Find 2 Unit Vectors at Angle π/3 with <3,4> using Dot Product

[nameVoid]

find 2 unit vectors that make an angle of pi/3 with <3,4>

<3,4>dot<a,b>=5/2=3a+4b
b=5/8-3/4 a

|<a,b>|=1
such that
a^2+25/64+15/16a+9/16 a^2=1
25/16 a^2+15/16/ a=39/64
100a^2+60a=39
a^2+3/5a=39/100
(a+3/10)^2=48/100
a=(4sqrt(3)+-3)/10
so
b=5/8-(12sqrt(3)+9)/40
=(200-96sqrt(3)-72)/320
=(128-96sqrt(3))/320
=(32-24sqrt(3))/80
=(8-6sqrt(3))/20
=(4-3sqrt(3))/10
so far so good

b=5/8-(12sqrt(3)-9)/40
=(200-96sqrt(3)+72)/320
=(272-96sqrt(3))/320
=(68-24sqrt(3))/80
=(34-12sqrt(3))/40
=(17-6sqrt(3))/20 // correct me if I am wrong

 

[tiny-tim]

hi nameVoid!

your method looks basically correct (but i haven't checked it)

however, it's really complicated

try first finding the vector perpendicular to (3,4) and with the same magnitude

 

[nameVoid]

The solution for b2 is incorrect by the book

 

[tiny-tim]

b=5/8-3/4 a

|<a,b>|=1
such that
a^2+25/64+15/16a+9/16 a^2=1

shouldn't it be minus 15/16 ?

 

[nameVoid]

Ah yes my mistake

 

[HallsofIvy]

Looking at <3, 4>, you should have seen that <4, -3> is perpendicular without needing to write anything!