 Hi and welcome to the session. Let's work out the following question. The question says the resultant of forces vector p and vector q acting at a particle is vector r. If vector q is doubled vector r is doubled. If vector q is reversed vector r is again doubled prove that p is to q is to r is equal to square root of 2 is to square root of 3 is to square root of 2 So, let's start with the solution to this question first of all let the angle between two forces q v theta this r is resultant of p and q therefore r square is equal to p square plus q square plus 2 p q cos theta and this we call equation 1 Now in the question, it's given that if q is doubled then r is also doubled so we will have 2r the whole square is equal to p square plus 2q the whole square plus 2p into 2q into cos theta This implies 4r square is equal to p square plus 4q square plus 4pq cos theta and this we call equation 2 now it's also given that if q is reversed r is again doubled so we will have now 2r the whole square is equal to p square plus q square plus 2p q cos pi minus theta This implies that 4r square is equal to p square plus q square minus 2p q cos theta and this we call equation 3 now adding equation 1 and equation 3 we get 5r square is equal to 2p square plus 2q square and this we call equation 4 Now from equation 1 if we multiply this equation by 2 we have 2r square is equal to 2p square plus 2q square plus 4pq cos theta and this we call equation 5 Now subtracting equation 5 from equation 2 that is equation 2 minus equation 5 will give us 4r square is equal to p square plus 4q square plus 4pq cos theta equation 5 is 2r square is equal to 2p square plus 2q square plus 4pq cos theta now subtracting them we get 2r square is equal to minus p square plus 2q square this is 0 now rewriting equation 4 and equation 6 this is equation 6 we have 2p square plus 2q square minus 5r square equals to 0 and p square minus 2q square Plus 2r square is equal to 0 we call this equation 7 and we call this equation 8 Now solving equation 7 and equation 8 by cross-pulteplication We have p square divided by 4 minus 10 is equal to q square divided by minus 5 minus 4 is equal to r square divided by minus 4 minus 2 this implies p square by minus 6 is equal to q square upon minus 9 is equal to r square upon minus 6 Now multiplying throughout by minus 3 we get p square by 2 is equal to q square by 3 is equal to r square by 2 this implies p by square root 2 is equal to q by square root 3 is equal to r by square root 2 This implies that p is to q is to r is equal to root 2 is to root 3 is to root 2 Now this is what we were supposed to prove in this question I hope that you understood the solution and enjoyed the session. Have a good day