MHB Check another visualise vector expression

[ognik]

$ \left( \vec{r} - \vec{a} \right). \vec{r} = 0 $ Please check this:

Then, $ r^2 - a.r = 0 $, then $ \left( r - \frac{a}{2} \right)^2 - \left(\frac{a}{2}\right)^2 = 0 $, then
$ \left( x - \frac{a}{2}\right)^2 + \left(y - \frac{a}{2}\right)^2 + \left( z - \frac{a}{2}\right)^2 = \left(\frac{a}{2}\right)^2$

This is a circle radius $ \frac{a}{2} $, centred at $\left(\frac{a}{2} , \frac{a}{2} , \frac{a}{2} \right) $?

 

[topsquark]

$ \left( \vec{r} - \vec{a} \right). \vec{r} = 0 $ Please check this:

Then, $ r^2 - a.r = 0 $, then $ \left( r - \frac{a}{2} \right)^2 - \left(\frac{a}{2}\right)^2 = 0 $, then
$ \left( x - \frac{a}{2}\right)^2 + \left(y - \frac{a}{2}\right)^2 + \left( z - \frac{a}{2}\right)^2 = \left(\frac{a}{2}\right)^2$

This is a circle radius $ \frac{a}{2} $, centred at $\left(\frac{a}{2} , \frac{a}{2} , \frac{a}{2} \right) $?

Close. You have forgotten that a is a vector so [math]r^2 - \vec{a} \cdot \vec{r} = 0 \implies (x^2 + y^2 + z^2) - (a_x \cdot x + a_y \cdot y + a_z \cdot z) = 0[/math], which says
[math]\left ( x - \frac{a_x}{2} \right ) ^2 + \left ( y - \frac{a_y}{2} \right ) ^2 + \left ( z - \frac{a_z}{2} \right ) ^2 = \frac{a^2}{4}[/math]

This is a circle with center [math]\left ( \frac{a_x}{2},~ \frac{a_y}{2},~ \frac{a_z}{2} \right )[/math] and radius [math]| \vec{a} \cdot \vec{a}|/2[/math].

-Dan

 

[ognik]

Thanks, needed that reminder.

Radius $ \frac{ |a ⃗ ⋅a ⃗ |}{2} $ or $ \frac{\sqrt{ |a ⃗ ⋅a ⃗ |}}{2} $?

 

[topsquark]

Thanks, needed that reminder.

Radius $ \frac{ |a ⃗ ⋅a ⃗ |}{2} $ or $ \frac{\sqrt{ |a ⃗ ⋅a ⃗ |}}{2} $?

[math]| \vec{a} \cdot \vec{b} | \equiv \sqrt{a_x b_x + a_y b_y +a_z b_z}[/math] so the "square root" of [math]\frac{a^2}{4}[/math] is [math] \frac{ | \vec{a} \cdot \vec{a} |}{2}[/math].

-Dan

 

[Deveno]

Please stop calling $\vec{r}\cdot \vec{r}$ by $r^2$. The dot product is not a binary operation on vectors, it's a bilinear functional. It's important (except in the case where one is talking about an extension field) to distinguish between vectors and scalars.

Also, it's not a circle. Circles live in planes, the word to properly describe this shape is a *sphere*.

It has radius $\sqrt{\dfrac{\vec{a}\cdot\vec{a}}{4}} = \dfrac{\sqrt{\vec{a}\cdot\vec{a}}}{2} = \dfrac{\|\vec{a}\|}{2}$.

 

[topsquark]

Please stop calling $\vec{r}\cdot \vec{r}$ by $r^2$. The dot product is not a binary operation on vectors, it's a bilinear functional. It's important (except in the case where one is talking about an extension field) to distinguish between vectors and scalars.

You are right and with all due apologies! My Physics background is coming into play here.

-Dan

 

[ognik]

Thanks, all useful