# Homework Help: Calc 3, vectors and length

1. Sep 11, 2010

### hover

1. The problem statement, all variables and given/known data

Find a vector that has the same direction as the given vector but has length 6.
‹4, 2, -2›

2. Relevant equations

L = ((x-i)^2+(y-j)^2+(z-k)^2)^.5

3. The attempt at a solution

For some reason I am just stuck. I don't know how to start. I assume L = ((x-i)^2+(y-j)^2+(z-k)^2)^.5 becomes L = (x^2+y^2+z^2)^.5 because of the 2 points (0,0,0) and (x,y,z) but how does one find something with the length of 6 in 3 dimensions? All I really have is this

36=x^2+y^2+z^2

I guess I could eliminate z by finding an equation for a line on points (0,0,0) and (4,2,-2). I found that equation to be z=x-3y but after plugging that in it seems that I have 2 independent variables(x and y). How do I deal with this?

Thanks!

2. Sep 11, 2010

### cronxeh

Ok so lets say a is scalar and v is your vector. The a*v has same direction but different magnitude than a, and so the magnitude of a*v is sqrt((ax)^2 + (ay)^2 + (az)^2)

sqrt((4a)^2 + (2a)^2 + (-2a)^2) = 6, and so a=sqrt(3/2). so a*v should equal to 6

Please refer to the vector calculus notes here http://tutorial.math.lamar.edu/Classes/CalcII/VectorArithmetic.aspx you need to know difference between addition, substraction, and multiplication of vectors and what it looks like

3. Sep 11, 2010

### hover

Ah yes yes yes. Ok now I see how you go about this. Simply find a scalar to multiply the original vector by. I found out that I multiply by (3/2)^.5 and the new vector is <4(3/2)^.5,2(3/2)^.5,-2(3/2)^.5> and this has a length 6.

Thanks!