Pengwuino
Gold Member
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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?
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?