Math Help - Simple question that I am too lazy to analyze...

1. Simple question that I am too lazy to analyze...

Find a vector that has the same direction as < -2, 4, 2> but has length 6. This probably has something to do with the magnitude and I get sqroot(24) but where do they get the 6?

Thank you.

2. What an honest title!
Here is a short answer: multiply by $\frac{{\sqrt 6 }}{2}$.

3. Originally Posted by Undefdisfigure
Find a vector that has the same direction as < -2, 4, 2> but has length 6. This probably has something to do with the magnitude and I get sqroot(24) but where do they get the 6?
So Plato essentially gave you the answer, but just in case you decide that you aren't too lazy to want to know how...

In order to get a vector with different magnitude but in the same direction, find the unit vector in that direction (i.e. multiply the original vector by 1/the magnitude), then multiply the new unit vector by the magnitude that you want it to be.

so in this case you had <-2,4,2>, to get the unit vector we take the magnitude, $\sqrt{(-2)^2+4^2+2^2}=\sqrt{24}=2\sqrt{6}$
Then mutliply your original vector by 1 over this, which is $\frac{1}{2\sqrt{6}}=\frac{\sqrt{6}}{12}$.

we are going to then multiply our vector by this to get the unit vector but since we are going to then multiply the unit vector by 6 to get the new magnitude we can just say $6(\frac{\sqrt{6}}{12})<-2,4,2>=\frac{\sqrt{6}}{2}<-2,4,2>=<-\sqrt{6},2\sqrt{6},\sqrt{6}>$

Hope that helps!!!

4. Originally Posted by Plato
What an honest title!
Here is a short answer: multiply by $\frac{{\sqrt 6 }}{2}$.
Indeed. Too lazy even to analyse that it involves no calculus whatsoever .....