Three-Dimensional Givens Rotations

By Joshua Burkholder July 14, 2019 Givens Rotation Matrices, Mathematics No Comments

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Remembering that $\cos (θ + \frac{π}{2}) = - \sin (θ)$ and $\sin (θ + \frac{π}{2}) = \cos (θ)$ , the following matrix is the three-dimensional Givens Rotation from the $y$ -axis to the $z$ -axis (rotation counter-clockwise around the $x$ -axis):

$\begin{matrix} (\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & \cos (θ_{y \to z} + 90^{\circ}) \\ 0 & \sin (θ_{y \to z}) & \sin (θ_{y \to z} + 90^{\circ}) \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & \cos (θ_{y \to z} + \frac{π}{2}) \\ 0 & \sin (θ_{y \to z}) & \sin (θ_{y \to z} + \frac{π}{2}) \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & - \sin (θ_{y \to z}) \\ 0 & \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix}) \end{matrix}$

This matrix rotates three-dimensional column vectors about the origin of the $y \to z$ plane:

$y \to z$	$x$	$y$	$z$
$x$	$1$	$0$	$0$
$y$	$0$	$\cos (θ_{y \to z})$	$- \sin (θ_{y \to z})$
$z$	$0$	$\sin (θ_{y \to z})$	$\cos (θ_{y \to z})$

Similarly, the following is the three-dimensional Givens Rotation from the $x$ -axis to the $z$ -axis (rotation clockwise around the $y$ -axis):

$\begin{matrix} (\begin{matrix} \cos (θ_{x \to z}) & 0 & \cos (θ_{x \to z} + 90^{\circ}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \sin (θ_{x \to z} + 90^{\circ}) \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to z}) & 0 & \cos (θ_{x \to z} + \frac{π}{2}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \sin (θ_{x \to z} + \frac{π}{2}) \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to z}) & 0 & - \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix}) \end{matrix}$

This matrix rotates three-dimensional column vectors about the origin of the $x \to z$ plane:

$x \to z$	$x$	$y$	$z$
$x$	$\cos (θ_{x \to z})$	$0$	$- \sin (θ_{x \to z})$
$y$	$0$	$1$	$0$
$z$	$\sin (θ_{x \to z})$	$0$	$\cos (θ_{x \to z})$

Notice that this is not the same as the traditional rotation matrix where we rotate around the $y$ axis using the right-hand rule (i.e. where we rotate in the opposite direction about the origin of the $x \to z$ plane). In other words, given $θ = - ϕ$ and remembering that $\cos (- ϕ) = \cos (ϕ)$ and $\sin (- ϕ) = - \sin (ϕ)$ , we have the following:

$\begin{matrix} (\begin{matrix} \cos (θ) & 0 & \cos (θ + 90^{\circ}) \\ 0 & 1 & 0 \\ \sin (θ) & 0 & \sin (θ + 90^{\circ}) \end{matrix}) \\ (\begin{matrix} \cos ((- ϕ)) & 0 & \cos ((- ϕ) + 90^{\circ}) \\ 0 & 1 & 0 \\ \sin ((- ϕ)) & 0 & \sin ((- ϕ) + 90^{\circ}) \end{matrix}) \\ (\begin{matrix} \cos (- ϕ) & 0 & \sin (- ϕ + \frac{π}{2}) \\ 0 & 1 & 0 \\ \sin (- ϕ) & 0 & \sin (- ϕ + \frac{π}{2}) \end{matrix}) \\ (\begin{matrix} \cos (- ϕ) & 0 & - \sin (- ϕ) \\ 0 & 1 & 0 \\ \sin (- ϕ) & 0 & \cos (- ϕ) \end{matrix}) \\ (\begin{matrix} \cos (ϕ) & 0 & \sin (ϕ) \\ 0 & 1 & 0 \\ - \sin (ϕ) & 0 & \cos (ϕ) \end{matrix}) \end{matrix}$

Lastly, the following is the three-dimensional Givens Rotation from the $x$ -axis to the $y$ -axis (rotation counter-clockwise around the $z$ -axis):

$\begin{matrix} (\begin{matrix} \cos (θ_{x \to y}) & \cos (θ_{x \to y} + 90^{\circ}) & 0 \\ \sin (θ_{x \to y}) & \sin (θ_{x \to y} + 90^{\circ}) & 0 \\ 0 & 0 & 1 \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to y}) & \cos (θ_{x \to y} + \frac{π}{2}) & 0 \\ \sin (θ_{x \to y}) & \sin (θ_{x \to y} + \frac{π}{2}) & 0 \\ 0 & 0 & 1 \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to y}) & - \sin (θ_{x \to y}) & 0 \\ \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix}) \end{matrix}$

This matrix rotates three-dimensional column vectors about the origin of the $x \to y$ plane:

$x \to y$	$x$	$y$	$z$
$x$	$\cos (θ_{x \to y})$	$- \sin (θ_{x \to y})$	$0$
$y$	$\sin (θ_{x \to y})$	$\cos (θ_{x \to y})$	$0$
$z$	$0$	$0$	$1$

Observe that given $n$ dimensions, we have the following number of distinct Givens Rotation matrices:

$\begin{matrix} (n - 1) + (n - 2) + \dots + 1 \\ \sum_{k = 1}^{n - 1} k \\ \frac{(n - 1) ((n - 1) + 1)}{2} \\ \frac{(n - 1) (n - 1 + 1)}{2} \\ \frac{(n - 1) (n)}{2} \\ \frac{n (n - 1)}{2} \end{matrix}$

For instance, when we’re in three dimensions, we have $n = 3$ and $\frac{n (n - 1)}{2} = \frac{(3) ((3) - 1)}{2} = \frac{3 (2)}{2} = 3$ distinct Givens Rotation matrices:

$y \to z$	$x$	$y$	$z$
$x$	$1$	$0$	$0$
$y$	$0$	$\cos (θ_{y \to z})$	$- \sin (θ_{y \to z})$
$z$	$0$	$\sin (θ_{y \to z})$	$\cos (θ_{y \to z})$

$x \to z$	$x$	$y$	$z$
$x$	$\cos (θ_{x \to z})$	$0$	$- \sin (θ_{x \to z})$
$y$	$0$	$1$	$0$
$z$	$\sin (θ_{x \to z})$	$0$	$\cos (θ_{x \to z})$

and

$x \to y$	$x$	$y$	$z$
$x$	$\cos (θ_{x \to y})$	$- \sin (θ_{x \to y})$	$0$
$y$	$\sin (θ_{x \to y})$	$\cos (θ_{x \to y})$	$0$
$z$	$0$	$0$	$1$

Moreover, notice the order in which we’re considering these distinct Givens Rotation matrices:

	$x$	$y$	$z$
$x$		$x \to y$	$x \to z$
$y$			$y \to z$
$z$

	$x$	$y$	$z$
$x$		$3$	$2$
$y$			$1$
$z$

We’ll leverage this order in a moment. Before that, let’s see what happens when we rotate the unit column vector $(1, 0, 0)$ using each of these Givens Rotations:

$\begin{matrix} (\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & \cos (θ_{y \to z} + 90^{\circ}) \\ 0 & \sin (θ_{y \to z}) & \sin (θ_{y \to z} + 90^{\circ}) \end{matrix}) (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & \cos (θ_{y \to z} + \frac{π}{2}) \\ 0 & \sin (θ_{y \to z}) & \sin (θ_{y \to z} + \frac{π}{2}) \end{matrix}) (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & - \sin (θ_{y \to z}) \\ 0 & \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix}) (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \\ (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \end{matrix}$ ,	$\begin{matrix} (\begin{matrix} \cos (θ_{x \to z}) & 0 & \cos (θ_{x \to z} + 90^{\circ}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \sin (θ_{x \to z} + 90^{\circ}) \end{matrix}) (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to z}) & 0 & \cos (θ_{x \to z} + \frac{π}{2}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \sin (θ_{x \to z} + \frac{π}{2}) \end{matrix}) (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to z}) & 0 & - \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix}) (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to z}) \\ 0 \\ \sin (θ_{x \to z}) \end{matrix}) \end{matrix}$ ,
and
$\begin{matrix} (\begin{matrix} \cos (θ_{x \to y}) & \cos (θ_{x \to y} + 90^{\circ}) & 0 \\ \sin (θ_{x \to y}) & \sin (θ_{x \to y} + 90^{\circ}) & 0 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to y}) & \cos (θ_{x \to y} + \frac{π}{2}) & 0 \\ \sin (θ_{x \to y}) & \sin (θ_{x \to y} + \frac{π}{2}) & 0 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to y}) & - \sin (θ_{x \to y}) & 0 \\ \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to y}) \\ \sin (θ_{x \to y}) \\ 0 \end{matrix}) \end{matrix}$ .

