[SOLVED]vector equation of a spherical surface

dwsmith

Well-known member
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 $b_i = x_i$?

Klaas van Aarsen

MHB Seeker
Staff member
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:
$(\mathbf x - \frac 12 \mathbf b)^2 \overset{?}{=} (\frac 12 b)^2$​

Last edited:

dwsmith

Well-known member
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?

Klaas van Aarsen

MHB Seeker
Staff member
$(\mathbf{x} -\frac{1}{2}\mathbf{b})^2 = \mathbf{x}\cdot\mathbf{x} - \mathbf{x}\cdot\mathbf{b} + \frac{1}{4}\mathbf{b}\cdot\mathbf{b}$
How do I go from here?
What do you get from $(\mathbf{x} - \mathbf{b})\cdot\mathbf{x} = 0$?