The greatest resultant of two vectors `vec P` and `vec Q` is (n) times their least reast resultant. Fiven
`| vec P | gt |vec Q|`. When `theta` is the angle between the two vectors, their resultant is half the sum of the two vectors. Show that,
`cos theta =- (n^(2) +2) // (n n^(2) -1)`.

Explanation

The greatest resultant of given two vectors
`= (P+Q)`
and the least resultant of given two vectors
` =(P-Q)`
According to questions ,
` (P+Q) =n (P-Q) or Q = (n-1)/(n-1)P`
If anle between the vectors is `theta` then the resultant is given by
`R^(2)=P^(2) + Q^(2) + 2 PQ cos theta` ...(i)
Given, ` R= (P+Q)/2 =1/2 [P+ (n-1)/(n+1) P]`
`=(nP)/((n+1))`
Putting the values in (i), we get
` (n^(2) p^(2))/((n_1))^(2) =P^(2) +((n-1)_^(2)/((n+1))^(2)`
`+ 2 P ((n-1))/((n+ 1)) P cos theta`
On solving we get,
` cos theta =- ((n^(2)+2))/((n^(2)-1))`.