The angle between two vector of equal magnitude is 120* .Prove that th...
Proof:
Let us consider two equal magnitude vectors A and B.
Step 1: Draw a diagram
A diagram can help in visualizing the problem. Draw two vectors of equal magnitude, A and B, making an angle of 120 degrees between them.
Step 2: Draw a parallelogram
Draw a parallelogram using A and B as two adjacent sides.
Step 3: Draw the diagonal
Draw the diagonal of the parallelogram that passes through the point where A and B meet. This diagonal represents the resultant vector, R.
Step 4: Use the law of cosines
Apply the law of cosines to the triangle formed by A, B, and R.
cos(120) = (A^2 + B^2 - R^2) / (2AB)
Since A and B have equal magnitudes, we can simplify this equation to:
cos(120) = (2A^2 - R^2) / (2A^2)
Simplify further to get:
R^2 = 4A^2 - 4A^2cos(120)
R^2 = 4A^2 - 4A^2(-1/2)
R^2 = 4A^2 + 2A^2
R^2 = 6A^2
Step 5: Conclusion
We can see that the magnitude of the resultant vector, R, is equal to the square root of 6 times the magnitude of either A or B. Therefore, the magnitude of the resultant vector is always greater than the magnitude of either A or B.
The angle between two vector of equal magnitude is 120* .Prove that th...
Let the 2 vectors be A and B and their resultant be R. To prove :- A=B=R. . R^2 = A^2 + B^2 + 2ABCostheta . A^2 = B^2 = R^2 , R^2 = R^2 + R^2 + 2R^2 (Costheta) . R^2 = 2R^2 + 2R^2 (Costheta) Cos theta = (R^2 - 2R^2) / 2R^2 = (-R^2) / 2R^2 Cos theta = -1/2 theta = 120 degrees. . hence proved.
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