Find 2 Unit Vectors at 60 Degrees with <3,4> - Dot Product Calculation

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SUMMARY

The discussion focuses on finding two unit vectors that form a 60-degree angle with the vector <3,4>. The calculations involve setting b=1 and using the cosine of 60 degrees (1/2) to derive the equations. The resulting quadratic equation, 11a^2 + 96a = -39, leads to the solution for 'a' as ±(sqrt(2265)-48)/sqrt(11). Additionally, a suggestion is made to impose another condition on 'a' and 'b' to ensure the vectors remain unit vectors, indicating that further calculations are necessary to meet this requirement.

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nameVoid
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Find 2 unit vectors that make a 60 degree angle with <3,4>

Vector <a,b>
Taking b=1 cos60=1/2

3a+4=(5/2 )sqrt(a^2+1)

36a^2+96a+64=25a^2+25
11a^2+96a=-39
(sqrt(11)a+48)^2= 2265
+- a=(sqrt(2265)-48)/sqrt(11)
 
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nameVoid said:
Find 2 unit vectors that make a 60 degree angle with <3,4>

Vector <a,b>
Taking b=1 cos60=1/2

3a+4=(5/2 )sqrt(a^2+1)

36a^2+96a+64=25a^2+25
11a^2+96a=-39
(sqrt(11)a+48)^2= 2265
+- a=(sqrt(2265)-48)/sqrt(11)

I don't see a question here.

But let me preempt it with a question of my own. You're asked to find unit vectors. Can you think of another condition you can impose on a and b to meet this requirement?

It's fine to do it the way you did (letting b = 1), but then you still have to find unit vectors in those directions. More work.
 

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