As expected, this simply extracts the first column vector from each Givens Rotation matrix. Because we’re extracting the first column vector, we’ll only see $θ$ in those vectors … we’ll never see $θ + 90^{\circ}$ ; so for the Givens Rotation matrices that modify $(1, 0, 0)$ , $\cos (θ)$ will always be in the $x$ position and $\sin (θ)$ will walk its way up the remaining dimensions based on the order established above. For the Givens Rotation matrices that do not modify $(1, 0, 0)$ (it’s just that first matrix in three dimensions), those rotations do not impact the $x$ position. Looking back at our established order:

	$x$	$y$	$z$
$x$		$3$	$2$
$y$			$1$
$z$

we see that the non- $x$ rows cannot impact the $x$ position. In this three-dimensional case, we see that the Givens Rotation matrix associated with the $y$ row cannot modify the $x$ position. To make this a little clearer, let’s rotate the arbitrarily positioned column vector $(x, y, z)$ :

$\begin{matrix} (\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & \cos (θ_{y \to z} + 90^{\circ}) \\ 0 & \sin (θ_{y \to z}) & \sin (θ_{y \to z} + 90^{\circ}) \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & \cos (θ_{y \to z} + \frac{π}{2}) \\ 0 & \sin (θ_{y \to z}) & \sin (θ_{y \to z} + \frac{π}{2}) \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & - \sin (θ_{y \to z}) \\ 0 & \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) \\ (\begin{matrix} x \\ y \cos (θ_{y \to z}) - z \sin (θ_{y \to z}) \\ y \sin (θ_{y \to z}) + z \cos (θ_{y \to z}) \end{matrix}) \end{matrix}$ ,	$\begin{matrix} (\begin{matrix} \cos (θ_{x \to z}) & 0 & \cos (θ_{x \to z} + 90^{\circ}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \sin (θ_{x \to z} + 90^{\circ}) \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to z}) & 0 & \cos (θ_{x \to z} + \frac{π}{2}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \sin (θ_{x \to z} + \frac{π}{2}) \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to z}) & 0 & - \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) \\ (\begin{matrix} x \cos (θ_{x \to z}) - z \sin (θ_{x \to z}) \\ y \\ x \sin (θ_{x \to z}) + z \cos (θ_{x \to z}) \end{matrix}) \end{matrix}$ ,
and
$\begin{matrix} (\begin{matrix} \cos (θ_{x \to y}) & \cos (θ_{x \to y} + 90^{\circ}) & 0 \\ \sin (θ_{x \to y}) & \sin (θ_{x \to y} + 90^{\circ}) & 0 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to y}) & \cos (θ_{x \to y} + \frac{π}{2}) & 0 \\ \sin (θ_{x \to y}) & \sin (θ_{x \to y} + \frac{π}{2}) & 0 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to y}) & - \sin (θ_{x \to y}) & 0 \\ \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) \\ (\begin{matrix} x \cos (θ_{x \to y}) - y \sin (θ_{x \to y}) \\ x \sin (θ_{x \to y}) + y \cos (θ_{x \to y}) \\ z \end{matrix}) \end{matrix}$ .

We immediately see that rotating $(x, y, z)$ about the $y \to z$ plane (multiplying by the first Givens Rotation matrix) does not impact the $x$ position … and only impacts the $y$ and $z$ positions if $y \neq 0$ or $z \neq 0$ … which only happens when we have a column vector off of the $x$ -axis. For the column vector $(r, 0, 0)$ on the $x$ -axis, rotating about the $y \to z$ plane (multiplying by that first Givens Rotation matrix) has no effect … we’ll still have $(r, 0, 0)$ after that multiplication.

This brings up the question: given the column vector $(r, 0, 0)$ where $r = \sqrt{x^{2} + y^{2} + z^{2}}$ and $r \geq 0$ , can all column vectors $(x, y, z)$ be recovered using just these Givens Rotation matrices?

First, we’ll start by using all the Givens Rotation matrices in our established order … and notice that the first Givens Rotation matrix doesn’t really help out that much:

$\begin{array}{r} \overset{3}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to y}) & - \sin (θ_{x \to y}) & 0 \\ \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}} \overset{2}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to z}) & 0 & - \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix})}} \overset{1}{\overset{︷}{(\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & - \sin (θ_{y \to z}) \\ 0 & \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix})}} (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) & = & (\begin{matrix} x \\ y \\ z \end{matrix}) \\ \overset{3}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to y}) & - \sin (θ_{x \to y}) & 0 \\ \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}} \overset{2}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to z}) & 0 & - \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix})}} (\overset{1}{\overset{︷}{(\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & - \sin (θ_{y \to z}) \\ 0 & \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix})}} (\begin{matrix} r \\ 0 \\ 0 \end{matrix})) & = & (\begin{matrix} x \\ y \\ z \end{matrix}) \\ \overset{3}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to y}) & - \sin (θ_{x \to y}) & 0 \\ \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}} \overset{2}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to z}) & 0 & - \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix})}} (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) & = & (\begin{matrix} x \\ y \\ z \end{matrix}) \end{array}$

Then, we have:

For $r = \sqrt{x^{2} + y^{2} + z^{2}}$ , $r = 0$ implies $x = 0$ , $y = 0$ , and $z = 0$ , so given $r = 0$ , we have:

$\begin{array}{r} (\begin{matrix} r \cos (θ_{x \to y}) \cos (θ_{x \to z}) \\ r \sin (θ_{x \to y}) \cos (θ_{x \to z}) \\ r \sin (θ_{x \to z}) \end{matrix}) & = & (\begin{matrix} x \\ y \\ z \end{matrix}) \\ (\begin{matrix} (0) \cos (θ_{x \to y}) \cos (θ_{x \to z}) \\ (0) \sin (θ_{x \to y}) \cos (θ_{x \to z}) \\ (0) \sin (θ_{x \to z}) \end{matrix}) & = & (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) \\ (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) & = & (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) \end{array}$

So, we can recover the column vector $(0, 0, 0)$ .

Given $r = \sqrt{x^{2} + y^{2} + z^{2}}$ with $r > 0$ and $- 90^{\circ} \leq θ_{x \to z} \leq 90^{\circ}$ , we have:

$\begin{matrix} r \sin (θ_{x \to z}) & = & z \\ (\sqrt{x^{2} + y^{2} + z^{2}}) \sin (θ_{x \to z}) & = & z \\ \sin (θ_{x \to z}) & = & \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} \\ θ_{x \to z} & = & \arcsin (\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}}) \end{matrix}$

Moreover, since $\arcsin (w) = \arctan (\frac{w}{\sqrt{1 - w^{2}}})$ , we also have:

$\begin{array}{l} θ_{x \to z} & = & \arcsin (\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}}) \\ = & \arctan (\frac{(\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}})}{\sqrt{1 - {(\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}})}^{2}}}) \\ = & \arctan (\frac{\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}}}{\sqrt{\frac{x^{2} + y^{2} + z^{2}}{x^{2} + y^{2} + z^{2}} - \frac{z^{2}}{x^{2} + y^{2} + z^{2}}}}) \\ = & \arctan (\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}} \sqrt{\frac{x^{2} + y^{2} + z^{2} - z^{2}}{x^{2} + y^{2} + z^{2}}}}) \\ = & \arctan (\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}} \frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2} + z^{2}}}}) \\ = & \arctan (\frac{z}{\sqrt{x^{2} + y^{2}}}) \end{array}$

Remember: $\arctan (\frac{y}{x})$ is atan2( y, x ) in C/C++ and ArcTan[ x, y ] in Mathematica.

