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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.

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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 $u_2$

$$u_2=-\frac{3}{4}u_1$$

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 $(u_1,u_3) \neq (0,0)$.

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  • $\begingroup$ 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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