Find the difference in the value of the vector, if the two with equal magnitude �P� are inclined at an angle such that the difference in the magnitude of the resultant and magnitude of either vector is 0.732 times either vector. Also, the angle between them is increased to half of its initial.

\(left | underset{R}{rightarrow} right |\) – P = 0.732P

\(left | underset{R}{rightarrow} right | = sqrt{P^{2}+P^{2}+2P^{2}cos;theta }\)

\(left | underset{R}{rightarrow} right |=sqrt{2}Psqrt{1+cos;theta }\)

\(left | underset{R}{rightarrow} right |=2Pcosfrac{theta }{2}\)

\(2Pcosfrac{theta }{2}-P=0.732P\)

\(2;cosfrac{theta }{2}=1.732\)

\(cosfrac{theta }{2}=frac{sqrt{3}}{2}\)

Ө = 60°

\(left | underset{R’}{rightarrow} right |=sqrt{P^{2}+P^{2}-2PP;cos90^{circ}}\) \(left | underset{R’}{rightarrow} right |=sqrt{2}P\)

BOOK

Free Class