Question

$\LARGE{ \orange\bigstar \; \red {\underline{ \mathcal{Q}\sf{UESTION :}} }}$
Two vectors having equal magnitudes A make an angle theta with eahc other. Find the magnitude and direction of the resultant.

$\LARGE{ \orange\bigstar \; \green {\underline{ \mathcal{A}\sf{NSWERS :}}}}$
$\sf{1. \; 2Acos \dfrac{\theta}{2}}$
$\sf{2. \; \dfrac{\theta}{2}}$

$\LARGE{ \orange\bigstar \; \blue {\underline{ \mathcal{R}\sf{EQUEST :}}}}$
Please explain it step-by-step and please explain the trigonometry part in detail and mention the trigonometric identities that you have used in solving.​​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

$\red{\rm :\longmapsto\:\vec{a} \: and \: \vec{b} \: are \: two \: vectors \: inclined \: at \: angle \: \theta}$

Further given that,

$\red{\rm :\longmapsto\: |\vec{a}| = A}$

and

$\red{\rm :\longmapsto\: |\vec{b}| = A}$

Let assume that magnitude of resultant vector be R.

We know,

$\red{\rm :\longmapsto\:R = \sqrt{ { |\vec{a}| }^{2} + { |\vec{b}| }^{2} + 2 |\vec{a}| \: |\vec{b}| \: cos\theta}}$

So, on substituting the values, we get

$\rm :\longmapsto\:R = \sqrt{ {A}^{2} + {A}^{2} + 2A \times A \times cos\theta}$

$\rm :\longmapsto\:R = \sqrt{2 {A}^{2} + {2A}^{2} cos\theta}$

$\rm :\longmapsto\:R = \sqrt{2 {A}^{2}(1 + cos\theta)}$

We know,

$\boxed{ \bf \:1 + cos2x = {2cos}^{2}x}$

So, using this, we get

$\rm :\longmapsto\:R = \sqrt{2 {A}^{2} \times {2cos}^{2} \dfrac{\theta}{2} }$

$\rm :\longmapsto\:R = \sqrt{4{A}^{2} {cos}^{2} \dfrac{\theta}{2} }$

$\red{\rm :\longmapsto\:R = 2Acos\dfrac{\theta}{2}}$

Now,

Let assume that,

$\red{\rm :\longmapsto\:Resultant \: R \: makes \: an \: angle \: x \: with \: \vec{a}}$

So,

$\boxed{ \rm \:tanx = \frac{ |\vec{b}| \: sin\theta }{ |\vec{b}| cos\theta + |\vec{a}| } }$

So, on substituting the values, we get

$\rm :\longmapsto\:tanx = \dfrac{Asin\theta}{Acos\theta + A}$

$\rm :\longmapsto\:tanx = \dfrac{Asin\theta}{A(cos\theta + 1)}$

We know,

$\boxed{ \bf \:sin2x = 2sinx \: cosx}$

and

$\boxed{ \bf \:1 + cos2x = {2cos}^{2}x}$

$\rm :\longmapsto\:tanx = \dfrac{2sin\bigg[\dfrac{\theta}{2} \bigg]cos\bigg[\dfrac{\theta}{2} \bigg]}{2 {cos}^{2}\bigg[\dfrac{\theta}{2} \bigg] }$

$\rm :\longmapsto\:tanx = tan\bigg[\dfrac{\theta}{2} \bigg]$

$\bf\implies \:x = \dfrac{\theta}{2}$