Vector equation of a spherical surface

[Dustinsfl]

Let the position vector of an arbitrary point $P(x_1x_2x_3)$ be $\mathbf{x} =x_i\hat{\mathbf{e}}_i$, and let $\mathbf{b} = b_i\hat{\mathbf{e}}_i$ be a constant vector.
Show that $(\mathbf{x} - \mathbf{b})\cdot\mathbf{x} = 0$ is the vector equation of a spherical surface having its center at $\mathbf{x} = \frac{1}{2}\mathbf{b}$ with radius of $\frac{1}{2}b$.
\begin{alignat}{3}
(x_i\hat{\mathbf{e}}_i - b_i\hat{\mathbf{e}}_i)\cdot x_i\hat{\mathbf{e}}_i & = & x_i^2-b_ix_i
\end{alignat}
How am I supposed to obtain that $b_i = x_i$?

 

[I like Serena]

Let the position vector of an arbitrary point $P(x_1x_2x_3)$ be $\mathbf{x} =x_i\hat{\mathbf{e}}_i$, and let $\mathbf{b} = b_i\hat{\mathbf{e}}_i$ be a constant vector.
Show that $(\mathbf{x} - \mathbf{b})\cdot\mathbf{x} = 0$ is the vector equation of a spherical surface having its center at $\mathbf{x} = \frac{1}{2}\mathbf{b}$ with radius of $\frac{1}{2}b$.
\begin{alignat}{3}
(x_i\hat{\mathbf{e}}_i - b_i\hat{\mathbf{e}}_i)\cdot x_i\hat{\mathbf{e}}_i & = & x_i^2-b_ix_i
\end{alignat}
How am I supposed to obtain that $b_i = x_i$?

You're not. That would mean $\mathbf b = \mathbf x$, but that is not what you need to prove.

You need to find the distance of $\mathbf x$ to $\frac 12 \mathbf b$ and proof that it is $\frac 12 b$.
That is, can you prove:

$(\mathbf x - \frac 12 \mathbf b)^2 \overset{?}{=} (\frac 12 b)^2$​

 

[Dustinsfl]

You're not. That would mean $\mathbf b = \mathbf x$, but that is not what you need to prove.

You need to find the distance of $\mathbf x$ to $\frac 12 \mathbf b$ and proof that it is $\frac 12 b$.
That is, can you prove:

$(\mathbf x - \frac 12 \mathbf b)^2 \overset{?}{=} (\frac 12 b)^2$​

$(\mathbf{x} -\frac{1}{2}\mathbf{b})^2 = \mathbf{x}\cdot\mathbf{x} - \mathbf{x}\cdot\mathbf{b} + \frac{1}{4}\mathbf{b}\cdot\mathbf{b}$
How do I go from here?

 

[I like Serena]

$(\mathbf{x} -\frac{1}{2}\mathbf{b})^2 = \mathbf{x}\cdot\mathbf{x} - \mathbf{x}\cdot\mathbf{b} + \frac{1}{4}\mathbf{b}\cdot\mathbf{b}$
How do I go from here?

What do you get from $(\mathbf{x} - \mathbf{b})\cdot\mathbf{x} = 0$?

 

[CellCoree]

To obtain $b_i = x_i$, we can use the fact that the center of a spherical surface is defined as the midpoint between any two points on the surface. In this case, we can choose two points on the surface, $\mathbf{x}$ and $\mathbf{b}$. Since the center is at $\frac{1}{2}\mathbf{b}$, we can set the position vector of this point as $\mathbf{c} = \frac{1}{2}\mathbf{b} = \frac{1}{2}b_i\hat{\mathbf{e}}_i$. We can then use the distance formula to find the distance between $\mathbf{x}$ and $\mathbf{c}$, which is equal to the radius of the spherical surface.

\begin{align}
r &= \sqrt{(x_1-\frac{1}{2}b_1)^2 + (x_2-\frac{1}{2}b_2)^2 + (x_3-\frac{1}{2}b_3)^2} \\
&= \sqrt{(x_i-\frac{1}{2}b_i)(x_i-\frac{1}{2}b_i)} \\
&= \sqrt{(x_i-\frac{1}{2}b_i)^2} \\
&= \frac{1}{2}b_i
\end{align}

Since we know that the radius of the spherical surface is $\frac{1}{2}b$, we can equate this to the value we just found and solve for $b_i$.

\begin{align}
\frac{1}{2}b_i &= \frac{1}{2}b \\
b_i &= x_i
\end{align}

Therefore, we have shown that $b_i = x_i$, which satisfies the condition for the center of the spherical surface. This also means that the vector equation $(\mathbf{x} - \mathbf{b})\cdot\mathbf{x} = 0$ is equivalent to the equation of a spherical surface with its center at $\frac{1}{2}\mathbf{b}$ and radius of $\frac{1}{2}b$.