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The maximum resultant of two vectors bar(A) and vec B(AgtB) is n times their lesultant.If theta be the and between the vectors and their resultant be half the sum of the vectors.Then show that cos theta=-(n^(2) 2)/(2(n^(2)-1))?
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The maximum resultant of two vectors bar(A) and vec B(AgtB) is n times...
Solution:
Given:
- The maximum resultant of two vectors bar(A) and vec B(AgtB) is n times their resultant.
- The angle between the vectors is theta.
- The resultant of the vectors is half the sum of the vectors.
To prove:
cos theta=-(n^(2) 2)/(2(n^(2)-1))
Solution:
Let vector A and vector B be two vectors, such that the angle between them is theta.
Finding the Resultant:
The resultant of the two vectors, R = (A + B)/2.
Given that the resultant, R is half the sum of the vectors, i.e., R = (A + B)/2, we can write:
A + B = 2R
Finding the Maximum Resultant:
The maximum resultant occurs when the two vectors are in the same direction, i.e., when they are parallel to each other.
Let the magnitude of vector A be a, and the magnitude of vector B be b.
Then, the maximum resultant, R(max) = a + b
Given that R(max) is n times the resultant of the vectors, R, we can write:
R(max) = nR
Substituting the values of R and R(max), we get:
a + b = 2nR
(a + b)/2 = nR/2
But, we also know that R = (A + B)/2
Substituting the value of R in the above equation, we get:
(a + b)/2 = n(A + B)/4
(a + b) = 2n(A + B)/2
(a + b) = 2n(A + B)/2
(a + b) = n(A + B)
a + b = na + nb
a(n - 1) = b(n - 1)
a/b = (n - 1)/(n - 1)
a/b = 1
Therefore, the two vectors are of equal magnitude.
Finding cos theta:
Let the magnitude of the vectors A and B be a.
Then, the maximum resultant, R(max) = 2a
Also, the resultant of the vectors, R = a
Substituting these values in the equation, R(max) = nR, we get:
2a = na
a(2 - n) = 0
This gives us two possible values of n:
n = 2, or n = -1
But, n cannot be negative.
Therefore, n = 2.
Substituting this value of n in the equation for cos theta:
cos theta=-(n^(2) 2)/(2(n^(2)-1))
cos theta = -(2^2 - 2)/(2(2^2 - 1))
cos theta = -6/14
cos theta = -3/7
Therefore, cos theta = -(n^(2) 2)/(2(n^(2)-1)) is proved.
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The maximum resultant of two vectors bar(A) and vec B(AgtB) is n times their lesultant.If theta be the and between the vectors and their resultant be half the sum of the vectors.Then show that cos theta=-(n^(2) 2)/(2(n^(2)-1))?
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