \quad z- z_1] \Big\Vert $ = 1 Using the norm definition, we have, $\sqrt{(x-x_1)^2+(y-y_1)^2+(z- z_1)^2} = 1$ or ${(x-x_1)^2+(y-y_1)^2+(z- z_1)^2} = 1$, which is the equation of a sphere of centre $(x_1,y_1,z_1)$ and radius $1$.

Answer 2

Because a sphere is defined exactly as:

the locus of points that has a constant distance (called the radius) from a fixed point that is called the center

and the norm $||\mathbf{r}-\mathbf{r_1}||$ is just the distance between $\mathbf{r}$ and $\mathbf{r_1}$.

Answer 3

A sphere is the set of points $S$ that are all at the same distance $R$ called the radius from a given point called the center in three-dimensional space Euicledean space $\Bbb R^3$.

This definition is not related to vectors or vector spaces!

Let $r_1=(x_1,y_1,z_1)$ be the center and $R=1$ be the radius. If $r=(x,y,z)\in S$ then $$d(r,r_1)=1.\tag1$$ By definition of Euclidean distance we have $$d(r,r_1)=\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}.\tag2$$ From $(1)$ and $(2)$, after squaring both sides which does no harm, $S$ is described by the equation $$(x-x_1)^2+(y-y_1)^2+(z-z_1)^2=1.$$ $S$ is called a unit sphere since its radius is $1$.

Note: If we see $\Bbb R^3$ as a normed vector space with Euicledean norm, then $d(r,r_1)=\Vert\vec r-\vec r_1\Vert$. Then, OP's equation also describes $S$.