Find the sum of the vectors u and v if theta is the angle between them

AI Thread Summary
The discussion revolves around finding the sum of two vectors, u and v, with given magnitudes and an angle between them. The user believes their math book's answer is incorrect, as they calculated the sum using the cosine law and arrived at approximately 9.1, while the book states 11.6. They explore the possibility that the book's answer might actually represent the difference between the vectors rather than their sum. Suggestions include using graphical methods or resolving vectors into components to find the resultant. The user remains uncertain about the correct approach, especially given the obtuse angle involved.
DevilTemptations
Messages
3
Reaction score
0
:eek: After spending about an hour on this problem, I've become completely convinced that my math book is wrong in the answer it provides. Am I wrong or right? This is from the Harcourt Geometry and Discrete Mathematics Text.

Question:
Find the sum of the vectors u and v if theta is the angle between them.

Given:
u=3 (magnitude)
v=10 (magnitude)
theta=115 degrees

Obviously, the angles in a parallelogram equal to 360. This means the other vertexs are 65 degrees. Then I would use the cosine law to solve for the sum of the vectors. (I wish I could draw a diagram but I can't). So...

u+v^2=(10)^2+(3)^2-2(10)(3)Cos 65 degrees
therefore, u+v=9.1 (approx)

However, the back of the book says u+v=11.6

I've found that to get 11.6, instead of putting Cos 65, I'd have to put Cos 115 but that would not follow according to the laws. Another way is instead of finding the sum, I'd actually be finding the DIFFERENCE between the vectors.

Then I thought maybe I'm doing something wrong... because in the previous question, the angle between the vectors was 70 degrees. The angle between the vectors was not REALLY 70, instead it was divided into two angles and 70 degrees was one of them. Thus, an exterior angle from 70 which is 110 is given... so this proves the vertex opposite from it would be 70. Use Cosine law, I got the right answer.

However, this is an obtuse angle. There is obviously a different method but maybe I don't know it.

I'm confused

Any ideas within the next hour or two would be appreciated...
 
Physics news on Phys.org
Well, one method is to make a scaled drawing and determine the resultant graphically. The other is to work with components.Consider the u vector along the x-axis. You would then resolve v vector into it's x-and y components. Add up the x and y components to get the components of the resultant. It's magnitude can determined by applying pythagoras' theorem to the components. The direction of the resultant is determined with the tangent of it's (the resultant's) components.
 
If the angle between the vetors is greater than 90 degrees, then the length of their sum will be smaller than the length of either one. Looks to me like the book's answer is the difference of the two vectors.
 

Attachments

  • a.jpg
    a.jpg
    1.7 KB · Views: 1,070
Last edited:
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top