I would like to know how I should visualize and intuitively imagine the shapes these vector equations represent:

  1. $ \textbf{a} \cdot \hat{\textbf{b}} = m|\textbf{a}| $

  2. $|\textbf{a} - (\textbf{a} \cdot \hat{\textbf{b}}) \hat{\textbf{b}}| = k$

Shouldn't 1. be a plane that is tilted at some angle? I have completely no idea what 2. represents.

  • $\begingroup$ a is variable and b is a constant vector? $\endgroup$ – velut luna Oct 13 '17 at 23:24

If ${\bf a}$ is the variable vector and ${\bf b}$ is a constant vector, then $${\bf a}\cdot \hat{{\bf b}}=m|{\bf a}| \iff \left({\bf a}={\bf0}\right) \lor \left(\hat{{\bf a}}\cdot \hat{{\bf b}}=m\right)$$ which is a conical surface with axis in the direction of $\hat{{\bf b}}$ and angle $\cos^{-1}m$, assuming $|m| \le 1$.

The second one is a cylindrical surface with axis in the direction of $\hat{\bf b}$ and radius $k$.

For analytical proof, since the equations are in vector form, we can WLOG choosing our coordinate system so that $\hat{\bf b}=\hat{\bf z}$, and let ${\bf a}=(x,y,z)$.

Then for (1):


In cylindrical coordinates $$z=z$$ $$x=\rho \cos\phi$$ $$y=\rho \sin\phi$$ it is $$z^2=m^2(x^2+y^2+z^2)$$ $$(1-m^2)z^2=m^2(x^2+y^2)=m^2(\rho^2\cos^2\phi+\rho^2\sin^2\phi)=m^2\rho^2$$ $$(1-m^2)z^2=m^2\rho^2$$ or $$\rho=\sqrt{\frac{1-m^2}{m^2}}|z|$$ which is clearly a conical surface.

For (2):

$$|(x,y,z)-(0,0,z)|=k$$ $$\sqrt{x^2+y^2}=k$$ or in cylindrical coordinates $$\rho = k$$ which is clearly a cylindrical surface.

  • $\begingroup$ Sorry I forgot to mod the left hand side term.. $\endgroup$ – oldselflearner1959 Oct 14 '17 at 8:50
  • $\begingroup$ @bigvision99 Edited my answer. $\endgroup$ – velut luna Oct 14 '17 at 8:55
  • $\begingroup$ Could you explain how you developed this intuition and visualize it? Is there a proof for these derivations? $\endgroup$ – oldselflearner1959 Oct 14 '17 at 9:11
  • $\begingroup$ I just draw some sketches. I am not surely I can explain the "intuition". Anyway, I've added kind of analytically proof. $\endgroup$ – velut luna Oct 14 '17 at 9:26
  • $\begingroup$ thanks for the proof..I don't get how you changed the equation in cartesian form into cylindrical coordinate form. Could you explain how this works? I got lost at the (1-m^2)z^2=.... step unfortunately. $\endgroup$ – oldselflearner1959 Oct 14 '17 at 11:44

As far as intuition goes, maybe this will help. For any two vectors $\bf a$ and $\bf b$, the dot product $\bf a \cdot \hat b$ is the component of $\bf a$ in the direction of $\bf b$. If you "remove" that component from $\bf a$ you would be left with the part of it which is perpendicular to $\bf b$:

$$ \begin{align} {\bf a}_{\parallel} &= ({\bf a} \cdot \hat {\bf b}) \hat {\bf b} \\ {\bf a}_\perp &= {\bf a} - ({\bf a} \cdot \hat {\bf b}) \hat {\bf b} \end{align} $$

[Note that to do any subtraction, first we need to make that "component" into an actual vector, hence the $\bf \hat b$'s multiplying the dot products in the above.]

Thus we have obtained a decomposition of the original vector $\bf a$ as $$ {\bf a} = {\bf a}_\parallel + {\bf a}_\perp $$

where the two "parts" are vectors that are parallel and perpendicular to $\bf b$, respectively. Now you can see quite easily that your second vector equation is equivalent to $|{\bf a}_\perp| = k$, which is exactly what a cylinder is: a set of points that are a given perpendicular distance from an axis (defined here by the vector $\hat {\bf b}$).

The first one is not that straightforward, but you can see it is equivalent to $|{\bf a}_\parallel| = m|\bf a|$ which, if you try to visualize all possible vectors $\bf a$, tells you that when each of them is "joined" to $\bf b$ (tail-to-tail at the origin), all the triangles formed are similar; that is what a cone is.


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