Given $r = \sqrt{x^{2} + y^{2} + z^{2}}$ with $r > 0$ , $- 90^{\circ} \leq θ_{x \to z} \leq 90^{\circ}$ , and remembering that $\cos (\arctan (w)) = \frac{1}{\sqrt{1 + w^{2}}}$ and $\sin (\arctan (w)) = \frac{w}{\sqrt{1 + w^{2}}}$ , if $θ_{x \to z} = \arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})$ , we have:

$\begin{matrix} \cos (θ_{x \to z}) \\ \cos (\arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})) \\ \frac{1}{\sqrt{1 + {(\frac{z}{\sqrt{x^{2} + y^{2}}})}^{2}}} \\ \frac{1}{\sqrt{\frac{x^{2} + y^{2}}{x^{2} + y^{2}} + \frac{z^{2}}{x^{2} + y^{2}}}} \\ \frac{1}{(\frac{\sqrt{x^{2} + y^{2} + z^{2}}}{\sqrt{x^{2} + y^{2}}})} \\ \frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2} + z^{2}}} \end{matrix}$

and

$\begin{matrix} \sin (θ_{x \to z}) \\ \sin (\arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})) \\ \frac{(\frac{z}{\sqrt{x^{2} + y^{2}}})}{\sqrt{1 + {(\frac{z}{\sqrt{x^{2} + y^{2}}})}^{2}}} \\ \frac{\frac{z}{\sqrt{x^{2} + y^{2}}}}{\sqrt{\frac{x^{2} + y^{2}}{x^{2} + y^{2}} + \frac{z^{2}}{x^{2} + y^{2}}}} \\ \frac{z}{\sqrt{x^{2} + y^{2}} \frac{\sqrt{x^{2} + y^{2} + z^{2}}}{\sqrt{x^{2} + y^{2}}}} \\ \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} \end{matrix}$ .

Given $r = \sqrt{x^{2} + y^{2} + z^{2}}$ with $r > 0$ and $- 180^{\circ} \leq θ_{x \to y} \leq 180^{\circ}$ , since $\cos (\arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})) = \frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2} + z^{2}}}$ , if $θ_{x \to z} = \arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})$ , we have:

$\begin{matrix} r \sin (θ_{x \to y}) \cos (θ_{x \to z}) & = & y \\ r \sin (θ_{x \to y}) \cos (\arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})) & = & y \\ (\sqrt{x^{2} + y^{2} + z^{2}}) \sin (θ_{x \to y}) (\frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2} + z^{2}}}) & = & y \\ \sqrt{x^{2} + y^{2} + z^{2}} \sin (θ_{x \to y}) \frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2} + z^{2}}} & = & y \\ \sin (θ_{x \to y}) & = & \frac{y}{\sqrt{x^{2} + y^{2}}} \\ θ_{x \to y} & = & \arcsin (\frac{y}{\sqrt{x^{2} + y^{2}}}) \end{matrix}$

Moreover, since $\arcsin (w) = \arctan (\frac{w}{\sqrt{1 - w^{2}}})$ , we also have:

$\begin{array}{l} θ_{x \to y} & = & \arcsin (\frac{y}{\sqrt{x^{2} + y^{2}}}) \\ = & \arctan (\frac{(\frac{y}{\sqrt{x^{2} + y^{2}}})}{\sqrt{1 - {(\frac{y}{\sqrt{x^{2} + y^{2}}})}^{2}}}) \\ = & \arctan (\frac{y}{\sqrt{x^{2} + y^{2}} \sqrt{\frac{x^{2} + y^{2}}{x^{2} + y^{2}} - \frac{y^{2}}{x^{2} + y^{2}}}}) \\ = & \arctan (\frac{y}{\sqrt{x^{2} + y^{2}} \sqrt{\frac{x^{2} + y^{2} - y^{2}}{x^{2} + y^{2}}}}) \\ = & \arctan (\frac{y}{\sqrt{x^{2} + y^{2}} \frac{\sqrt{x^{2}}}{\sqrt{x^{2} + y^{2}}}}) \\ = & \arctan (\frac{y}{\sqrt{x^{2}}}) \\ = & \arctan (\frac{y}{x}) \end{array}$

Given $r = \sqrt{x^{2} + y^{2} + z^{2}}$ with $r > 0$ , $- 180^{\circ} \leq θ_{x \to y} \leq 180^{\circ}$ , and remembering that $\cos (\arctan (w)) = \frac{1}{\sqrt{1 + w^{2}}}$ and $\sin (\arctan (w)) = \frac{w}{\sqrt{1 + w^{2}}}$ , we have:

$\begin{matrix} \cos (θ_{x \to y}) \\ \cos (\arctan (\frac{y}{x})) \\ \frac{1}{\sqrt{1 + {(\frac{y}{x})}^{2}}} \\ \frac{1}{\sqrt{\frac{x^{2}}{x^{2}} + \frac{y^{2}}{x^{2}}}} \\ \frac{1}{(\frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2}}})} \\ \frac{\sqrt{x^{2}}}{\sqrt{x^{2} + y^{2}}} \\ \frac{x}{\sqrt{x^{2} + y^{2}}} \end{matrix}$

and

$\begin{matrix} \sin (θ_{x \to y}) \\ \sin (\arctan (\frac{y}{x})) \\ \frac{(\frac{y}{x})}{\sqrt{1 + {(\frac{y}{x})}^{2}}} \\ \frac{\frac{y}{x}}{\sqrt{\frac{x^{2}}{x^{2}} + \frac{y^{2}}{x^{2}}}} \\ \frac{y}{x \frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2}}}} \\ \frac{y}{x \frac{\sqrt{x^{2} + y^{2}}}{x}} \\ \frac{y}{\sqrt{x^{2} + y^{2}}} \end{matrix}$

Given $r = \sqrt{x^{2} + y^{2} + z^{2}}$ with $r > 0$ , $- 180^{\circ} \leq θ_{x \to y} \leq 180^{\circ}$ , and $- 90^{\circ} \leq θ_{x \to z} \leq 90^{\circ}$ , since $\cos (\arctan (\frac{y}{x})) = \frac{x}{\sqrt{x^{2} + y^{2}}}$ and $\cos (\arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})) = \frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2} + z^{2}}}$ , if $θ_{x \to y} = \arctan (\frac{y}{x})$ and $θ_{x \to z} = \arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})$ , we have:

$\begin{matrix} r \cos (θ_{x \to y}) \cos (θ_{x \to z}) \\ r \cos (\arctan (\frac{y}{x})) \cos (\arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})) \\ (\sqrt{x^{2} + y^{2} + z^{2}}) (\frac{x}{\sqrt{x^{2} + y^{2}}}) (\frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2} + z^{2}}}) \\ \sqrt{x^{2} + y^{2} + z^{2}} \frac{x}{\sqrt{x^{2} + y^{2}}} \frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2} + z^{2}}} \\ x \end{matrix}$

So, we have $r \cos (θ_{x \to y}) \cos (θ_{x \to z}) = x$ .

