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

In summary, Homework Equations state that the angle between u and v is equal to the dot product of the lengths of the two vectors. If v is greater than u, then the dot product is negative and that is why the homework says to check if the dot product is positive.
  • #1
phosgene
146
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!
 
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  • #2
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)
 
  • #3


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!
 

What does it mean to find k such that u and v form {\pi}/3 Radians?

Finding k such that u and v form {\pi}/3 Radians means finding the constant value that, when multiplied by u and v, will result in an angle of {\pi}/3 radians between the two vectors. This value of k is often referred to as the scalar or dot product.

Why is it important to find k for u and v to form {\pi}/3 Radians?

Finding the value of k is important for understanding the relationship between two vectors and their angle. It can also be used in various mathematical calculations and in solving problems in physics, engineering, and other fields.

How is k calculated for u and v to form {\pi}/3 Radians?

The value of k can be calculated by using the formula k = u*v*cos({\theta}), where u and v are the magnitudes of the two vectors and {\theta} is the angle between them. This formula is derived from the definition of the dot product.

Can k be negative when u and v form {\pi}/3 Radians?

Yes, k can be negative when u and v form {\pi}/3 Radians. This indicates that the two vectors are pointing in opposite directions, resulting in an angle of {\pi} radians or 180 degrees between them.

What are some real-world applications of finding k for u and v to form {\pi}/3 Radians?

Some real-world applications of finding k include calculating the work done by a force on an object, determining the direction and magnitude of an electric field, and analyzing the forces acting on an object in three-dimensional space. It is also used in computer graphics and animation to create realistic movements and interactions between objects.

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