Finding k such that u and v form {\pi}/3 Radians

  • Thread starter Thread starter phosgene
  • Start date Start date
  • Tags Tags
    Form Radians
Click For Summary
SUMMARY

The discussion centers on finding the value of k in vectors u = (1, k) and v = (3, 4) such that the angle between them is π/3 radians. The user correctly calculates the lengths of the vectors and their dot product, leading to the equation 39k² + 96k + 11 = 0. Upon applying the quadratic formula, two potential solutions for k are found: -2.341058209 and -0.1204802515. The conversation highlights the importance of verifying the sign of the dot product to determine the correct angle between the vectors.

PREREQUISITES
  • Understanding of vector operations, including dot product and magnitude
  • Familiarity with trigonometric functions, specifically cosine
  • Knowledge of solving quadratic equations
  • Basic understanding of angles in radians
NEXT STEPS
  • Study the properties of the dot product in relation to angles between vectors
  • Learn about the implications of squaring equations in algebra
  • Explore vector geometry and its applications in physics
  • Review the use of the quadratic formula and its applications in solving real-world problems
USEFUL FOR

Students studying linear algebra, particularly those working with vector mathematics and trigonometry, as well as educators seeking to clarify concepts related to vector angles and dot products.

phosgene
Messages
145
Reaction score
1

Homework Statement



Let u = (1,k) and v = (3,4). Find k such that the angle between u and v is [itex]{\pi}/3[/itex] radians.

Homework Equations



[itex]u{\bullet}v=||u|| ||v|| cos{\theta}=x_{1}x_{2}+y_{1}y_{2}[/itex]

[itex]||u||=\sqrt{x^2 + y^2}[/itex]

The Attempt at a Solution



Firstly I calculate the length of v and find an expression for the length of u:

[itex]||u||=\sqrt{1 + k^2}[/itex]

[itex]||v||=\sqrt{3^2 + 4^2}[/itex]
[itex]||v||=5[/itex]

Then I find an expression for the dot product:

[itex]u{\bullet}v=3+4k[/itex]

I plug my expressions for the dot product and lengths into the definition of the dot product, and set [itex]\theta[/itex] to [itex]\pi/3[/itex], giving me:

[itex]3+4k=5cos{(\pi/3)}\sqrt{1+k^2}[/itex]

as [itex]cos{(\pi/3)}=1/2[/itex], I can substitute [itex]cos{(\pi/3)}[/itex] in my equation for 1/2, giving:

[itex]3+4k=5/2\sqrt{1+k^2}[/itex]

I rearrange and expand brackets to get:

[itex]6+8k=5\sqrt{1+k^2}[/itex]

I then square both sides to get rid of the square root, then expand brackets:

[itex](6+8k)^2=25+25k^2[/itex]
[itex]36 + 96k + 64k^2=25+25k^2[/itex]

I move everything to one side:

[itex]39k^2+96k+11=0[/itex]

Using the quadratic formula, I get the answer that k= -2.341058209 or -0.1204802515. Clearly one of these is wrong (or both), as there can't be two angles in the same quadrant that make an angle of [itex]\pi/3[/itex] with a vector. But both of these values do satisfy the equation for the dot product of these two angles when the angle between them is [itex]\pi/3[/itex]. Have I done something wrong?

PS: I wasn't sure if this was the right forum, as both maths forums seem to be calculus-orientated. Sorry if it's not in the right place!
 
Physics news on Phys.org
hi phosgene! :smile:
phosgene said:
Using the quadratic formula, I get the answer that k= -2.341058209 or -0.1204802515. Clearly one of these is wrong (or both), as there can't be two angles in the same quadrant that make an angle of [itex]\pi/3[/itex] with a vector. But both of these values do satisfy the equation for the dot product of these two angles when the angle between them is [itex]\pi/3[/itex]. Have I done something wrong?

no, you did everything fine :smile:

however, when you squared both sides (which was correct), you automatically introduced an extra solution (for cos = -1/2, ie 2π/3), and you now need to check which of your two solutions is for π/3 ! :wink:

(ie just check that the dot-product is positive)
 


Thanks:) I just found out using google that the dot product is negative if the angle between the vectors is greater than 90 degrees. I had no idea that this was the case (it wasn't mentioned in the lectures). So it's crystal clear now. Thanks again!
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
9
Views
2K
Replies
1
Views
2K
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K