How to find a vector that has the same direction

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Homework Statement


Can anyone help me with his problem:
how to find a vector that has the same direction as <-2,4,2> but has length 6


Homework Equations





The Attempt at a Solution


the length of that vector is: 6= (-2)^2*a^2+(4)^2*b^2+(2)^2*c^2
Then I don't know what to do next?
 
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Suppose I have a unit vector (a,b,c). What is the length of k(a,b,c)?
 
divide the lengh from the original vector, that will give you the unit vector. What do you do with the unit vector to get a lengh of 6 and same direction?
 
k(a,b,c)=(ka,kb,kc)=(-2k,4k,2k)
lenght= root(4k^2+16k^2+4k^2)=6
=> root( 24k^2) =6
=> k*root(24)=6
=>k= 6/root(24)= 3/root6
oh I got it, Thank you very much! Ziox AND Antineutron
 
Although you haven't done what we said, that works. Note how this scaling quantity is unique up to +/-. Which makes sense: the direction induced by (-2,4,2) is just a line in R^3.

The scaling value that you actually calculated is the ratio of new length to the original length - dividing by the old and multiplying by the new.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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