Question

2 vectors having equal magnitude s A make an angle theta with each other. Find the magnitude and direction of the resultant.

Answer 1

Answer:

magnitude

$|a + b| = \sqrt{ {a}^{2} + {b}^{2} + 2ab \cos( \theta) } \\ \\since \: |a| = |b| \\ \\ = \sqrt{a {}^{2} + {a}^{2} + 2(a)(a) \cos\theta } \\ \\ = \sqrt{2 {a}^{2}(1 + cos \theta) } \\ \\ = \sqrt{2 {a}^{2}.2 \cos {}^{2} ( \frac{ \theta}{2} ) } \\ \\ = 2acos \frac{ \theta}{2}$

Another way

Assume that vector a is along x axis and b is at angle theta with it . magnitude of b is a as well .

Take components of b along x and y axis .

the resultant vector is

$(a + a \cos( \theta) )i + a \sin( \theta) j \\ \\ |a + b| = \sqrt{ ({a + acos \theta)}^{2} + {a}^{2} \sin {}^{2} ( \theta) } \\ \\ |a + b| = \sqrt{ {a}^{2} (1 + 2cos \theta +cos {}^{2} ( \theta) + sin {}^{2} ( \theta) } \\ \\ = \sqrt{ {a}^{2} (1 +2 cos \theta + 1)} \\ \\ = a \sqrt{2(1 + cos \theta)}$

Again same results

Angle of resultant

$\tan( \alpha ) = \frac{a \sin( \theta) }{a + a \cos( \theta) } \\ \\ \tan( \alpha ) = \frac{ \sin( \theta) }{1 + \cos( \theta) } = \tan( \frac{ \theta}{2} ) \\ \\ \alpha = { \theta \over2}$