I have a problem of scaling a unit vector $u$ in such way that its scalar product (by itself) multiplied by a costant has to be equal to one. $$a u\cdot u=1$$ How do I do it?

Edit: the unit vector - it is a normalized vector which has the length equal to $1$: $||u||=1$

Edit2: I need to rescale this vector $u$ in such way that $a u\cdot u=1$

  • $\begingroup$ What's the definition of a unit vector? $\endgroup$ – lulu Jul 23 '15 at 13:57
  • $\begingroup$ $$ |u|=1\\au.u=1\\a |u| |u|cos (0^0) =1\\a*1*1*1=1$$ $\endgroup$ – Khosrotash Jul 23 '15 at 14:04
  • $\begingroup$ But if we already have $u \cdot u = 1$ then we can just take a = 1, no? I suspect you meant to start with a vector that was not a unit vector, though I am not certain. $\endgroup$ – lulu Jul 23 '15 at 14:08
  • 1
    $\begingroup$ I see the edits but what you are writing is not at all clear. The equation you wrote, $a\,u \cdot u = 1$ has the unique solution $a = 1$. I now believe you meant to write: "Given a unit vector $u$ and an arbitrary non-zero scalar $a$ find a scalar $\lambda$ such that $a\, (\lambda u) \cdot (\lambda u) = 1$.". Is this correct? But then we see that this has no solution if $a < 0$, trusting everything in sight is real, and if $a > 0$ then we just take $\lambda = \frac {1}{\sqrt {a}}$. Or did you mean something else? $\endgroup$ – lulu Jul 23 '15 at 14:14
  • $\begingroup$ $a u\cdot u=a(u\cdot u)=a||u||^2=1$ Since $||u||=1$, $a=1$. $\endgroup$ – user137035 Jul 23 '15 at 14:18

I think you mean that $u$ is a unit vector. If so, $u\cdot u = 1$, and $au\cdot u = 1$ is not possible in general.

But I think what what the question is asking is that you are given $u$ and you are supposed to find $v$ in the same direction as $u$ so that $$av\cdot v = 1.$$ Is that right?

If so, write $v=bu$ and then solve $$a(bu)\cdot (bu)=1$$ to find $b$, using the fact that $u$ is a unit vector so $u\cdot u = 1$.


You can divide vector $v$ by its own modulus and by $\sqrt a$ so that the dot product is 1


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.