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If vector $v_{xy}$ is a vector on the $xy$ plane of magnitude $r$, and $v_{yz}$ is a vector on the $yz$ plane also of magnitude $r$, then $v_{xy} + v_{yx}$ results in vector $v$ of magnitude $R$.

How, given any vector $v$ to start with, can you calculate the two vectors $v_{xy}$ and $v_{yx}$ (which both have equal magnitude $r$) that add together to produce $v$ ?

A solution can be expressed in either cartesian or preferably polar coordinates.

(For clarity, I am not seeking to resolve into unit vectors, but specifically find the 2 vectors of equal magnitude from perpendicular planes that add to produce the vector $v$ of magnitude $R$)

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3 Answers 3

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Did you mean to say that $v_{xy} + v_{yz}$ results in a vector $v$ of magnitude $R$? And did you mean to say the vector $v$ is given? (You referred to $R$ as a magnitude, and then as a vector). If the vector $v$ is given, and hence its magnitude $R$ as well, then you can find the vectors $v_{xy}$ and $v_{yz}$ lying in the $xy$-plane and $yz$-plane respectively as follows:

Let $v_{xy} = \left[ \begin{matrix} v_{xy,1} \\ v_{xy,2} \\ 0 \end{matrix} \right] \text{ and } v_{yz} = \left[ \begin{matrix} 0 \\ v_{yz,2} \\ v_{yz,3} \end{matrix} \right]$ . Because these sum to the known vector $v = \left[ \begin{matrix} v_1 \\ v_2 \\ v_3\end{matrix} \right]$, we can immediately see that $v_{xy,1} = v_1$ and $v_{yz,3} = v_3$, so we now know those coordinates. What is left is to find the two coordinates $v_{xy,2}$ and $v_{yz,2}$.

To solve for these, we use the fact that $v_2 = v_{xy,2} + v_{yz,2}$, and because both $v_{xy}$ and $v_{yz}$ have the same magnitude (and hence the same squared magnitude), we can write $r^2 = v_{1}^2 + v_{xy,2}^2 = v_{yz,2}^2 + v_{3}^2$.

To solve this system of two equations, begin by substituting $v_{xy,2} = v_2 - v_{yz, 2}$ into the second equation, so we have \begin{align*} v_{1}^2 + (v_2 - v_{yz, 2})^2 & = v_{yz,2}^2 + v_{3}^2 \\ v_{1}^2 + v_2^2 - 2v_2 v_{yz, 2} + v_{yz, 2}^2 & = v_{yz,2}^2 + v_{3}^2 \\ v_{yz, 2} & = \frac{v_{3}^2 - v_{1}^2 - v_2^2}{- 2v_2} \\ v_{yz, 2} & = \frac{v_{1}^2 + v_2^2 - v_{3}^2}{2v_2} \\ \end{align*} and \begin{align*} v_{xy, 2} & = v_2 + \frac{v_{3}^2 - v_{1}^2 - v_2^2}{2v_2} \\ v_{xy, 2} & = \frac{v_2^2 + v_{3}^2 - v_{1}^2}{2v_2} . \end{align*}

So the two vectors are $ v_{xy} = \left[ \begin{matrix} v_1 \\ \frac{v_2^2 + v_{3}^2 - v_{1}^2}{2v_2} \\ 0 \end{matrix} \right] \text{ and } v_{yz} = \left[ \begin{matrix} 0 \\ \frac{v_{1}^2 + v_2^2 - v_{3}^2}{2v_2} \\ v_3 \end{matrix} \right]$.

Note: I've implicitly assumed that $v_2 \ne 0$. If $v_2 = 0$, then there are many such pairs of vectors, as the image in @David_G._Stork's answer shows.

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  • $\begingroup$ That perfectly answers the question in cartesian. Thanks. Is there any way of doing the same just using polar? $\endgroup$
    – bba13
    Jun 7, 2020 at 17:28
  • $\begingroup$ Polar coordinates are a coordinate system for $\mathbb{R}^2$, but these vectors are in $\mathbb{R}^3$. You could use cylindrical or spherical coordinates for $\mathbb{R}^3$. $\endgroup$ Jun 7, 2020 at 20:58
  • $\begingroup$ Sorry yes I do mean spherical. Is it possible to take a vector in spherical coordinates and find the 2 component vectors as above without reverting to cartesian? $\endgroup$
    – bba13
    Jun 8, 2020 at 6:08
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Impossible.

Add the two vectors of the same color to see each sum gives the black vector.

enter image description here

Now interpret that result.

Same thing in three dimensions:

enter image description here

Or more generally:

enter image description here

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  • $\begingroup$ In your first example, all the vectors are in the same plane. I am specifically looking for two vectors from perpendicular planes. If you look at the attached image in OP there is an example. $\endgroup$
    – bba13
    May 31, 2020 at 9:17
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This is a supplement to erikpekerson’s answer rather than a complete answer in itself.

We can find the unknown $y$-coordinates of the two vectors by noting that they form the sides of a rhombus with the original vector as one diagonal. Thus, the orthogonal projections of $v_{xy}$ and $v_{yz}$ onto $v$ are both $\frac12v$, from which $v_{xy}\cdot v = \frac12v\cdot v = \frac{R^2}2.$ So, for $v_{xy}$ we have $$v_1^2+yv_2 = \frac{v_1^2+v_2^2+v_3^2}2 \\ y = {v_2^2+v_3^2-v_1^2\over 2v_2}$$ and similarly for $v_{yz}$.

For the same reason, the two vectors lie in the plane perpendicular to $v$ and halfway along that vector, i.e., the plane $v_1x+v_2y+v_3z=\frac{R^2}2$. The two vectors are the intersections of the lines $(v_1,s,0)$ and $(0,t,v_3)$ with this plane, which can be computed in various ways, including a direct computation using the Plücker matrices of these lines. For instance, for $v_{xy}$ we would get the homogeneous coordinates $$(v_1,0,0,1)\cdot(v_1,v_2,v_3,-R^2/2)\,(0,1,0,0) - (0,1,0,0)\cdot(v_1,v_2,v_3,-R^2/2)\, (v_1,0,0,1) = (-v_1v_2,v_1^2-R^2/2,0,-v_2),$$ which dehomogenizes to the same vector computed in other ways.

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