My question will be very similar to this question. However, I found either the solutions didn't answer all of my questions, or were overall too confusing to do so.

Say I set $a$ and $b$ to be real numbers. If I have a vector $\vec{v}=\langle{3,4,0}\rangle$, and want to find all unit vectors which are orthogonal to $\vec{v}$, I am interested in how to best go about determining this solution set, or even a single member of the solution set.

I know I could perhaps use the cross product to find a single vector which is orthogonal to $\vec{v}$, but this doesn't seem as though it will be effective in finding the full solution set.


In order for to vectors to be orthogonal, the dot product has to vanish.

Let's say we are looking for vectors $u=(u_1,u_2,u_3)^T$. Hence $$<u,v>=3u_1+4u_2+0u_3=0$$

now as you can see, all vectors u with this property can solve the problem. Lets solve this equation for $u_2$


Now lets plug this back into our formula for $u'$. We get $$u=(u_1,-\frac{3}{4}u_1,u_3)$$

To make these vectors $u$ unit vectors, you need to devide $u$ by its magnitude $$|u|=\sqrt{u_1^2+\frac{9}{16}u_1^2+u_3^2}=\sqrt{\frac{25}{16}u_1^2+u_3^2}$$ $$ u' = \frac{(u_1,-\frac{3}{4}u_1,u_3)}{\sqrt{\frac{25}{16}u_1^2+u_3^2}}$$ Where $u_1,u_3$ are arbitrary real numbers, with $(u_1,u_3) \neq (0,0)$.

  • $\begingroup$ You didn't address that they are unit vectors. Also, please explain yourself and your steps more thoroughly, if you would. $\endgroup$ – alxmke Sep 9 '15 at 8:05
  • $\begingroup$ Added the part with unit vectors into the answer. $\endgroup$ – MrYouMath Sep 9 '15 at 8:12
  • $\begingroup$ I don't understand what you mean by $\langle{u,v}\rangle$. $\endgroup$ – alxmke Sep 9 '15 at 8:25
  • $\begingroup$ $<u,v>$ is the dot product. $\endgroup$ – MrYouMath Sep 9 '15 at 8:37

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