Obviously, since $\sin (\arctan (\frac{y}{x})) = \frac{y}{\sqrt{x^{2} + y^{2}}}$ , $\cos (\arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})) = \frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2} + z^{2}}}$ , and $\sin (\arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})) = \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}}$ , if $θ_{x \to y} = \arctan (\frac{y}{x})$ and $θ_{x \to z} = \arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})$ , we also have:

$\begin{matrix} r \sin (θ_{x \to y}) \cos (θ_{x \to z}) \\ r \sin (\arctan (\frac{y}{x})) \cos (\arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})) \\ (\sqrt{x^{2} + y^{2} + z^{2}}) (\frac{y}{\sqrt{x^{2} + y^{2}}}) (\frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2} + z^{2}}}) \\ \sqrt{x^{2} + y^{2} + z^{2}} \frac{y}{\sqrt{x^{2} + y^{2}}} \frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2} + z^{2}}} \\ y \end{matrix}$

and

$\begin{matrix} r \sin (θ_{x \to z}) \\ r \sin (\arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})) \\ (\sqrt{x^{2} + y^{2} + z^{2}}) (\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}}) \\ \sqrt{x^{2} + y^{2} + z^{2}} \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} \\ z \end{matrix}$ .

So, we also have $r \sin (θ_{x \to y}) \cos (θ_{x \to z}) = y$ and $r \sin (θ_{x \to z}) = z$ such that:

Therefore, all column vectors $(x, y, z)$ can be recovered by applying the Givens Rotation matrices associated with the $x$ row to the column vector $(r, 0, 0)$ where $r = \sqrt{x^{2} + y^{2} + z^{2}}$ and $r \geq 0$ :

	$x$	$y$	$z$
$x$		$3$	$2$
$y$			$1$
$z$

Notice that the $(r, θ_{x \to y}, θ_{x \to z})$ coordinate system implied here is not the same as the traditional spherical coordinate system $(ρ, θ, ϕ)$ . In the traditional spherical coordinate system, we have:

$\begin{array}{l} x & = & ρ \cos (θ) \sin (ϕ) \\ y & = & ρ \sin (θ) \sin (ϕ) \\ z & = & ρ \cos (ϕ) \end{array}$

where $ρ = \sqrt{x^{2} + y^{2} + z^{2}}$ , $0^{\circ} \leq θ \leq 360^{\circ}$ where $θ$ is the angle measured from the positive $x$ -axis (toward the positive $y$ -axis), and $0^{\circ} \leq ϕ \leq 180^{\circ}$ where $ϕ$ is the angle measured from the positive $z$ -axis (toward the negative $z$ -axis); however, what we have here is the following:

$\begin{array}{l} x & = & r \cos (θ_{x \to y}) \cos (θ_{x \to z}) \\ y & = & r \sin (θ_{x \to y}) \cos (θ_{x \to z}) \\ z & = & r \sin (θ_{x \to z}) \end{array}$

where $r = \sqrt{x^{2} + y^{2} + z^{2}}$ , $- 180^{\circ} \leq θ_{x \to y} \leq 180^{\circ}$ where $θ_{x \to y}$ is the angle measured from the positive $x$ -axis (toward the positive $y$ -axis), and $- 90^{\circ} \leq θ_{x \to z} \leq 90^{\circ}$ where $θ_{x \to z}$ is the angle measured from the positive $x$ -axis (toward the positive $z$ -axis).

Now, let’s rotate a column vector $(x, y, z)$ back to the $x$ -axis so that we get the column vector $(r, 0, 0)$ . In order to do that, let’s first take a look at each inverse of these Givens Rotation matrices using augmentation and finding the reduced row echelon forms … which as you’ll see is very similar for each matrix. Remembering that $\cos^{2} (θ) + \sin^{2} (θ) = 1$ which implies that $1 - \sin^{2} (θ) = \cos^{2} (θ)$ , we have:

$\begin{matrix} (\begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & - \sin (θ_{y \to z}) & 0 & 1 & 0 \\ 0 & \sin (θ_{y \to z}) & \cos (θ_{y \to z}) & 0 & 0 & 1 \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & - \frac{\sin (θ_{y \to z})}{\cos (θ_{y \to z})} & 0 & \frac{1}{\cos (θ_{y \to z})} & 0 \\ 0 & \sin (θ_{y \to z}) & \cos (θ_{y \to z}) & 0 & 0 & 1 \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & - \frac{\sin (θ_{y \to z})}{\cos (θ_{y \to z})} & 0 & \frac{1}{\cos (θ_{y \to z})} & 0 \\ 0 & 0 & \cos (θ_{y \to z}) + \frac{\sin^{2} (θ_{y \to z})}{\cos (θ_{y \to z})} & 0 & - \frac{\sin (θ_{y \to z})}{\cos (θ_{y \to z})} & 1 \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & - \frac{\sin (θ_{y \to z})}{\cos (θ_{y \to z})} & 0 & \frac{1}{\cos (θ_{y \to z})} & 0 \\ 0 & 0 & \frac{\cos^{2} (θ_{y \to z}) + \sin^{2} (θ_{y \to z})}{\cos (θ_{y \to z})} & 0 & - \frac{\sin (θ_{y \to z})}{\cos (θ_{y \to z})} & 1 \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & - \frac{\sin (θ_{y \to z})}{\cos (θ_{y \to z})} & 0 & \frac{1}{\cos (θ_{y \to z})} & 0 \\ 0 & 0 & \frac{1}{\cos (θ_{y \to z})} & 0 & - \frac{\sin (θ_{y \to z})}{\cos (θ_{y \to z})} & 1 \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & - \frac{\sin (θ_{y \to z})}{\cos (θ_{y \to z})} & 0 & \frac{1}{\cos (θ_{y \to z})} & 0 \\ 0 & 0 & 1 & 0 & - \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & \frac{1}{\cos (θ_{y \to z})} - \frac{\sin^{2} (θ_{y \to z})}{\cos (θ_{y \to z})} & \sin (θ_{y \to z}) \\ 0 & 0 & 1 & 0 & - \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & \frac{1 - \sin^{2} (θ_{y \to z})}{\cos (θ_{y \to z})} & \sin (θ_{y \to z}) \\ 0 & 0 & 1 & 0 & - \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & \frac{\cos^{2} (θ_{y \to z})}{\cos (θ_{y \to z})} & \sin (θ_{y \to z}) \\ 0 & 0 & 1 & 0 & - \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & \cos (θ_{y \to z}) & \sin (θ_{y \to z}) \\ 0 & 0 & 1 & 0 & - \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix}) \end{matrix}$

$\begin{matrix} (\begin{matrix} \cos (θ_{x \to z}) & 0 & - \sin (θ_{x \to z}) & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) & 0 & 0 & 1 \end{matrix}) \\ (\begin{matrix} 1 & 0 & - \frac{\sin (θ_{x \to z})}{\cos (θ_{x \to z})} & \frac{1}{\cos (θ_{x \to z})} & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) & 0 & 0 & 1 \end{matrix}) \\ (\begin{matrix} 1 & 0 & - \frac{\sin (θ_{x \to z})}{\cos (θ_{x \to z})} & \frac{1}{\cos (θ_{x \to z})} & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & \cos (θ_{x \to z}) + \frac{\sin^{2} (θ_{x \to z})}{\cos (θ_{x \to z})} & - \frac{\sin (θ_{x \to z})}{\cos (θ_{x \to z})} & 0 & 1 \end{matrix}) \\ (\begin{matrix} 1 & 0 & - \frac{\sin (θ_{x \to z})}{\cos (θ_{x \to z})} & \frac{1}{\cos (θ_{x \to z})} & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & \frac{\cos^{2} (θ_{x \to z}) + \sin^{2} (θ_{x \to z})}{\cos (θ_{x \to z})} & - \frac{\sin (θ_{x \to z})}{\cos (θ_{x \to z})} & 0 & 1 \end{matrix}) \\ (\begin{matrix} 1 & 0 & - \frac{\sin (θ_{x \to z})}{\cos (θ_{x \to z})} & \frac{1}{\cos (θ_{x \to z})} & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & \frac{1}{\cos (θ_{x \to z})} & - \frac{\sin (θ_{x \to z})}{\cos (θ_{x \to z})} & 0 & 1 \end{matrix}) \\ (\begin{matrix} 1 & 0 & - \frac{\sin (θ_{x \to z})}{\cos (θ_{x \to z})} & \frac{1}{\cos (θ_{x \to z})} & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & - \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 & \frac{1}{\cos (θ_{x \to z})} - \frac{\sin^{2} (θ_{x \to z})}{\cos (θ_{x \to z})} & 0 & \sin (θ_{x \to z}) \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & - \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 & \frac{1 - \sin^{2} (θ_{x \to z})}{\cos (θ_{x \to z})} & 0 & \sin (θ_{x \to z}) \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & - \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 & \frac{\cos^{2} (θ_{x \to z})}{\cos (θ_{x \to z})} & 0 & \sin (θ_{x \to z}) \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & - \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 & \cos (θ_{x \to z}) & 0 & \sin (θ_{x \to z}) \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & - \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix}) \end{matrix}$

