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Addition of vectors (i) Triangle Rule [For vectors with a common point] A B C

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(ii) Parallelogram Rule [for vectors with same initial point] A B C D

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(iii) Extensions follow to three or more vectors p q r p+q+rp+q+r

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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.

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A few examples b a b a b – a

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p q Which vector is represented by p – q ? – q p – q

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A C B CB = CA + AB = - AC + AB = AB – AC

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Position Vectors Relative to a fixed point O [origin] the position of a Point P in space is uniquely determined by OP O P OP is a position vector of a point P. We usually associate p with OP p

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A very Important result! O b a AB = b - a A B

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The Midpoint of AB O B A b a OM = ½( b + a) M

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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. AB = AO + OB = – p + (2p – q) = p – q AB &BC are parallel (even though in opposite directions) and have a common point B BC = BO + OC = – (2p – q) + q = 2q – 2p = –2(p – q ) = –2AB C B A O Hence A, B and C are co-linear.

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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? Example c b a (i)BC = BO + OC = c – b (ii) NQ = NB + BQ = c (ii) MP = MA + AP = c MNPQ is a parallelogram as NQ and MP are equal and parallel. a)

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The 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. (i)AB = AO + OB = a + c (ii) CB = CO + OB = 2a (iii)Trapezium : CB is parallel to OA. P a) b) (i)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.

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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.

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M O C A b a B P (i)OM = (ii)AP = kAB (iii) OP = hOM OB + BM = b + a OP = OA + AP = a + kAB = a + k(b – a) = (1 – k)a +kb OP = h(b + a) a1 – k = h b k = h Hence 1 = h h = OP = a + b 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.

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