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I want to define the projection of a vector $\mathbf{v} \in \mathbb{R}^3$ onto the line $\mathbf{r} \in \mathbb{R}^3$ in terms of the components of $\mathbf{v} = (v_x, v_y, v_z)$. In 2D, this looks like the following:

enter image description here

2D component-wise vector projection

The magnitude of the component-wise projection of the velocity vector $\mathbf{v}$ onto $\mathbf{r}$ is the radial velocity $v_r$, and (in 2D) can be expressed as the following:

$$ v_r = \text{comp}_{\mathbf{r}} \mathbf{v}_x + \text{comp}_{\mathbf{r}} \mathbf{v}_y = v_x \cos \theta + v_y \sin \theta $$

or in vector form,

$$ v_r = \begin{bmatrix} \cos \theta & \sin \theta \end{bmatrix} \begin{bmatrix} v_x \\ v_y \end{bmatrix} $$

The 3D case becomes a bit more complicated, and this is where I haven't been able to understand the geometry of the component-wise projection. In the 3D case, we have the vector $\mathbf{r}$ expressed in a form of polar coordinates where the azimuth angle $\theta$ is defined in the x-y plane relative to the positive y-axis, and the elevation angle $\phi$ is defined in the w-z plane relative to the positive w-axis, where the direction of the w-axis is defined by the direction of the projection of $\mathbf{r}$ onto the x-y plane.

enter image description here

3D geometry

The 3D vector $\mathbf{v}$ is defined with its origin at the point $(x,y,x)$ and has components $(v_x, v_y, v_z)$. The magnitude of the component-wise projection of $\mathbf{v}$ onto $\mathbf{r}$ will be a function of both the azimuth ($\theta$) and elevation ($\phi$) angles of the form:

$$ v_r = \begin{bmatrix} f_1(\theta, \phi) & f_2(\theta, \phi) & f_3(\theta, \phi) \end{bmatrix} \begin{bmatrix} v_x \\ v_y \\v_z \end{bmatrix} $$

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  • $\begingroup$ I'm sorry, I find the picture hard to follow: What is the componentwise projection in $2D$? $\endgroup$ Apr 24 '19 at 16:39
  • $\begingroup$ In blue are the components of $\mathbf{v}$: $\mathbf{v}_x$ and $\mathbf{v}_y$. And in red are the projections of $\mathbf{v}_x$ and $\mathbf{v}_y$ onto $\mathbf{r}$. The negative signs can be ignored. $\endgroup$
    – Carl S.
    Apr 24 '19 at 16:48
  • $\begingroup$ Though there are two red vectors, their labels are the same: $proj_r(-v_x)$. Could you please clarify which of them is for $x$ and which for $y$? I presume the longer one is for $x$? So the output of the process in 3D should be three vectors along $r$ with magnitudes equal to their projections? $\endgroup$ Apr 24 '19 at 16:51
  • $\begingroup$ Yes, you're correct. I'm sorry for the error in the graphic. I can fix that. The longer red arrow represents the projection of $\mathbf{v}_x$ onto $\mathbf{r}$ (and is labeled correctly). The shorter red arrow represents the projection of $\mathbf{v}_y$ onto $\mathbf{r}$ (and is currently mislabeled). $\endgroup$
    – Carl S.
    Apr 24 '19 at 16:54
  • $\begingroup$ Thank you for the feedback. The 2D projection figure has been updated $\endgroup$
    – Carl S.
    Apr 24 '19 at 17:49
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In $\mathbb R^n$ in any number of dimensions, with the usual distance function, if you have a line through the origin in the direction of a vector $\mathbf r,$ as you have in your first figure, and any other vector $\mathbf v,$ the length of the projected vector you get by projecting $\mathbf v$ onto the line of $\mathbf r$ is just the inner product (aka dot product) $$ v_r = \left(\frac1{\lVert\mathbf r\rVert}\mathbf r\right) \cdot v. $$

In general, the vector $\frac1{\lVert\mathbf r\rVert}\mathbf r$ is simply a unit vector in the same direction as $\mathbf r.$ In two dimensions, with a vector $\mathbf r$ at an angle $\theta$ from the $x$ axis, it happens that the vector on the left side of that inner product is $$ \frac1{\lVert\mathbf r\rVert} \mathbf r= \begin{bmatrix} \cos\theta \\ \sin\theta \end{bmatrix},$$ and therefore the inner product is given by the matrix multiplication you showed.

In your three-dimensional case, you can set $$\frac1{\lVert\mathbf r\rVert} \mathbf r= \begin{bmatrix} f_1(\theta, \phi) \\ f_2(\theta, \phi) \\ f_3(\theta, \phi) \end{bmatrix},$$ that is, simply set $f_1(\theta, \phi),$ $f_2(\theta, \phi),$ and $f_3(\theta, \phi)$ to the three coordinates of the unit vector in the direction $\theta,\phi.$ You can read these coordinates off your diagram. (They are combinations of trigonometric functions of $\theta$ and $\phi$ which you've already written; just don't multiply by $R$.)