$\begin{matrix} (\begin{matrix} \cos (θ_{x \to y}) & - \sin (θ_{x \to y}) & 0 & 1 & 0 & 0 \\ \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{matrix}) \\ (\begin{matrix} 1 & - \frac{\sin (θ_{x \to y})}{\cos (θ_{x \to y})} & 0 & \frac{1}{\cos (θ_{x \to y})} & 0 & 0 \\ \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{matrix}) \\ (\begin{matrix} 1 & - \frac{\sin (θ_{x \to y})}{\cos (θ_{x \to y})} & 0 & \frac{1}{\cos (θ_{x \to y})} & 0 & 0 \\ 0 & \cos (θ_{x \to y}) + \frac{\sin^{2} (θ_{x \to y})}{\cos (θ_{x \to y})} & 0 & - \frac{\sin (θ_{x \to y})}{\cos (θ_{x \to y})} & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{matrix}) \\ (\begin{matrix} 1 & - \frac{\sin (θ_{x \to y})}{\cos (θ_{x \to y})} & 0 & \frac{1}{\cos (θ_{x \to y})} & 0 & 0 \\ 0 & \frac{\cos^{2} (θ_{x \to y}) + \sin^{2} (θ_{x \to y})}{\cos (θ_{x \to y})} & 0 & - \frac{\sin (θ_{x \to y})}{\cos (θ_{x \to y})} & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{matrix}) \\ (\begin{matrix} 1 & - \frac{\sin (θ_{x \to y})}{\cos (θ_{x \to y})} & 0 & \frac{1}{\cos (θ_{x \to y})} & 0 & 0 \\ 0 & \frac{1}{\cos (θ_{x \to y})} & 0 & - \frac{\sin (θ_{x \to y})}{\cos (θ_{x \to y})} & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{matrix}) \\ (\begin{matrix} 1 & - \frac{\sin (θ_{x \to y})}{\cos (θ_{x \to y})} & 0 & \frac{1}{\cos (θ_{x \to y})} & 0 & 0 \\ 0 & 1 & 0 & - \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 & \frac{1}{\cos (θ_{x \to y})} - \frac{\sin^{2} (θ_{x \to y})}{\cos (θ_{x \to y})} & \sin (θ_{x \to y}) & 0 \\ 0 & 1 & 0 & - \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 & \frac{1 - \sin^{2} (θ_{x \to y})}{\cos (θ_{x \to y})} & \sin (θ_{x \to y}) & 0 \\ 0 & 1 & 0 & - \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 & \frac{\cos^{2} (θ_{x \to y})}{\cos (θ_{x \to y})} & \sin (θ_{x \to y}) & 0 \\ 0 & 1 & 0 & - \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{matrix}) \\ (\begin{matrix} 1 & 0 & 0 & \cos (θ_{x \to y}) & \sin (θ_{x \to y}) & 0 \\ 0 & 1 & 0 & - \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{matrix}) \end{matrix}$

Therefore, the inverse of each three-dimensional Givens Rotation matrix is simply the transpose of that matrix:

$\begin{matrix} {(\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & - \sin (θ_{y \to z}) \\ 0 & \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix})}^{- 1} & = & {(\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & - \sin (θ_{y \to z}) \\ 0 & \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix})}^{T} & = & (\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & \sin (θ_{y \to z}) \\ 0 & - \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix}) \\ {(\begin{matrix} \cos (θ_{x \to z}) & 0 & - \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix})}^{- 1} & = & {(\begin{matrix} \cos (θ_{x \to z}) & 0 & - \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix})}^{T} & = & (\begin{matrix} \cos (θ_{x \to z}) & 0 & \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ - \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix}) \\ {(\begin{matrix} \cos (θ_{x \to y}) & - \sin (θ_{x \to y}) & 0 \\ \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}^{- 1} & = & {(\begin{matrix} \cos (θ_{x \to y}) & - \sin (θ_{x \to y}) & 0 \\ \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}^{T} & = & (\begin{matrix} \cos (θ_{x \to y}) & \sin (θ_{x \to y}) & 0 \\ - \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix}) \end{matrix}$

and it shouldn’t be too hard to see that the inverse of each $n$ -dimensional Givens Rotation matrix is also the transpose of that matrix … like it is for any other $n$ -dimensional rotation matrix.

Also, if we consider rotations in the opposite direction where $θ = - ϕ$ while remembering that $\cos (- ϕ) = \cos (ϕ)$ and $\sin (- ϕ) = - \sin (ϕ)$ , then we have:

$\begin{matrix} (\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & \sin (θ_{y \to z}) \\ 0 & - \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix}) & = & (\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (- ϕ_{y \to - z}) & \sin (- ϕ_{y \to - z}) \\ 0 & - \sin (- ϕ_{y \to - z}) & \cos (- ϕ_{y \to - z}) \end{matrix}) & = & (\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (ϕ_{y \to - z}) & - \sin (ϕ_{y \to - z}) \\ 0 & \sin (ϕ_{y \to - z}) & \cos (ϕ_{y \to - z}) \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to z}) & 0 & \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ - \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix}) & = & (\begin{matrix} \cos (- ϕ_{x \to - z}) & 0 & \sin (- ϕ_{x \to - z}) \\ 0 & 1 & 0 \\ - \sin (- ϕ_{x \to - z}) & 0 & \cos (- ϕ_{x \to - z}) \end{matrix}) & = & (\begin{matrix} \cos (ϕ_{x \to - z}) & 0 & - \sin (ϕ_{x \to - z}) \\ 0 & 1 & 0 \\ \sin (ϕ_{x \to - z}) & 0 & \cos (ϕ_{x \to - z}) \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to y}) & \sin (θ_{x \to y}) & 0 \\ - \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix}) & = & (\begin{matrix} \cos (- ϕ_{x \to - y}) & \sin (- ϕ_{x \to - y}) & 0 \\ - \sin (- ϕ_{x \to - y}) & \cos (- ϕ_{x \to - y}) & 0 \\ 0 & 0 & 1 \end{matrix}) & = & (\begin{matrix} \cos (ϕ_{x \to - y}) & - \sin (ϕ_{x \to - y}) & 0 \\ \sin (ϕ_{x \to - y}) & \cos (ϕ_{x \to - y}) & 0 \\ 0 & 0 & 1 \end{matrix}) \end{matrix}$

We see that each inverse amounts to just rotating in the opposite direction. In other words, since $\cos (- θ) = \cos (θ)$ and $\sin (- θ) = - \sin (θ)$ implies that $\cos (θ) = \cos (- θ)$ and $\sin (θ) = - \sin (- θ)$ , we have:

$\begin{matrix} {(\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & - \sin (θ_{y \to z}) \\ 0 & \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix})}^{- 1} & = & (\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & \sin (θ_{y \to z}) \\ 0 & - \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix}) & = & (\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (- θ_{y \to z}) & - \sin (- θ_{y \to z}) \\ 0 & \sin (- θ_{y \to z}) & \cos (- θ_{y \to z}) \end{matrix}) \\ {(\begin{matrix} \cos (θ_{x \to z}) & 0 & - \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix})}^{- 1} & = & (\begin{matrix} \cos (θ_{x \to z}) & 0 & \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ - \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix}) & = & (\begin{matrix} \cos (- θ_{x \to z}) & 0 & - \sin (- θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (- θ_{x \to z}) & 0 & \cos (- θ_{x \to z}) \end{matrix}) \\ {(\begin{matrix} \cos (θ_{x \to y}) & - \sin (θ_{x \to y}) & 0 \\ \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}^{- 1} & = & (\begin{matrix} \cos (θ_{x \to y}) & \sin (θ_{x \to y}) & 0 \\ - \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix}) & = & (\begin{matrix} \cos (- θ_{x \to y}) & - \sin (- θ_{x \to y}) & 0 \\ \sin (- θ_{x \to y}) & \cos (- θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix}) \end{matrix}$

Now, let’s use those inverse Givens Rotation matrices to rotate a column vector $(x, y, z)$ in those opposite directions back to the $x$ -axis so that we get the column vector $(r, 0, 0)$ .

