 Hi friends, I am Purva and today we will discuss the following question. Find the position vector of a point r which divides the line joining two points p and q whose position vectors are i cap plus 2 j cap minus k cap and minus i cap plus j cap plus k cap respectively in the ratio 2 is to 1 first internally, second externally. Let p and q be the two points represented by the position vectors o p and o q respectively with respect to the origin o and let r be a point which divides the point p and q internally in the ratio m is to n. Then the position vector of the point r which divides p and q internally in the ratio m is to n is given by vector o r is equal to m into vector b plus n into vector a upon m plus n that is we have vector r is equal to m into vector b plus n into vector a upon m plus n. Now if r divides p and q externally in the ratio m is to n then the position vector of the point r which divides line segment p q externally in the ratio m is to n is given by vector o r is equal to m into vector b minus n into vector a upon m minus n. This is called section formula and this is the key idea which we will use to solve this question. Let us now begin with the solution. Now we are given position vector of p is i cap plus 2 j cap minus k cap and position vector of q is minus i cap plus 2 j cap plus k cap. Therefore points p and q are given by 1 comma 2 comma minus 1 and minus 1 comma 2 comma 1 respectively. Now let r of x comma y comma z be the point which divides p q in the ratio 2 is to 1. Now in the first part we have to find the position vector of the point r which divides p q in the ratio 2 is to 1 internally. So we have the position vector of the point r which divides p q internally in the ratio 2 is to 1 is given by x is equal to 2 into minus 1 plus 1 into 1 upon 2 plus 1 y is equal to 2 into 1 plus 1 into 2 upon 2 plus 1 and z is equal to 2 into 1 plus 1 into minus 1 upon 2 plus 1 and this is by section formula that is we have x is equal to 2 into minus 1 minus 2 so we have minus 2 plus 1 upon 2 plus 1 and this is equal to minus 1 upon 3 y is equal to 2 into 1 is 2 plus 1 into 2 is 2 upon 2 plus 1 and this is equal to 4 upon 3 and z is equal to 2 into 1 is 2 1 into minus 1 is minus 1 upon 3 and this is equal to 1 upon 3. Therefore we have point r is minus 1 upon 3 comma 4 upon 3 comma 1 upon 3 hence the position vector of point r is minus 1 upon 3 i cap plus 4 upon 3 j cap plus 1 upon 3 k cap. Now in the second part we have to find the position vector of point r which divides p q in the ratio 2 is to 1 externally. So the position vector of r which divides p q externally in the ratio 2 is to 1 is given by x is equal to 2 into minus 1 minus 1 into 1 upon 2 minus 1 y is equal to 2 into 1 minus 1 into 2 upon 2 minus 1 minus 1 minus 1 and z is equal to 2 into 1 minus 1 into minus 1 upon 2 minus 1 and this is again by section formula that is we have x is equal to 2 into minus 1 gives minus 2 minus 1 into 1 gives 1 upon 2 minus 1 and this is equal to minus 3 y is equal to 2 into 1 gives 2 minus 1 into 2 gives 2 upon 2 minus 1 and this is equal to 0 and z is equal to 2 into 1 is 2 minus 1 into minus 1 is minus 1 upon 2 minus 1 and this is equal to 3. Therefore we have point r is given by minus 3 comma 0 comma 3 hence the position vector of point r is minus 3 i cap plus 0 j cap plus 3 k cap that is we have minus 3 i cap plus 3 k cap thus the answer for the first part is minus 1 upon 3 i cap plus 4 upon 3 j cap plus 1 upon 3 k cap and the answer for the second part is minus 3 i cap plus 3 k cap. Hope you have understood the solution. Bye and take care.