Question

what is the angle between the two vectors
(R+S) and (R-S) So that their resultant is√3R²+s²?​

Answer 1

Answer :-

60°

Given :-

• Two vectors (R+S) and (R-S) resultant is√3R²+s²

To find :-

• Angle between them

Solution :-

We have formula to find the resultant forces that is

$F_r = \sqrt{F{}^{2} + R {}^{2} + 2FRcos \theta }$

Substituting the values ,

$\sqrt{3R {}^{2} + S {}^{2} } = \sqrt{(R + S) {}^{2} + (R - S) {}^{2} + 2(R + S)(R - S)cos \theta }$

Squaring on both sides ,

$\bigg( \sqrt{3R {}^{2} + S {}^{2} } \bigg) {}^{2} = \bigg(\sqrt{(R + S) {}^{2} + (R - S) {}^{2} + 2(R+ S)(R- S)cos \theta } \bigg) {}^{2}$

$3R {}^{2} + S {}^{2} = (R + S) {}^{2} + (R- S) {}^{2} + 2(R {}^{2} - S {}^{2}) cos\theta)$

Simplifying the equation (r +s)² +(r-s)² by algebraic identity

(a + b)² +(a-b)² = 2a² + 2b²

$3R {}^{2} + S {}^{2} = 2R {}^{2} + 2S {}^{2} + 2(R{}^{2} - S {}^{2} )cos \theta$

Transpose 2R² + 2S² into L.H.S

$3R {}^{2} + S {}^{2} - 2R {}^{2} - 2S {}^{2} = 2(R {}^{2} - S {}^{2} )cos \theta$

$R {}^{2} - S {}^{2} = 2(R {}^{2} - S {}^{2} )cos \theta$

$2cos \theta = 1$

$cos \theta = \dfrac{1}{2}$

$cos \theta = cos60 ^{ \circ}$

So, theta = 60°

Angle between the given vectors is 60°