First, we’ll start by using all the Givens Rotation matrices in our established order … and get rid of the ones that don’t matter here:

$\begin{array}{l} (\begin{matrix} x \\ y \\ z \end{matrix}) & = & \overset{3}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to y}) & - \sin (θ_{x \to y}) & 0 \\ \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}} \overset{2}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to z}) & 0 & - \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix})}} \overset{1}{\overset{︷}{(\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & - \sin (θ_{y \to z}) \\ 0 & \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix})}} (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \\ (\begin{matrix} x \\ y \\ z \end{matrix}) & = & \overset{3}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to y}) & - \sin (θ_{x \to y}) & 0 \\ \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}} \overset{2}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to z}) & 0 & - \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix})}} (\overset{1}{\overset{︷}{(\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & - \sin (θ_{y \to z}) \\ 0 & \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix})}} (\begin{matrix} r \\ 0 \\ 0 \end{matrix})) \\ (\begin{matrix} x \\ y \\ z \end{matrix}) & = & \overset{3}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to y}) & - \sin (θ_{x \to y}) & 0 \\ \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}} \overset{2}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to z}) & 0 & - \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix})}} (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \end{array}$

Next, remembering that the inverse of a rotation matrix is its transpose, then we have:

$\begin{matrix} (\begin{matrix} x \\ y \\ z \end{matrix}) = \overset{3}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to y}) & - \sin (θ_{x \to y}) & 0 \\ \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}} \overset{2}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to z}) & 0 & - \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix})}} (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \\ \overset{Inverse of 3}{\overset{︷}{{(\begin{matrix} \cos (θ_{x \to y}) & - \sin (θ_{x \to y}) & 0 \\ \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}^{- 1}}} (\begin{matrix} x \\ y \\ z \end{matrix}) = \overset{2}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to z}) & 0 & - \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix})}} (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \\ \overset{Inverse of 2}{\overset{︷}{{(\begin{matrix} \cos (θ_{x \to z}) & 0 & - \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix})}^{- 1}}} \overset{Inverse of 3}{\overset{︷}{{(\begin{matrix} \cos (θ_{x \to y}) & - \sin (θ_{x \to y}) & 0 \\ \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}^{- 1}}} (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \\ \overset{Inverse of 2}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to z}) & 0 & \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ - \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix})}} \overset{Inverse of 3}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to y}) & \sin (θ_{x \to y}) & 0 \\ - \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}} (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \end{matrix}$

Keeping in mind that $\cos (θ) = \cos (- θ)$ and $\sin (θ) = - \sin (- θ)$ , we have:

$\begin{array}{r} \overset{Inverse of 2}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to z}) & 0 & \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ - \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix})}} \overset{Inverse of 3}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to y}) & \sin (θ_{x \to y}) & 0 \\ - \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}} (\begin{matrix} x \\ y \\ z \end{matrix}) & = & (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \\ \overset{Inverse of 2}{\overset{︷}{(\begin{matrix} \cos (- θ_{x \to z}) & 0 & - \sin (- θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (- θ_{x \to z}) & 0 & \cos (- θ_{x \to z}) \end{matrix})}} \overset{Inverse of 3}{\overset{︷}{(\begin{matrix} \cos (- θ_{x \to y}) & - \sin (- θ_{x \to y}) & 0 \\ \sin (- θ_{x \to y}) & \cos (- θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}} (\begin{matrix} x \\ y \\ z \end{matrix}) & = & (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \\ \overset{2 for - θ_{x \to z}}{\overset{︷}{(\begin{matrix} \cos ((- θ_{x \to z})) & 0 & - \sin ((- θ_{x \to z})) \\ 0 & 1 & 0 \\ \sin ((- θ_{x \to z})) & 0 & \cos ((- θ_{x \to z})) \end{matrix})}} \overset{3 for - θ_{x \to y}}{\overset{︷}{(\begin{matrix} \cos ((- θ_{x \to y})) & - \sin ((- θ_{x \to y})) & 0 \\ \sin ((- θ_{x \to y})) & \cos ((- θ_{x \to y})) & 0 \\ 0 & 0 & 1 \end{matrix})}} (\begin{matrix} x \\ y \\ z \end{matrix}) & = & (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \end{array}$

Since $θ = - ϕ$ implies $- θ = ϕ$ , we also have:

$\begin{array}{r} \overset{2 for - θ_{x \to z}}{\overset{︷}{(\begin{matrix} \cos ((- θ_{x \to z})) & 0 & - \sin ((- θ_{x \to z})) \\ 0 & 1 & 0 \\ \sin ((- θ_{x \to z})) & 0 & \cos ((- θ_{x \to z})) \end{matrix})}} \overset{3 for - θ_{x \to y}}{\overset{︷}{(\begin{matrix} \cos ((- θ_{x \to y})) & - \sin ((- θ_{x \to y})) & 0 \\ \sin ((- θ_{x \to y})) & \cos ((- θ_{x \to y})) & 0 \\ 0 & 0 & 1 \end{matrix})}} (\begin{matrix} x \\ y \\ z \end{matrix}) & = & (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \\ \overset{2 for ϕ_{x \to - z}}{\overset{︷}{(\begin{matrix} \cos (ϕ_{x \to - z}) & 0 & - \sin (ϕ_{x \to - z}) \\ 0 & 1 & 0 \\ \sin (ϕ_{x \to - z}) & 0 & \cos (ϕ_{x \to - z}) \end{matrix})}} \overset{3 for ϕ_{x \to - y}}{\overset{︷}{(\begin{matrix} \cos (ϕ_{x \to - y}) & - \sin (ϕ_{x \to - y}) & 0 \\ \sin (ϕ_{x \to - y}) & \cos (ϕ_{x \to - y}) & 0 \\ 0 & 0 & 1 \end{matrix})}} (\begin{matrix} x \\ y \\ z \end{matrix}) & = & (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \end{array}$

In other words, given a column vector $(x, y, z)$ , we can rotate that vector back to the $x$ -axis by applying Givens Rotation matrices in the opposite order using the opposite angles. Instead of the following:

	$x$	$y$	$z$
$x$		$x \to y$	$x \to z$
$y$			$y \to z$
$z$

	$x$	$y$	$z$
$x$		$3$	$2$
$y$			$1$
$z$

repeated again with subscripts to keep things clear:

	$x$	$y$	$z$
$x$		$3_{forward}$	$2_{forward}$
$y$			$1_{forward}$
$z$

we now have the following:

	$- x$	$- y$	$- z$
$x$		$x \to - y$	$x \to - z$
$y$			$y \to - z$
$z$

	$- x$	$- y$	$- z$
$x$		$1_{backward}$	$2_{backward}$
$y$			$3_{backward}$
$z$

In terms of this new backward ordering and keeping in mind that $ϕ = - θ$ , we now have the following:

