Question

Physics - Vectors class 11​

Answer 1

Answer :-

The Angle Between A and B is 150°

Explanation:

Given :-

$\sf{•|R|=\frac{|\vec{B}|}{2}}$

$\sf{•\frac{|R|}{|\vec{B}|}=\frac{\vec{1}}{2}}$

$\sf{•\vec{R}\perp {\vec{A}}}$

To Find :-

The angle between A and B

Solution :-

In ∆OXY

$\sf{\cos \theta= \frac{\vec{R}}{\vec{B}}=\frac{1}{2}}$

$\qquad\sf{\boxed{\blue{\theta = 60°}}}$

Thus Angle Between A & B is :-

$</p><p>\: \: \sf{\large{\boxed{90+60=\bold{\red{150°}}}}}$

Henceforth the angle between A & B is 150°

Answer 2

magnitude of resultant of A and B is given by,

$\sf{R = \sqrt{ (A² + B² + 2AB \: cos \theta) }}$

Question's Concept:-

Magnitude of resultant = half of Magnitude of B

$\sf{ \sqrt{ (A² + B² + 2A.B \: cos \theta) = \dfrac{B}{2} }}$

Taking square both sides,

$➢ \: \sf{A² + B² + 2A.B \: cos \theta = \dfrac{ B²}{4} }$

$➢ \: \sf{A² + 2AB \: cos \theta + \dfrac{3B²}{4}= 0 \xrightarrow{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }(1)}$

Also, A and R is perpendicular upon each other ,

$➟ \: \sf{A.(A + B) = 0}$

$➟ \: \sf{A.A + A.B = 0}$

$➟ \: \sf{A² + A.B \: cos \theta = 0}$

$\bigstar\boxed{ \sf{cos \theta= - \dfrac{A}{B} , put \: it \: in \: equation \: (1)}} \bigstar$

$➸ \: \sf{A² + 2A.B \bigg(- \dfrac{A}{B} \bigg) + \dfrac{ 3B²}{4 }= 0}$

$➸ \: \sf{A² - 2A² + \dfrac{ 3B²}{4 }= 0}$

$➸ \: \sf{ A = \dfrac{ \sqrt{ 3B}}{2 }}$

$➸ \: \sf{Now, cos \theta = \dfrac{-A}{B} = \dfrac{ - \sqrt{ -3B}}{{2B} }= \dfrac{- \sqrt{3}}{2 }}$

$➸ \: \sf{cos \theta= cos150° \implies\theta= 150°}$

Hence angle between A and B = 150°