MHB I want to visualise this vector expression

[ognik]

The vector r, starting at the origin, terminates at and specifies the point in space (x, y, z). A surface is swept out by the tip of r if $ (\vec{r} −\vec{a}) · \vec{a} = 0 $. $\vec{a}$ is a constant vector...

1) I tried $ (r - a).a = 0, \therefore r.a = a^2$, but still can't 'see' what shape that might produce. Somewhat uncertainly, I reasoned that this looks like r is proportional to a constant which would give a plane? But r is varying, so that's not quite right ...?
Also, I am used to the dot product being a projection, so r varying, but projected onto a ...would really like some help to think this through (end of course revision)

2) Please also suggest graphical software I could sketch this with - if it can be done with mathematica, how would I do that? I always struggle with mathematica with general expressions instead of values...

Much appreciated

 

[Euge]

Hi ognik,

The equation $(\vec{r}-\vec{a})\cdot \vec{a} = 0$ gives a plane perpendicular to $\vec{a}$ and passing through $a$ (by $a$ I mean $\vec{a}$ viewed as a point in three-space).

If $\vec{r} = \langle x,y,z\rangle$ and $\vec{a} = \langle a_1, a_2, a_3\rangle$, then $\vec{r} - \vec{a} = \langle x - a_1, y - a_2, z - a_3\rangle$. So then $$0 = (\vec{r} - \vec{a})\cdot \vec{a} = (x - a_1)a_1 + (y - a_2)a_2 + (z - a_3)a_3,$$
or $$a_1 x + a_2 y + a_3 z = d,$$ where $$d = a_1^2 + a_2^2 + a_3^2.$$ Perhaps in this form you can tell more easily that this is an equation for a plane.

With a TI 89 Calculator you could graph the equation for given values of $a_1, a_2, a_3$.

 

[ognik]

nice explanation, thanks Euge. I don't have a graphical calculator, is there any online software that might do this?

 

[I like Serena]

Wolfram|Alpha can plot it.

 

[Euge]

Hi ognik,

You can use a 3D grapher here, or on the Google search bar, write something like "plot z = y - x" and enter -- you will get an animated graph of the function.

 

[ognik]

Thanks guys, I was hoping to plot both the constant vector a and the relative plane together. I am still learning Wolfram - I see we can plot several function with Plot3D

I assume (tell me if wrong) I can only do this for known coefficients, not in the general form, so choosing an arbitrary \vec{a} = {1,2,3} I set a_1,_2_3 and d , rewrote the plane eqtn in terms of z and plotted the plane with

Plot3D[{Subscript[a, 1]/Subscript[a, 3] x + Subscript[a, 2]/Subscript[a, 3] y - d}, {x, -10, 10}, {y, -10, 10}] ... which works.

Now I want to add the vector a into the above, need some help pls - I tried using [Line[{0, 0, 0, Null}, {Subscript[a, 1], Subscript[a, 2], Subscript[a, 3]}] - but can't figure out how to combine it into Plot3D - keeps on asking for more arguments, but its just a line segment with no variables?

 

[ognik]

Just to close this, Gianluca Gorni did this really cool solution:

Manipulate[With[{a = {a1, a2, a3}, r = {x, y, z}},
Show[ContourPlot3D[(r - a).a == 0,
{x, -2, 2}, {y, -2, 2}, {z, -2, 2}],
Graphics3D[{PointSize[Large], Arrow[{{0, 0, 0}, a}], Point[a]}]]],
{a1, -1, 1}, {a2, -1, 1}, {a3, -1, 1}]