$\begin{array}{r} \overset{3_{backward} for ϕ_{y \to - z}}{\overset{︷}{(\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (ϕ_{y \to - z}) & - \sin (ϕ_{y \to - z}) \\ 0 & \sin (ϕ_{y \to - z}) & \cos (ϕ_{y \to - z}) \end{matrix})}} \overset{2_{backward} for ϕ_{x \to - z}}{\overset{︷}{(\begin{matrix} \cos (ϕ_{x \to - z}) & 0 & - \sin (ϕ_{x \to - z}) \\ 0 & 1 & 0 \\ \sin (ϕ_{x \to - z}) & 0 & \cos (ϕ_{x \to - z}) \end{matrix})}} \overset{1_{backward} for ϕ_{x \to - y}}{\overset{︷}{(\begin{matrix} \cos (ϕ_{x \to - y}) & - \sin (ϕ_{x \to - y}) & 0 \\ \sin (ϕ_{x \to - y}) & \cos (ϕ_{x \to - y}) & 0 \\ 0 & 0 & 1 \end{matrix})}} (\begin{matrix} x \\ y \\ z \end{matrix}) & = & (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \\ \overset{3_{backward} for - θ_{y \to z}}{\overset{︷}{(\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (- θ_{y \to z}) & - \sin (- θ_{y \to z}) \\ 0 & \sin (- θ_{y \to z}) & \cos (- θ_{y \to z}) \end{matrix})}} \overset{2_{backward} for - θ_{x \to z}}{\overset{︷}{(\begin{matrix} \cos (- θ_{x \to z}) & 0 & - \sin (- θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (- θ_{x \to z}) & 0 & \cos (- θ_{x \to z}) \end{matrix})}} \overset{1_{backward} for - θ_{x \to y}}{\overset{︷}{(\begin{matrix} \cos (- θ_{x \to y}) & - \sin (- θ_{x \to y}) & 0 \\ \sin (- θ_{x \to y}) & \cos (- θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}} (\begin{matrix} x \\ y \\ z \end{matrix}) & = & (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \\ \overset{1_{forward} for - θ_{y \to z}}{\overset{︷}{(\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (- θ_{y \to z}) & - \sin (- θ_{y \to z}) \\ 0 & \sin (- θ_{y \to z}) & \cos (- θ_{y \to z}) \end{matrix})}} \overset{2_{forward} for - θ_{x \to z}}{\overset{︷}{(\begin{matrix} \cos (- θ_{x \to z}) & 0 & - \sin (- θ_{x \to z}) \\ 0 & 1 & 0 \\ \sin (- θ_{x \to z}) & 0 & \cos (- θ_{x \to z}) \end{matrix})}} \overset{3_{forward} for - θ_{x \to y}}{\overset{︷}{(\begin{matrix} \cos (- θ_{x \to y}) & - \sin (- θ_{x \to y}) & 0 \\ \sin (- θ_{x \to y}) & \cos (- θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}} (\begin{matrix} x \\ y \\ z \end{matrix}) & = & (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \\ \overset{{Inverse of 1}_{forward} for θ_{y \to z}}{\overset{︷}{(\begin{matrix} 1 & 0 & 0 \\ 0 & \cos (θ_{y \to z}) & \sin (θ_{y \to z}) \\ 0 & - \sin (θ_{y \to z}) & \cos (θ_{y \to z}) \end{matrix})}} \overset{{Inverse of 2}_{forward} for θ_{x \to z}}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to z}) & 0 & \sin (θ_{x \to z}) \\ 0 & 1 & 0 \\ - \sin (θ_{x \to z}) & 0 & \cos (θ_{x \to z}) \end{matrix})}} \overset{{Inverse of 3}_{forward} for θ_{x \to y}}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to y}) & \sin (θ_{x \to y}) & 0 \\ - \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}} (\begin{matrix} x \\ y \\ z \end{matrix}) & = & (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to z}) & 0 & \sin (θ_{x \to z}) \\ - \sin (θ_{x \to z}) \sin (θ_{y \to z}) & \cos (θ_{y \to z}) & \cos (θ_{x \to z}) \sin (θ_{y \to z}) \\ - \sin (θ_{x \to z}) \cos (θ_{y \to z}) & - \sin (θ_{y \to z}) & \cos (θ_{x \to z}) \cos (θ_{y \to z}) \end{matrix}) \overset{{Inverse of 3}_{forward} for θ_{x \to y}}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to y}) & \sin (θ_{x \to y}) & 0 \\ - \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}} (\begin{matrix} x \\ y \\ z \end{matrix}) & = & (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to z}) & 0 & \sin (θ_{x \to z}) \\ - \sin (θ_{x \to z}) \sin (θ_{y \to z}) & \cos (θ_{y \to z}) & \cos (θ_{x \to z}) \sin (θ_{y \to z}) \\ - \sin (θ_{x \to z}) \cos (θ_{y \to z}) & - \sin (θ_{y \to z}) & \cos (θ_{x \to z}) \cos (θ_{y \to z}) \end{matrix}) \overset{{Inverse of 3}_{forward} for θ_{x \to y}}{\overset{︷}{(\begin{matrix} \cos (θ_{x \to y}) & \sin (θ_{x \to y}) & 0 \\ - \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ 0 & 0 & 1 \end{matrix})}} (\begin{matrix} x \\ y \\ z \end{matrix}) & = & (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \end{array}$

So, we have:

$(\begin{matrix} \cos (θ_{x \to y}) \cos (θ_{x \to z}) & \sin (θ_{x \to y}) \cos (θ_{x \to z}) & \sin (θ_{x \to z}) \\ - \cos (θ_{x \to y}) \sin (θ_{x \to z}) \sin (θ_{y \to z}) - \sin (θ_{x \to y}) \cos (θ_{y \to z}) & - \sin (θ_{x \to y}) \sin (θ_{x \to z}) \sin (θ_{y \to z}) + \cos (θ_{x \to y}) \cos (θ_{y \to z}) & \cos (θ_{x \to z}) \sin (θ_{y \to z}) \\ - \cos (θ_{x \to y}) \sin (θ_{x \to z}) \cos (θ_{y \to z}) + \sin (θ_{x \to y}) \sin (θ_{y \to z}) & - \sin (θ_{x \to y}) \sin (θ_{x \to z}) \cos (θ_{y \to z}) - \cos (θ_{x \to y}) \sin (θ_{y \to z}) & \cos (θ_{x \to z}) \cos (θ_{y \to z}) \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} r \\ 0 \\ 0 \end{matrix})$

Given $r = \sqrt{x^{2} + y^{2} + z^{2}}$ , we choose $θ_{x \to y} = \arctan (\frac{y}{x})$ , $θ_{x \to z} = \arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})$ , and $θ_{y \to z} = 0$ .

After applying $θ_{y \to z} = 0$ , we have:

