Question

Modulus of Vector A equals to modulus of vector B and the value of modulus Vector A and modulus of Vector B is 'a' then find out the resultant of the given vectors.​

Answer 1

Answer:

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

Explanation:

Given

$| \vec{A}| = a \\ | \vec{B}| = a$

We have to find

Resultant Vector = ???

$R = \sqrt{ {a}^{2} + {b}^{2} + 2ab \cos \theta } \\ = > R = \sqrt{ {a}^{2} + {a}^{2} + 2 \times a \times a \cos \theta } \\ = > R = \sqrt{ {2a}^{2} + {2a}^{2} cos \theta} \\ = > R = \sqrt{ {2a}^{2}(1 + \cos\theta ) } \\ = > R = \sqrt{ {2a}^{2}.2( {\cos \frac{ \theta}{2}})^{2} } \\ = > R = \sqrt{4 {a}^{2}. {(\cos \frac{ \theta}{2})}^{2} }$

$= > \boxed{R = 2a \cos( \frac{ \theta}{R} ) }$

Used identity in the solution

$\boxed{ \cos 2 \theta = 2 { \cos }^{2} \theta - 1}$

Answer 2

Answer:-

$\mathsf{R = 2aCos\dfrac{\theta}{2}}$

Given :-

$|\vec{A}| = |\vec{B}| = a$

To find :-

The resultant of the given vectors.

Solution:-

The formula of resultant of two vector is given by :-

→${R = \sqrt{|\vec{A}|^2+ |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}|Cos \theta}}$

• put the magnitude of the vectors.

→$\mathsf{R = \sqrt{a^2 + a^2 + 2 a\times a Cos \theta }}$

→$\mathsf{R = \sqrt{2a^2 + 2a^2Cos \theta }}$

• Take 2a² as common.

→$\mathsf{R = \sqrt{2a^2 (1+Cos\theta)}}$

→$\mathsf{R = a \sqrt{2(1+Cos\theta)}}$

→$\mathsf{R = a \sqrt{2(2Cos^2\dfrac{ \theta}{2}})}$

→$\mathsf{R = a \sqrt{4 Cos^2 \dfrac{\theta}{2}}}$

→$\mathsf{ R = a \times 2 Cos \dfrac{\theta}{2}}$

→$\mathsf{R = 2aCos\dfrac{\theta}{2}}$

hence,

The magnitude of resultant is

$\mathsf{2aCos\dfrac{\theta}{2}}$

• By using Sub multiple formula

$\huge { \boxed{Cos \theta + 1 = 2Cos^2 \dfrac{\theta}{2}}}$