OP P , we have Using the properties of vector addition, the above equation becomes i.e. x i y j z k x i y j z k ˆ x x i y y z z k The magnitude of vector is given by | | = x x y y z z Fig . Example Find the vector joining the points P( , , ) and Q(– , – , – ) directed from P to Q. Solution Since the vector is to be directed from P to Q, clearly P is the initial point and Q is the terminal point.
So, the required vector joining P and Q is the vector given by ( ) ( ) ( ) −− + −− + −− i.e. . . .
Section formula Let P and Q be two points represented by the position vectors , respectively, with respect to the origin O. Then the line segment joining the points P and Q may be divided by a third point, say R, in two ways – internally (Fig . ) and externally (Fig . ).
Here, we intend to find the position vector for the point R with respect to the origin O. We take the two cases one by one. Case I When R divides PQ internally (Fig . ).
If R divides such that where m and n are positive scalars, we say that the point R divides internally in the ratio of m : n . Now from triangles ORQ and OPR, we have Therefore, we have (Why?) or (on simplification) Hence, the position vector of the point R which divides P and Q internally in the ratio of m : n is given by Fig . Case II When R divides PQ externally (Fig . ).
We leave it to the reader as an exercise to verify that the position vector of the point R which divides the line segment PQ externally in the ratio m : n PR i.e. QR m n is given by Remark If R is the midpoint of