Vector of a certain length given unit vector

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SUMMARY

The discussion centers on calculating a vector of length 4 using a given unit vector, specifically the unit vector \((- \frac{2}{\sqrt{89}}, \frac{7}{\sqrt{89}}, -\frac{6}{\sqrt{89}})\). The solution involves multiplying the unit vector by 4, resulting in the new vector \((- \frac{8}{\sqrt{89}}, \frac{28}{\sqrt{89}}, -\frac{24}{\sqrt{89}})\). This method confirms that scaling a unit vector by its desired length directly yields the required vector. The mathematical principle utilized is straightforward: a unit vector's length is 1, and scaling it by a factor produces a vector of the desired length.

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Pengwuino
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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?
 
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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
 
haha damn I'm dumb

Moments like these make me question my understanding of math.
 

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