Algebra

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

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  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.

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  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

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  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.

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  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.

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

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