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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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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))? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 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))? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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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