# Vector of a certain length given unit vector

1. Nov 26, 2005

### Pengwuino

I have a unit vector

$$frac{{{\rm - 2}}}{{\sqrt {{\rm 89}} }},\frac{{\rm 7}}{{\sqrt {{\rm 89}} }},\frac{{{\rm - 6}}}{{\sqrt {{\rm 89}} }}$$

I need to figure out a vector with a length of 4 with that as a unit vector. Now I'm thinking that I can just do…

$$t(\frac{{{\rm - 2}}}{{\sqrt {{\rm 89}} }},\frac{{\rm 7}}{{\sqrt {{\rm 89}} }},\frac{{{\rm - 6}}}{{\sqrt {{\rm 89}} }})$$

Distribute T and then do the $$\sqrt {a^2 + b^2 + c^2 }$$

Thing to get the length to equal 4?

$$4 = \sqrt {(\frac{{{\rm - 2t}}}{{\sqrt {{\rm 89}} }})^2 + (\frac{{{\rm 7t}}}{{\sqrt {{\rm 89}} }})^2 + (\frac{{{\rm - 6t}}}{{\sqrt {{\rm 89}} }})^2 }$$

??? Is that how you would figure it out?

2. Nov 27, 2005

### alephnought

???
You have a unit vector right? and you need a vector in the same direction as this and of length 4?

Since the unit vector has length 1, all you need to do is multiply the unit vector by 4!!! Its as simple as that.

unit vector= ai + bj + ck
root of( a^2 + b^2 + c^2) = 1
new vector= 4ai + 4bj + 4ck
its length= root of( 4^2 * 1) = 4

3. Nov 27, 2005

### Pengwuino

haha damn i'm dumb

Moments like these make me question my understanding of math.