Two ships P and Q are travelling at night with constant velocities. At midnight,P is at point with position vector (20i + 10j) km relative to fixed origin 0. At the same time, q is at the point with position vectro (14i - 6J)Km. Three hours later, P is at point with position vector (29i + 34j) Km . the ship Q travels with velocity 12jkm h^-1 . At time t hours after midnight, the position vector of P and Q are p km and qKm respectively. find a) the velocity of P, in terms of i and j. i got t=3= (20i + 10j)Km + 3( i + J) t = 3 = (20i + 10j)Km + 3( i + J) = (29 i + 34j) = 20 + 3i = 29 i = 3 then 10 + 3b = 34j so b = 8j Then velocity will be = 3i + 8j (ok now i need someone to see if this part is right please.) B) Then i got to find an expression for P and Q in terms of t, i and j I got P= (20i + 10j)+ ( 3i + 8J)t and For Q = (14i - 6j) + (12j)t then the questions goes : At time t hours after midnight, the disatnce between p and Q is d Km. c) by finding an expression For PQ show that d^2 = 25t^2 - 92t + 292. Now for this one i know that i have to take away Q from P but does it mean i take away the following : Q = (14i - 6j) + (12j)t - P= (20i + 10j)+ ( 3i + 8J)t Because i get the worng answer, so what i want from someone is to kindly, see if i have done part A and B right and to show me how to do C becuase i ahve been goign in circles, and no hope, so can someone please help thank you.