Question

Hi, I'm struggling to do two questions.
1) If ||u||=2, ||v||=root 3 and u dot v = 1, find ||u+v||

2)Show that there are no vectors u and v such that ||u|| = 1, ||v||=2 and u dot v = 3

Please help if you can. Thanks a million

Answer 1

1) If ||u||=2, ||v||=root 3 and u dot v = 1, find ||u+v||
-------------------------------
T is angle between u and v
|u||v| cos T = 1

2 sqrt 3 cos T = 1
so
cos T = 1/2sqrt3

from now on using u for |u} and v for |v|
component of u in direction of v = u cos T
component of sum perpendicular to v = u sin T

total in direction of v = v + u cos T
total perpendicular to v = u sin T


magnitude of resultant = sqrt(u^2 sin^2 t + v^2 + 2 u v cosT + u^2 cos^2T)

= sqrt( u^2 + 2 u v +v^2)

You know everything in there.

Answer 2

2)Show that there are no vectors u and v such that ||u|| = 1, ||v||=2 and u dot v = 3
1*2*cos T = 3
cos T = 1.5
cosines range from -1 to +1, not to 1 1/2

Answer 3

Make a diagram, with |u| and |v| as 2 and √3 and the angle between them as θ.

recall that |u| ∙ |v| = |u| |v|cosθ.

so 1 = 2√3cosθ
cosθ = 1/(2√3), (θ = appr. 73.2213º)

Now complete the parallelogram from your triangle diagram above, │u+v│ will be the diagonal.
Look at the triangle with sides │u│, │v│ and │u+v│ . The angle between them will be 180-θ

But remember that cos(180-θ) = -cosθ

by the cosine law
│u+v│^2 = │u│^2 + │v│^2 - 2│u││v│cos(180-θ)
= 4 + 3 - 2(2)(√3)(-1/(2√3)
= 9
so │u+v│ = 3

For the second one, you should get the cosθ greater than one, which is not possible.

Answer 4

magnitude of resultant = sqrt(u^2 sin^2 t + v^2 + 2 u v cosT + u^2 cos^2T)

= sqrt( u^2 + 2 u v cos T +v^2)

(I left the cos T out)

You know everything in there.

Answer 5

Thanks Damon and Reiny, it helped a lot. Really appreciate it :)