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Given an arbitrary vector, I can find any orthogonal vector by solving $ax + by + cz = 0$.

I want to find the "most upright" orthogonal (unit) vector, where $z$ is maximized. There must be a straightforward closed form solution to this, right?

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Using the observation made by dtldarek in his answer, you’re looking for the unit vector in the plane generated by $w=(a,b,c)$ and $e_3=(0,0,1)$ that is perpendicular to $w$ and has a non-negative $z$-coordinate. Observe that such a vector is also perpendicular to $e_3\times w$, so the vector you seek is $w\times(e_3\times w)$, normalized.

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  • $\begingroup$ Elegant! $\ddot\smile$ $\endgroup$ – dtldarek May 16 '17 at 19:56
  • $\begingroup$ Yup, I have to say, that was pretty darn simple, and so obvious in retrospect. This one will be tucked away in my toolkit forever after. Thank you! $\endgroup$ – David Parks May 16 '17 at 21:17
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    $\begingroup$ This one is actually used a bunch in the 3d graphics world as part of creating orthonormal matrices which represent the effects of looking in a direction. $\endgroup$ – Cort Ammon May 17 '17 at 0:41
  • $\begingroup$ @CortAmmon Yep. That’s what led me to look at the problem from this point of view, so to speak. $\endgroup$ – amd May 17 '17 at 0:43
  • $\begingroup$ This was exactly my use case, I needed to compute the point of gaze as the head rotates left/right while focused at an angle. This vector provides the axis of rotation which then allowed me to apply Rodrigues' rotation formula to perform the rotation. $\endgroup$ – David Parks May 17 '17 at 17:20
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Restating your problem (and adding a condition), you are given a vector $\vec{v}=\langle a,b,c\rangle$ and you want to find a vector $\vec{w}=\langle x,y,z\rangle$ so that

  1. $\vec{w}\cdot\vec{v}=0$,
  2. $\|\vec{w}\|=1$
  3. $\vec{w}\cdot\vec{e}_3$ is maximized

As you state, you want to maximize $z$ subject to \begin{align*} ax+by+cz&=0\\ x^2+y^2+z^2&=1 \end{align*}

Perhaps a Lagrange multiplier approach would be good here: \begin{align*} 0&=\lambda a+2\mu x\\ 0&=\lambda b+2\mu y\\ 1&=\lambda c+2\mu z\\ 0&=ax+by+cz\\ 1&=x^2+y^2+z^2 \end{align*}

We should really assume that $a$ and $b$ are not both zero because otherwise $\vec{v}$ points vertically and there are no orthogonal, upward pointing vectors.

  • If $\mu=0$, then we get that $\lambda=\frac{1}{c}$ (and $c\not=0$), but then the first two equations are $0=\frac{a}{c}$ and $0=\frac{b}{c}$, so $a=0=b$ This contradicts our assumption, so we're ok here.

  • If $\mu\not=0$, then we can solve for $x$, $y$, and $z$: \begin{align*} x&=-\frac{\lambda a}{2\mu}\\ y&=-\frac{\lambda b}{2\mu}\\ z&=\frac{1-\lambda c}{2\mu} \end{align*} We can substitute these into the fourth equation (and cancel the $2\mu$'s) to get $$ -\lambda a^2-\lambda b^2-\lambda c^2+c=0. $$ Now, we make things simpler by adding the assumption that $\langle a,b,c\rangle $ is a unit vector, so we could rewrite this equation as $c=\lambda$. If not, we use $c=\lambda\|\vec{v}\|$ throughout the rest of this problem.

Therefore, \begin{align*} x&=-\frac{ac}{2\mu}\\ y&=-\frac{bc}{2\mu}\\ z&=\frac{1-c^2}{2\mu} \end{align*} Substituting all of this into the final equation gives $$ 4\mu^2=a^2c^2+b^2c^2+1-2c^2+c^4=(a^2+b^2+c^2)c^2+1-2c^2. $$ Therefore $$ 4\mu^2=1-c^2 $$ or $$ \mu=\pm\sqrt{\frac{1-c^2}{4}} $$ Substituting these into the formulae for $x,y,z$ gives that \begin{align*} x&=\mp\frac{ac}{\sqrt{1-c^2}}\\ y&=\mp\frac{bc}{\sqrt{1-c^2}}\\ z&=\pm\frac{1-c^2}{\sqrt{1-c^2}} \end{align*} Depending on the signs, you should get the largest and smallest that $z$ could be (provided I made no errors).

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  • $\begingroup$ Ahhh, lagrange multipliers... there's still scar tissue in my brain from the last time I had to deal with them. I'm really happy to see a straightforward application that I can get my head around. This is useful beyond the particular problem I'm solving right now. Thank you! $\endgroup$ – David Parks May 16 '17 at 18:32
  • $\begingroup$ @amd Yes, I added that assumption, but it doesn't really matter since orthogonality is independent of length. So, if you wanted to drop it, you would have $c=\lambda\|v\|$, and carry this through the computation. $\endgroup$ – Michael Burr May 16 '17 at 19:55
  • $\begingroup$ @amd Agreed, I made a change to the text. $\endgroup$ – Michael Burr May 16 '17 at 20:54
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Assuming $\mathbb{R}^3$, a geometric approach would be to observe that the solution would belong to plane generated by $(a,b,c)$ and $(0,0,1)$. That forces the ratio between $x$ and $y$ to be the same as between $a$ and $b$. In other words, we have two equations:

$$\begin{cases}ax+by+cz=0\\ay-bx=0\end{cases}$$

If $c = 0$, then $(0,0,z)$ is your solution for any $z>0$. Otherwise, if $x=ka$ and $y=kb$, then $k(a^2+b^2)+cz=0$ and $z=-k\frac{a^2+b^2}{c}$. To maximize $z$ make sure $z\geq 0$, that is, $k\cdot c \leq 0$.

I hope this helps $\ddot\smile$

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  • $\begingroup$ Might be worth noting that the second equation in the system is that of the plane spanned by $(a,b,c)$ and the $z$-axis. $\endgroup$ – amd May 16 '17 at 22:14

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