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
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
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.
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.
Thanks Damon and Reiny, it helped a lot. Really appreciate it :)
Of course, I'd be happy to help you with these questions!
1) To find ||u+v||, we are given the magnitudes of u and v, as well as their dot product. We can start by using the definition of the dot product of two vectors:
u · v = ||u|| ||v|| cos θ,
where θ is the angle between u and v.
Given that u · v = 1, ||u|| = 2, and ||v|| = √3, we can substitute these values into the equation:
1 = 2 * √3 * cos θ.
To find cos θ, we can rearrange the equation and solve for it:
cos θ = 1 / (2 * √3) = √3 / 6.
Now, we can use the law of cosines to find the magnitude of u + v:
||u+v||^2 = ||u||^2 + ||v||^2 - 2 ||u|| ||v|| cos θ.
Substituting the given values:
||u+v||^2 = 2^2 + (√3)^2 - 2 * 2 * √3 * (√3 / 6).
Simplifying the equation:
||u+v||^2 = 4 + 3 - 2 = 5.
Finally, we can take the square root of both sides to find the magnitude:
||u+v|| = √5.
Therefore, ||u+v|| is equal to the square root of 5.
2) To show that there are no vectors u and v such that ||u|| = 1, ||v|| = 2, and u · v = 3, we can use the Cauchy-Schwarz inequality.
The Cauchy-Schwarz inequality states that for any two vectors u and v in an inner product space:
|u · v| ≤ ||u|| ||v||.
Given ||u|| = 1, ||v|| = 2, and u · v = 3, we can substitute these values:
|3| ≤ 1 * 2,
which simplifies to:
3 ≤ 2.
However, this is a contradiction since 3 is not less than or equal to 2. Therefore, there are no vectors u and v such that ||u|| = 1, ||v|| = 2, and u · v = 3.
I hope this helps! Let me know if you have any further questions.
1) If ||u||=2, ||v||=root 3 and u dot v = 1, find ||u+v||
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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.