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  • $\begingroup$ This doesn't answer the question I posed. Which is to define the magnitude of the projection of $\mathbf{v}$ onto $\mathbf{r}$ in terms of the components of $\mathbf{v}$ ($v_x, v_y, v_z$) and the azimuth and elevation angles, $\theta$ and $\phi$ respectively. $\endgroup$
    – Carl S.
    Apr 24 '19 at 17:53
  • $\begingroup$ Admittedly I did not just write the formula out explicitly for you. But all the pieces of the solution are either already in your question or in this answer. I think it's better if you recognize them for yourself rather than having it written out as a formula without understanding it. $\endgroup$
    – David K
    Apr 24 '19 at 17:55
  • $\begingroup$ Do you know what I meant by "set $f_1(\theta, \phi),$ $f_2(\theta, \phi),$ and $f_3(\theta, \phi)$ to the three coordinates of the unit vector in the direction $\theta,\phi.$"? You drew a vector of length $R$ in that direction, you showed the coordinates of that vector, now imagine we set the length to $1.$ $\endgroup$
    – David K
    Apr 24 '19 at 18:00
  • $\begingroup$ The reason I'm making such a big deal out of finding the coordinates of the unit vector and taking a dot product is that if you're doing a problem like this next week but someone has labeled the angles differently (maybe $\theta$ measured from the $x$ axis and $\phi$ downward from the $z$ axis), the formula with the trig functions will be different, but as long as you can draw a figure like the one you drew in the question you'll be able to write the correct formula. $\endgroup$
    – David K
    Apr 24 '19 at 18:06
  • $\begingroup$ So, given the coordinate frame defined in the 3D image, we would have: $f_1(\theta, \phi) = \cos \phi \sin \theta$, $f_2(\theta, \phi) = \cos \phi \cos \theta$, and $f_3(\theta, \phi) = \sin \phi$ $\endgroup$
    – Carl S.
    Apr 24 '19 at 18:42
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We want to create vectors along the vector $\vec r$ with components being that of the $x, y, z$ projections of the vector $\vec v = (v_x, v_y, v_z)$.

To create a vector in some direction with some magnitude, we multiply the unit vector along a direction with the magnitude we wish for. So, for example, to create the projection of the $x$ value along the direction by $\vec r$, we will:

  1. Create a unit vector $\hat r = \vec r / |r|$, which is the unit vector in the direction along $\vec r$
  2. Make the new vector $\vec r_x = v_x \hat r$, where $v_x$ is the x-component of the vector $v$.

Similarly, for $y$, we can create a vector $\vec r_y = v_y \hat r$. And for $z$, we can create $\vec z = v_z \hat r$.

Since we have the vector $\vec r$ in spherical coordinates, we can write the unit vector along the direction $\vec r$ in cartesian coordinates as:

$\hat r = (\cos \phi \sin \theta, \cos \phi \cos \theta, \sin \phi)$.

So the vectors will be:

  1. $\vec r_x = v_x \hat r = (v_x\cos \phi \sin \theta, v_x \cos \phi \cos \theta,v_x \sin \phi)$
  2. $\vec r_y = v_y \hat r = (v_y\cos \phi \sin \theta, v_y \cos \phi \cos \theta,v_y \sin \phi)$
  3. $\vec r_z = v_z \hat r = (v_z\cos \phi \sin \theta, v_z \cos \phi \cos \theta,v_z \sin \phi)$

If I completely misunderstood the question, please do tell me!

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    $\begingroup$ So are you implying that $\text{proj}_{\mathbf{r}} \mathbf{v} = \vec{r}_x + \vec{r}_y + \vec{r}_z$ ? $\endgroup$
    – Carl S.
    Apr 24 '19 at 17:33
  • $\begingroup$ I mean, in the picture at least, there were two separate vectors for the 2D case, so I presume you want 3 vectors for the 3D case? The three vectors will be $(\vec r_x, \vec r_y, \vec r_z)$. And to follow the notation from the 2D picture, the function $proj_r(v) = \hat r |v|$ $\endgroup$ Apr 24 '19 at 17:35
  • $\begingroup$ No, the goal is to define the magnitude of the projection of $\mathbf{v}$ onto $\mathbf{r}$ in terms of the components of $\mathbf{v}$ ($v_x, v_y, v_z$) and the azimuth and elevation angles, $\theta$ and $\phi$ respectively. $\endgroup$
    – Carl S.
    Apr 24 '19 at 17:55

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