Example – 9 SECTION FORMULA
Let A( ) and B( ) be two fixed points. Find the position vectors of the points lying on the (extended) line AB
which divide the segment internally and externally in the ratio m : n.
Solution: We consider internal division; the external division case follows analogously.
Let C( ) be the point which divides AB internally in the ratio m : n.
We have,
Similarly, the point D( ) which divides AB externally in the ratio m : n is given by
A particular case of internal division is the mid-point of A() and B(): the mid-point is ,(( +) /2)
Example – 10
Show by vector methods that the angular bisectors of a triangle are concurrent and find the position of the point of concurrency in terms of the position vectors of the vertices.
Solution: Let the vertices of the triangle by A(), B() andC(). We use the geometrical fact that an angle bisector divides the opposite side in the ratio of the sides containing the angle.
We thus have
(BD/DC) = c/b
Thus, D is given by (the internal division formula):
In ΔABD, since BI is the angle bisector, we have
Thus, we now have the position vectors of A and D we know what ratio I divides AD in. I can now be easily determined using the internal division formula:
 …….(1)
The symmetrical nature of this expression proves that the bisector of C will also pass through I. The angle bisectors will therefore be concurrent at I, called the incentre. The position vector of the incentre is given by (1).
Example – 11
In a parallelogram ABCD, let M be the mid-point of AB. AC and MD meet in E. Prove that both AC and MD are trisected at E.
Solution:
There.s no loss of generality in assuming A to be the origin ( ). Let B and C have the position vectors and . The position vector of D is given by  where
D is therefore the point - .
Since M is the mid-point of AB, M's position vector is /2. Let us first find the position vector of a point E' which lies on AC and trisects it. We.ll then show that same point lies on DM and trisects on DM and trisects it too, proving the stated assertion.
A point E’’ which trisects MD is
Since E’ and E’’ are the same-point, say E, we see that AC and MD are trisected at E.
Example – 12
Prove that the lines joining the vertices of a tetrahedron to the centroids of the opposite faces are concurrent.
Solution: It is in this example that the powerful nature of vector algebra will become apparent; this is a 3-D problem and any other methods of proving the assertion will be extremely cumbersome. For those of you who are more mathematically inclined, you can try proving the assertion using already known methods. First of all, we need to know how to write the position vector of the centroid of a triangle in terms of the position vectors of its vertices.
Since D is the mid-point of BC, we have
Dº (( + ) /2 )
Since AG : GD = 2 : 1, we can now determine G:
We now consider a tetrahedron, say, with the vertices A(), B(), C( )and D().
Consider the centroid of ΔBCD, say G, which will be given by
Now, pause and think about the assertion we are required to prove. If the four lines are indeed concurrent, the point of concurrency must be given by a position vector which has a .symmetrical. expression with respect to all the four vertices. Can you find a point on the segment AG which has a symmetrical expression with respect to the four vertices? A little thought will show that the answer is yes: consider the point P which divides AG in the ratio 3: 1. P will be given by
It is immediately apparent now that P lies on each of the four lines joining the vertices to centroids of the opposite faces! In addition, we.ve also been able to find the position vector of P in terms of the position vectors of the four vertices of the tetrahedron.
Example – 13
In a quadrilateral PQRS,  = , = and  = - . If M is the mid-point of QR and X is a point on SM such that SX : SM = 4 : 5, prove that P, X and R collinear.
Solution: Since no position vectors have been specified in the question (only the sides have been specified), there is no loss of generality in assuming that P is the origin .
We have,
Thus,  = (3/5) 
implying that P, X and R are collinear.
Example – 14
It is known that in a ΔABC with centroid G, circumcentre O and orthocentre H,
OG:GH =1: 2
Let P be any point in the plane of ΔABC. Prove the following assertions:
Solution:
(Since G lies on AD and divides it in the ratio 2 : 1)
(c) For any arbitrary point P in the plane of ΔABC, we have
=  + 2 = 3
Go over the solution again if you find any part of it confusing.
Example – 15
Justify the following tests for collinearity and coplanarity
(a) Three points with position vectors , , are collinear iff there exist scalars p, q, r not all zero such that
(b) Four points with position vectors , , , are coplanar iff there exist scalars p, q, r, s not all zero such that
where p + q + r + s = 0
Solution: There is nothing new in these tests; we.ve already seen their justification in the discussion preceeding the examples. These tests are just the same facts put into slightly different language.
(a) Since p + q + r = 0, we have r = - ( p + q )
Assume r ¹?0
Now, p + q + r =
This implies that is the position vector of a point which divides the points and in the ratio
p : q. Thus, the points , and are collinear.
Now, we prove the other way implication, i.e, we first assume that points , and are collinear:
This completes our proof.
(b) Again, first assume the existence of scalars p, q, r, s such that
p + q + r + s = 0
Þ s = -( p + q + r )
Now,
Assuming p ¹?0 we have
That ( - ) can be written as a linear combination of the vectors ( - )and ( - ) implies
that ( - ), ( - )
and ( - ) are coplanar.
Þ Points with position vectors ,, , are coplanar.
Now lets prove the other way implication. Assume that A(), B( ),C( )and D( ) are coplanar points. Thus, there must exist scalars l,m such that
where p + q + r + s = (l +m -1)+ (1)+ (-l )+ (-m ) = 0
This completes the second proof.
Example – 16
If any point O inside or outside a tetrahedron ABCD is joined to the vertices and AO, BO, CO, DO are produced so as to cut the planes of the opposite faces in P, Q, R, S respectively, prove that
(OP/AP) + (OQ/BQ) + OR/CR) + OS/DS) = 1
Solution: Assume O to be the origin, and the position vectors of A, B, C, D to be , , , respectively. Since , , , are non-coplanar vectors, we must have scalars l1 ,l2 ,l3 ,l4 such that (Page 17)
l1 +l2 +l3 +l4 = 0 ... (1)
Since AO is produced to meet the plane of the opposite face in P,  and  must be collinear vectors. Thus,
This when used in (1) gives
However, since B, C, D and P will be coplanar, we have, using the result of the last example,
We similarly have,
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