Question

Two vectors having equal magnitude A make an angle thetha with each other. Find the magnitude and direction of resultants.

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Answer 1

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Answer 2

Q. Two vectors having equal magnitude A make an angle thetha with each other. Find the magnitude and direction of resultants.

Ans:-

$magnitude = 2a \cos( \frac{ theta }{2} )$

$direction \: of \: resultant = \alpha = \frac{theta}{2}$

Explanation:-

The magnitude of resultant will be

$b = \sqrt{ {a}^{2} + {b}^{2} + 2ab \cos(theta) }$

Since, the magnitude of both vectors are equal A=B.

$b = \sqrt{ {a}^{2} + {a}^{2} + 2 {a}^{2} \cos(theta) }$

$= \sqrt{2 {a}^{2}(1 + \cos(theta) }$

$= \sqrt{4 {a}^{2} {cos}^{2} \frac{theta}{2} }$

The resultant will make an angle alpha with the first vector where

$\tan( \alpha ) = \frac{a \sin(theta) }{a + a \cos(theta) }$

$= \frac{2a \: \sin( \frac{theta}{2} ) \cos( \frac{theta}{2} ) }{2a {cos}^{2} \frac{theta}{2} }$

$= \tan( \frac{theta}{2} )$

$\alpha = \frac{theta}{2}$

Thus, the resultant of two vectors bisect the angle between them.