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Given the vectors $\mathbf{u,v}$ in R³, determine all vectors that are vertical to $\mathbf{u}$ and $\mathbf{v}$ with length = 1

Every vector $\mathbf{x'}$ that is to be found must meet these two conditions:

  • $\mathbf{x'} \cdot \mathbf{u} = \mathbf{x'} \cdot \mathbf{v}=0$
  • $\mathbf{x'} \times \mathbf{u}= \mathbf{x'} \times \mathbf{v}=1$

My approach is to insert the given vector components into the first condition and get something like this
$ ax_1' + bx_2' + cx_3' =0 $

since the length of the vectors should be 1, there's

$||x'||=\sqrt{( x' \cdot x')}=\sqrt{x_1'^2+x_2'^2+x_3^3}$

is that the right way to go? I don't really know how to tackle this

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    $\begingroup$ i think your second condition might be $ \mathbf{x'} \times \mathbf{x'}=1$ Otherwise the question would have been something like: determine all vectors that are vertical to $\mathbf{u}$ and$\mathbf{v}$ such that the crossproduct between those vectors and $\mathbf{u},\mathbf{v}$ is $1$ $\endgroup$
    – sigmatau
    May 16, 2013 at 11:24
  • $\begingroup$ oh boy, you are right! Well that makes things a bit easier I suppose $\endgroup$
    – deemel
    May 16, 2013 at 11:27
  • $\begingroup$ edited my question $\endgroup$
    – deemel
    May 16, 2013 at 12:18

1 Answer 1

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If you want it to be perpendicular to both $\mathbf{u}$ and $\mathbf{v}$ then take $ \mathbf{u} \times \mathbf{v}$. And then to normalize it (so that it has length 1) just divide this vector by its norm to get

$$\mathbf{x'}=\frac{\mathbf{u} \times \mathbf{v}}{|| \mathbf{u} \times \mathbf{v}||}$$

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  • $\begingroup$ please check the edit of my question, I made a mistake in the initial one $\endgroup$
    – deemel
    May 16, 2013 at 12:55
  • $\begingroup$ This answers the question you put in the box. What's your objection to it? $\endgroup$ May 16, 2013 at 13:12

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