Finding values to make a unit vector

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SUMMARY

The discussion centers on determining the values of 'a' for the vector w = ai + (a/8)j to be a unit vector. The key equation derived is 1 = √(a² + (a/8)²), leading to the simplification 1 = a² + (a²/64). This results in the equation 65a²/64 = 1, ultimately yielding the solutions a = ±(8/√65). The critical mistake identified in the calculations was the incorrect multiplication of terms, which misled the initial solution attempt.

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Sammy600
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Homework Statement


Find all values of a such that w=ai+[itex]\frac{a}{8}[/itex]j is a unit vector.

Homework Equations


unit vector has length of 1. and for a vector v unit vectors would be v/magv

The Attempt at a Solution


1=magw=\sqrt (a2+(a/8)2)
1=a2+(a2/64)
64=2a2
32=a2
a=[itex]\sqrt{32}[/itex]

i know that the solution is: +/- [itex]\frac{8}{\sqrt{65}}[/itex] but am at a loss as to how it was obtained. any help is appreciated.
 
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sqrt(a^2+a^2/64)=1
so this means sqrt((64a^2+a^2)/64)=1
so sqrt(65a^2/64)=1
this means 65a^2/64=1
and then you get the result
 
Sammy600 said:

Homework Statement


Find all values of a such that w=ai+[itex]\frac{a}{8}[/itex]j is a unit vector.

Homework Equations


unit vector has length of 1. and for a vector v unit vectors would be v/magv

The Attempt at a Solution


1=magw=\sqrt (a2+(a/8)2)
1=a2+(a2/64)
64=2a2
Your mistake is above. Multiply each term on the right side by 64. You don't get 2a2.
Sammy600 said:
32=a2
a=[itex]\sqrt{32}[/itex]

i know that the solution is: +/- [itex]\frac{8}{\sqrt{65}}[/itex] but am at a loss as to how it was obtained. any help is appreciated.
 

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