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Addition of vectors (i) Triangle Rule [For vectors with a common point] C AB BC AC B A (ii) Parallelogram Rule [for vectors with same initial point] D C AB AC AD B A (iii) Extensions follow to three or more vectors r p+q+r q p Subtraction First we need to understand what is meant by the vector – a a –a a and – a are vectors of the same magnitude, are parallel, but act in opposite senses. A few examples a b b–a a b q p p–q –q Which vector is represented by p – q ? B CB = CA + AB = - AC + AB = AB – AC C A Position Vectors Relative to a fixed point O [origin] the position of a Point P in space is uniquely determined by OP P p O OP is a position vector of a point P. We usually associate p with OP A very Important result! B AB = b - a b A a O The Midpoint of AB A M OM = ½(b + a) a B b O An 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 direction Example Given OA = p, OC = q and OB = 2p – q , show that A, B and C are co-linear. A AB = AO + OB BC = BO + OC = – p + (2p – q) = – (2p – q) + q =p–q = B C 2q – 2p = –2(p – q ) = –2AB O AB &BC are parallel (even though in opposite directions) and have a common point B Hence A , B and C are co-linear. Example M, N, P and Q are the mid-points of OA, OB, AC and BC. B OA = a, OB = b, OC = c a) (i) BC = BO + OC = c–b A (ii) NQ = NB + BQ N M a Q b = c P (ii) MP = MA + AP O c 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? = c MNPQ is a parallelogram as NQ and MP are equal and parallel. The diagram shows quadrilateral OABC. OA = a, OC = c and OB = 2a + c B C Not drawn accurately a) (i) AB = AO + OB = a+c c 2a + c (ii) CB = CO + OB P = 2a O a A b) (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. (iii) Trapezium : CB is parallel to OA. (i) OP = OA + AP =a +c (ii) OB = 3 x (OP) They are parallel and have a common point, hence O, P & B are colinear. Example OACB is a parallelogram with OA = a and OB = b M is the midpoint of AC P is the intersection of OM with AB (i) Obtain the position vector of M (ii) Given that AP = kAB use the triangle OAP to obtain an expression for OP in terms of a, b and k. (iii) Deduce the position vector P. Example (i) OM = OB + BM OACB is a parallelogram with OA = a and OB = b = b +a M is the midpoint of AC P is the intersection of OM with AB (i) Obtain the position vector of M (ii) Given that AP = kAB use the triangle OAP to obtain an expression for OP in terms of a, b and k. (iii) Deduce the position vector P. A C (ii) AP = kAB OP = OA + AP = a + kAB = a + k(b – a) = (1 – k)a +kb (iii) OP = hOM OP = h(b + a) a O M P b B a 1–k=h b k=h Hence 1 = h h = OP = a + b