$\begin{matrix} (\begin{matrix} \cos (θ_{x \to y}) \cos (θ_{x \to z}) & \sin (θ_{x \to y}) \cos (θ_{x \to z}) & \sin (θ_{x \to z}) \\ - \cos (θ_{x \to y}) \sin (θ_{x \to z}) \sin (θ_{y \to z}) - \sin (θ_{x \to y}) \cos (θ_{y \to z}) & - \sin (θ_{x \to y}) \sin (θ_{x \to z}) \sin (θ_{y \to z}) + \cos (θ_{x \to y}) \cos (θ_{y \to z}) & \cos (θ_{x \to z}) \sin (θ_{y \to z}) \\ - \cos (θ_{x \to y}) \sin (θ_{x \to z}) \cos (θ_{y \to z}) + \sin (θ_{x \to y}) \sin (θ_{y \to z}) & - \sin (θ_{x \to y}) \sin (θ_{x \to z}) \cos (θ_{y \to z}) - \cos (θ_{x \to y}) \sin (θ_{y \to z}) & \cos (θ_{x \to z}) \cos (θ_{y \to z}) \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to y}) \cos (θ_{x \to z}) & \sin (θ_{x \to y}) \cos (θ_{x \to z}) & \sin (θ_{x \to z}) \\ - \cos (θ_{x \to y}) \sin (θ_{x \to z}) \sin (0) - \sin (θ_{x \to y}) \cos (0) & - \sin (θ_{x \to y}) \sin (θ_{x \to z}) \sin (0) + \cos (θ_{x \to y}) \cos (0) & \cos (θ_{x \to z}) \sin (0) \\ - \cos (θ_{x \to y}) \sin (θ_{x \to z}) \cos (0) + \sin (θ_{x \to y}) \sin (0) & - \sin (θ_{x \to y}) \sin (θ_{x \to z}) \cos (0) - \cos (θ_{x \to y}) \sin (0) & \cos (θ_{x \to z}) \cos (0) \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to y}) \cos (θ_{x \to z}) & \sin (θ_{x \to y}) \cos (θ_{x \to z}) & \sin (θ_{x \to z}) \\ - \cos (θ_{x \to y}) \sin (θ_{x \to z}) (0) - \sin (θ_{x \to y}) (1) & - \sin (θ_{x \to y}) \sin (θ_{x \to z}) (0) + \cos (θ_{x \to y}) (1) & \cos (θ_{x \to z}) (0) \\ - \cos (θ_{x \to y}) \sin (θ_{x \to z}) (1) + \sin (θ_{x \to y}) (0) & - \sin (θ_{x \to y}) \sin (θ_{x \to z}) (1) - \cos (θ_{x \to y}) (0) & \cos (θ_{x \to z}) (1) \end{matrix}) \\ (\begin{matrix} \cos (θ_{x \to y}) \cos (θ_{x \to z}) & \sin (θ_{x \to y}) \cos (θ_{x \to z}) & \sin (θ_{x \to z}) \\ - \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ - \cos (θ_{x \to y}) \sin (θ_{x \to z}) & - \sin (θ_{x \to y}) \sin (θ_{x \to z}) & \cos (θ_{x \to z}) \end{matrix}) \end{matrix}$

Then after applying $θ_{x \to y} = \arctan (\frac{y}{x})$ and $θ_{x \to z} = \arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})$ , since $\cos (\arctan (\frac{y}{x})) = \frac{x}{\sqrt{x^{2} + y^{2}}}$ , $\sin (\arctan (\frac{y}{x})) = \frac{y}{\sqrt{x^{2} + y^{2}}}$ , $\cos (\arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})) = \frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2} + z^{2}}}$ , and $\sin (\arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})) = \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}}$ , we have:

$\begin{matrix} (\begin{matrix} \cos (θ_{x \to y}) \cos (θ_{x \to z}) & \sin (θ_{x \to y}) \cos (θ_{x \to z}) & \sin (θ_{x \to z}) \\ - \sin (θ_{x \to y}) & \cos (θ_{x \to y}) & 0 \\ - \cos (θ_{x \to y}) \sin (θ_{x \to z}) & - \sin (θ_{x \to y}) \sin (θ_{x \to z}) & \cos (θ_{x \to z}) \end{matrix}) \\ (\begin{matrix} \cos (\arctan (\frac{y}{x})) \cos (\arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})) & \sin (\arctan (\frac{y}{x})) \cos (\arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})) & \sin (\arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})) \\ - \sin (\arctan (\frac{y}{x})) & \cos (\arctan (\frac{y}{x})) & 0 \\ - \cos (\arctan (\frac{y}{x})) \sin (\arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})) & - \sin (\arctan (\frac{y}{x})) \sin (\arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})) & \cos (\arctan (\frac{z}{\sqrt{x^{2} + y^{2}}})) \end{matrix}) \\ (\begin{matrix} (\frac{x}{\sqrt{x^{2} + y^{2}}}) (\frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2} + z^{2}}}) & (\frac{y}{\sqrt{x^{2} + y^{2}}}) (\frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2} + z^{2}}}) & (\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}}) \\ - (\frac{y}{\sqrt{x^{2} + y^{2}}}) & (\frac{x}{\sqrt{x^{2} + y^{2}}}) & 0 \\ - (\frac{x}{\sqrt{x^{2} + y^{2}}}) (\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}}) & - (\frac{y}{\sqrt{x^{2} + y^{2}}}) (\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}}) & (\frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2} + z^{2}}}) \end{matrix}) \\ (\begin{matrix} \frac{x}{\sqrt{x^{2} + y^{2} + z^{2}}} & \frac{y}{\sqrt{x^{2} + y^{2} + z^{2}}} & \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} \\ - \frac{y}{\sqrt{x^{2} + y^{2}}} & \frac{x}{\sqrt{x^{2} + y^{2}}} & 0 \\ - \frac{x z}{\sqrt{x^{2} + y^{2}} \sqrt{x^{2} + y^{2} + z^{2}}} & - \frac{y z}{\sqrt{x^{2} + y^{2}} \sqrt{x^{2} + y^{2} + z^{2}}} & \frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2} + z^{2}}} \end{matrix}) \end{matrix}$

Therefore, since $r = \sqrt{x^{2} + y^{2} + z^{2}}$ , we have:

$\begin{array}{r} (\begin{matrix} \frac{x}{\sqrt{x^{2} + y^{2} + z^{2}}} & \frac{y}{\sqrt{x^{2} + y^{2} + z^{2}}} & \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} \\ - \frac{y}{\sqrt{x^{2} + y^{2}}} & \frac{x}{\sqrt{x^{2} + y^{2}}} & 0 \\ - \frac{x z}{\sqrt{x^{2} + y^{2}} \sqrt{x^{2} + y^{2} + z^{2}}} & - \frac{y z}{\sqrt{x^{2} + y^{2}} \sqrt{x^{2} + y^{2} + z^{2}}} & \frac{\sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2} + z^{2}}} \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) & = & (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \\ (\begin{matrix} \frac{x^{2}}{\sqrt{x^{2} + y^{2} + z^{2}}} + \frac{y^{2}}{\sqrt{x^{2} + y^{2} + z^{2}}} + \frac{z^{2}}{\sqrt{x^{2} + y^{2} + z^{2}}} \\ - \frac{x y}{\sqrt{x^{2} + y^{2}}} + \frac{x y}{\sqrt{x^{2} + y^{2}}} + 0 \\ - \frac{x^{2} z}{\sqrt{x^{2} + y^{2}} \sqrt{x^{2} + y^{2} + z^{2}}} - \frac{y^{2} z}{\sqrt{x^{2} + y^{2}} \sqrt{x^{2} + y^{2} + z^{2}}} + \frac{(\sqrt{x^{2} + y^{2}}) z}{\sqrt{x^{2} + y^{2} + z^{2}}} \end{matrix}) & = & (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \\ (\begin{matrix} \frac{x^{2} + y^{2} + z^{2}}{\sqrt{x^{2} + y^{2} + z^{2}}} \\ 0 \\ \frac{- x^{2} z - y^{2} z + (x^{2} + y^{2}) z}{\sqrt{x^{2} + y^{2}} \sqrt{x^{2} + y^{2} + z^{2}}} \end{matrix}) & = & (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \\ (\begin{matrix} \sqrt{x^{2} + y^{2} + z^{2}} \\ 0 \\ \frac{- x^{2} z - y^{2} z + x^{2} z + y^{2} z}{\sqrt{x^{2} + y^{2}} \sqrt{x^{2} + y^{2} + z^{2}}} \end{matrix}) & = & (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \\ (\begin{matrix} \sqrt{x^{2} + y^{2} + z^{2}} \\ 0 \\ 0 \end{matrix}) & = & (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \\ (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) & = & (\begin{matrix} r \\ 0 \\ 0 \end{matrix}) \end{array}$

Three-Dimensional Givens Rotations
Three-Dimensional Givens Rotations (Word Document)
Three-Dimensional Givens Rotations (Print to PDF)

Tagged on: Givens Rotation Matrices, Mathematics

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Joshua Burkholder

Mathematics and Computer Science

Three-Dimensional Givens Rotations

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