The maximum magnitude of the resultant of two vectors, `vec(P) and vec(Q) ("where" Pgt Q)` is x times the minimum magnitude ot the resultant. When the angle between `vec(P) and vec(Q)" ""is" " "theta` , the magnitude of the resultant is equal to half the sum of the magnitudes of the two vectors. Prove that, `cos theta =(x^2+2)/(2(1-x^2))*`

Explanation

P+Q =x(P-Q) (given ) or, `Q=(x-1)/(x+1)*P`
If R is the resultant of the two vectors when the angle between them is `theta`, then
`R^2=P^2+Q^2+2PQ cos theta`
Given, `R=(P+Q)/(2)=(P+(x-1)/(x+1)*P)/(2)=(xP)/(x+1)`
Putting in (1) ,
`therefore (x^2P^2)/((x+1)^2)=P^2+((x-1)^2)/((x+1)^2)P^2+2P^2((x-1)/(x+1)) cos theta`
`or, (x^2)/((x+1)^2)=1+((x-1)^2)/((x+1)^2)+(2(x-1))/(x+1) cos theta`
`or, (-(x^2+2))/((x+1)^2)=(2(x-1))/((x+1)) cos theta`
`or,cos theta =(-(x^2+2))/(2(x^2-1))=(x^2+2)/(2(1-x^2))*`