# Component vectors from a 3D vector

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 $$R$$)

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.

• That perfectly answers the question in cartesian. Thanks. Is there any way of doing the same just using polar? Jun 7, 2020 at 17:28
• 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 $\mathbb{R}^3$. Jun 7, 2020 at 20:58
• 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? Jun 8, 2020 at 6:08

Impossible.

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

Now interpret that result.

Same thing in three dimensions:

Or more generally:

• 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. May 31, 2020 at 9:17

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.