The position vector of four points A , B , C and D are a → , b → , 2 a → + 3 b → and a → - 2 b → , respectively. Express the vectors A C → , D B → , B C → and C A → in terms of a → and b → .

The position vector has one end at the origin and the other end describes it. To obtain the vector A C → , we need to consider vectors O A → as a → and O C → as 2 a → + 3 b → , where O is the origin. Vector A C → can be found by using the relation given as, A C → = O C → - O A → (i.e., Position vector of C - Position vector of A ) Therefore, we get, A C → = 2 a → + 3 b → - a → A C → = a → + 3 b → Similarly, we can find other vectors as, D B → = O B → - O D → D B → = b → - a → - 2 b → D B → = - a → + 3 b → B C → = O C → - O B → B C → = 2 a → + 3 b → - b → B C → = 2 a → + 2 b → C A → = O A → - O C → C A → = a → - 2 a → + 3 b → C A → = - a → - 3 b →

If A D , B E and C F are the medians of a △ A B C , prove that A D → + B E → + C F → = 0 .

In order to prove that A D → + B E → + C F → = 0 , we need to consider one median at a time and formulate the equations for the same, using the triangle law of vector addition. Considering A D → , we can draw the figure as below, taking into account all the vectors from vertex A . Now, using the triangle law of vector addition, we get, A B → + B D → = A D → A C → + C D → = A D → Adding both the above equations, we get, A B → + B D → + A C → + C D → = 2 A D → From the figure, vectors B D → and C D → have the same magnitude but are opposite in direction. So, they cancel each other. We have the first equation as, A B → + A C → = 2 A D → . . . 1 Now, following the same steps as above for median B E → , we get the second equation as, B A → + B C → = 2 B E → . . . 2 And, the third equation obtained by considering median C F → , is given as, C A → + C B → = 2 C F → . . . 3 Adding equations ( 1 ) , ( 2 ) and ( 3 ) , we get A B → + A C → + B A → + B C → + C A → + C B → = 2 A D → + 2 B E → + 2 C F → . . . 4 We know that, A B → = - B A → A C → = - C A → B C → = - C B → Substituting the above values in equation ( 4 ) and simplifying the equation, we get 0 = 2 A D → + B E → + C F → ⇒ A D → + B E → + C F → = 0 Hence, proved.