- #1
Pengwuino
Gold Member
- 5,123
- 20
I have a unit vector
[tex]frac{{{\rm - 2}}}{{\sqrt {{\rm 89}} }},\frac{{\rm 7}}{{\sqrt {{\rm 89}} }},\frac{{{\rm - 6}}}{{\sqrt {{\rm 89}} }}[/tex]
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…
[tex]t(\frac{{{\rm - 2}}}{{\sqrt {{\rm 89}} }},\frac{{\rm 7}}{{\sqrt {{\rm 89}} }},\frac{{{\rm - 6}}}{{\sqrt {{\rm 89}} }})[/tex]
Distribute T and then do the [tex]\sqrt {a^2 + b^2 + c^2 } [/tex]
Thing to get the length to equal 4?
[tex]4 = \sqrt {(\frac{{{\rm - 2t}}}{{\sqrt {{\rm 89}} }})^2 + (\frac{{{\rm 7t}}}{{\sqrt {{\rm 89}} }})^2 + (\frac{{{\rm - 6t}}}{{\sqrt {{\rm 89}} }})^2 } [/tex]
? Is that how you would figure it out?
[tex]frac{{{\rm - 2}}}{{\sqrt {{\rm 89}} }},\frac{{\rm 7}}{{\sqrt {{\rm 89}} }},\frac{{{\rm - 6}}}{{\sqrt {{\rm 89}} }}[/tex]
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…
[tex]t(\frac{{{\rm - 2}}}{{\sqrt {{\rm 89}} }},\frac{{\rm 7}}{{\sqrt {{\rm 89}} }},\frac{{{\rm - 6}}}{{\sqrt {{\rm 89}} }})[/tex]
Distribute T and then do the [tex]\sqrt {a^2 + b^2 + c^2 } [/tex]
Thing to get the length to equal 4?
[tex]4 = \sqrt {(\frac{{{\rm - 2t}}}{{\sqrt {{\rm 89}} }})^2 + (\frac{{{\rm 7t}}}{{\sqrt {{\rm 89}} }})^2 + (\frac{{{\rm - 6t}}}{{\sqrt {{\rm 89}} }})^2 } [/tex]
? Is that how you would figure it out?