Presentation on theme: "Addition of vectors (i) Triangle Rule [For vectors with a common point] C B A."— Presentation transcript:
1Addition of vectors(i) Triangle Rule [For vectors with a common point]CBA
2(ii) Parallelogram Rule [for vectors with same initial point] DCBA
3(iii) Extensions follow to three or more vectors p+q+rqp
4First we need to understand what is meant by the vector – a SubtractionFirst we need to understand what is meant by the vector – aa– aa and – a are vectors of the same magnitude, are parallel, but act in opposite senses.
8OP is a position vector of a point P. We usually associate p with OP Position VectorsRelative to a fixed point O [origin] the position of a Point P in space is uniquely determined by OPPpOP is a position vector of a point P. We usually associate p with OPO
11An Important technique To establish or express the co-linearity of three points [Lie in a straight line]Choose any two line segments, AB, AC or BC.For the points to be co-linear AB, AC or BC must lie in the same directionExampleGiven OA = p, OC = q and OB = 2p – q , show that A, B and C are co-linear.AB = AO + OB= – p + (2p – q)= p – qBABC = BO + OC= – (2p – q) + q= 2q – 2p= –2(p – q ) = –2ABCHence A , B and C are co-linear.OAB &BC are parallel (even though in opposite directions) and have a common point B
12Example M, N, P and Q are the mid-points of OA, OB, AC and BC. OA = a, OB = b, OC = c(a) Find, in terms of a, b and c expressions for(i) BC (ii) NQ (iii) MP(b) What can you deduce about the quadrilateral MNQP?a)BC = BO + OC= c – b(ii) NQ = NB + BQb= ca(ii) MP = MA + APc= cMNPQ is a parallelogram as NQ and MP are equal and parallel.
13The diagram shows quadrilateral OABC. OA = a, OC = c and OB = 2a + c (a) Find expressions, in terms of a and c, for(i) AB (ii) CB(iii) What kind of quadrilateral is OABC?Give a reason for your answer.(b) Point P lies on AC and AP = AC.(i) Find an expression for OP in terms of a and c.Write your answer in its simplest form.(ii) Describe, as fully as possible, the position of P.a)AB = AO + OB= a + c(ii) CB = CO + OBP= 2aTrapezium :CB is parallel to OA.b)OP = OA + AP= a + c(ii) OB = 3 x (OP)They are parallel and have a common point, hence O, P & B are co-linear.
14ExampleOACB is a parallelogram with OA = a and OB = bM is the midpoint of ACP is the intersection of OM with ABObtain the position vector of MGiven that AP = kAB use the triangle OAP to obtain an expression for OP in terms of a, b and k.Deduce the position vector P.
15A C a M P O B b OM = OB + BM = b + a AP = kAB OP = OA + AP = a + kAB ExampleOACB is a parallelogram with OA = a and OB = bM is the midpoint of ACP is the intersection of OM with ABObtain the position vector of MGiven that AP = kAB use the triangle OAP to obtain an expression for OP in terms of a, b and k.Deduce the position vector P.OM =OB + BM= b + aAP = kABOP = OA + AP= a + kAB= a + k(b – a)= (1 – k)a +kbOCAbaB(iii) OP = hOMOP = h(b + a)MPa 1 – k = hb k = hHence 1 = h h = OP = a + b