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I am looking for a vector of length $3$ that is perpendicular to both u=(-2,0,1) and v=(1,2,-3). I know that the cross product of u and v will give me a vector perpendicular to both of them, but I don't know how to get it to be specifically $3$ units of length. Could anybody help me out with this?

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$$ n = 3 \cdot \frac{u \times v}{\|u \times v\|}$$

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Let $(a,b,c)$ the needed vector.

Thus, $-2a+c=0$,$a+2b-3c=0$ and $a^2+b^2+c^2=9$.

Hence, $b=\frac{5}{2}a$, which gives, $a^2=\frac{4}{5}$ and we obtain two vectors:

$\left(\frac{2}{\sqrt5},\sqrt5,\frac{4}{\sqrt5}\right)$ and $\left(-\frac{2}{\sqrt5},-\sqrt5,-\frac{4}{\sqrt5}\right)$

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