Question

let there be two vectors of equal magnitude and inclind at an angle of 120 the ratio of magnitude their sum and difference is​

Answer 1

Given :

▪ Angle between two equal vectors = 120°

To Find :

▪ The ratio of their sum and difference.

Solution :

➡ By triangle law or parallelogram law of vector addition, the magnitude of resultant vector R at two vectors P and Q inclined to each other at angle Φ , is given by

☞ R^2 = P^2 + Q^2 + 2PQcosΦ

➡ Formula of vector substraction :

☞ R^2 = P^2 + Q^2 - 2PQcosΦ

____________________________

ATQ,

• P = Q = x

$\implies\sf\:R^2=x^2+x^2+2(x)(x)\cos120\degree\\ \\ \implies\sf\:R^2=2x^2+2x^2(-\dfrac{1}{2})\\ \\ \implies\sf\:R^2=2x^2-x^2\\ \\ \implies\sf\:R^2=x^2\\ \\ \implies\underline{\boxed{\bf{\pink{R=x}}}}\\ \\ \leadsto\sf\:{R'}^2=x^2+x^2-2(x)(x)\cos120\degree\\ \\ \leadsto\sf\:{R'}^2=2x^2-2x^2(-\dfrac{1}{2})\\ \\ \leadsto\sf\:{R'}^2=2x^2+x^2\\ \\ \leadsto\sf\:{R'}^2=3x^2\\ \\ \leadsto\underline{\boxed{\bf{\orange{R'=\sqrt{3}x}}}}\\ \\ \longrightarrow\underline{\boxed{\bf{\purple{\large{R:R'=1:\sqrt{3}}}}}}$

Answer 2

QuEsTi0N -

Let there be two vectors of equal magnitude , inclined at an angle of 120 .

Find the ratio of magnitude of their sum and difference .

S0LUTI0N -

Let there be two vectors of equal magnitude , inclined at an angle of 120 .

So, the given angle of inclination is 120°

These Vectors after equal in Magnitude .

So, let them be $\vec{a} \: and \: \vec{a}$

If there are two Vectors P and Q inclined at an angle of 120°

Sum Of Vectors -

$\sf{ Sum \: = \: \sqrt{ { \vec{P}} ^2 + {\vec{Q}}^2 + 2{\vec{P}}{\vec{Q}} \cos( \theta ) } }$

Substituting the given Values in the formula -

$\sf{ Sum \: = \: \sqrt{ ( \vec{a}) ^2 + ( \vec{a})^2 + 2( \vec{a}) . ( \vec{a}). \cos( 120 ) } } \\ \\ \sf{ \cos( \120 ) = - \cos( 30 ) = - \dfrac{1}{2} } \\ \\ \sf{Sum = \: \sqrt{ 2( \vec{a}) ^{2} - 2 \times \dfrac{1}{2} . ( {\vec{a})^2 } } } \\ \\ \sf{ Sum = \vec{a} }$

Difference Of Vectors -

$\sf{ Difference \: = \: \sqrt{ { \vec{P}} ^2 + {\vec{Q}}^2 - 2{\vec{P}}{\vec{Q}} \cos( \theta ) } }$

Substituting the given Values in the formula -

$\sf{ Difference \: = \: \sqrt{ ( \vec{a}) ^2 +( \vec{a})^2 - 2( \vec{a}) . ( \vec{a}) \cos( 120 ) } } \\ \\ \sf{ \cos( \120 ) = - \cos( 30 ) = - \dfrac{1}{2} } \\ \\ \sf{Difference = \: \sqrt{ 2( \vec{a})^2 + 2 \times \dfrac{1}{2} ( \vec{a})^2 } } \\ \\ \sf{ Difference = \sqrt{ 3( \vec{a})^2 } } \\ \\ \sf{ Difference = ( \vec{a}) \times \sqrt{3} }$

$\sf{\bold{ Required \: Ratio \: = \: \dfrac{Sum}{Difference} }} \\ \\ \sf{ = \dfrac{( \vec{a})}{ ( \vec{a}) . \sqrt{3} } } \\ \\ \\ \sf{ = \dfrac{1}{ \sqrt{3}} } ........ [ A ]$