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

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**(ii) Parallelogram Rule [for vectors with same initial point]**

D C B A

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**(iii) Extensions follow to three or more vectors**

p+q+r q p

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**First we need to understand what is meant by the vector – a**

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 a a b b

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

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

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**OP is a position vector of a point P. We usually associate p with OP**

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

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**A very Important result!**

B AB = b - a b A a O

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

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

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**Example 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 + BQ b = c a (ii) MP = MA + AP c = c MNPQ is a parallelogram as NQ and MP are equal and parallel.

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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. a) AB = AO + OB = a + c (ii) CB = CO + OB P = 2a Trapezium : 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.

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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 Obtain the position vector of M Given 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.

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**A C a M P O B b OM = OB + BM = b + a AP = kAB OP = OA + AP = a + kAB**

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 Obtain the position vector of M Given 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 + a AP = kAB OP = OA + AP = a + kAB = a + k(b – a) = (1 – k)a +kb O C A b a B (iii) OP = hOM OP = h(b + a) M P a 1 – k = h b k = h Hence 1 = h h = OP = a + b

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