Q4 Cambridge International Examinations

Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level May-June 2015 MATHEMATICS (US) Paper 1 Pure Mathematics 1 (P1) shahbaz ahmed June 2024 1 Relevant formulas and skills required to solve the problem Reference to fig A(a1 , a2 , a3 ) is the position of the point A with respect to the origin O in 2 −→ three dimensional space.The position vector OA is defined as:   a1   −→   OA =  a2    a3 (1) Similarly B(b1 , b2 , b3 ) is the position of the point B with respect to the origin −−→ O in three dimensional space.The position vector OB is defined as:   b1   −−→   OB =  b2    b3 (2) −→ −−→ The scalar product of vectors OA and OB is defined as: −→ −−→ OA.OB = a1 b1 + a2 b2 + a3 b3 |OA||OB| cos ∠AOB = a1 b1 + a2 b2 + a3 b3 +a2 b2 +a3 b3 cos ∠AOB = a1 b1|OA||OB| −→ −−→ −−→ Also OA + AB = OB −−→ −−→ −→ AB = OB − OA −−→ −→ −−→ if OB = b, OA = a, AB = d Then −−→ AB = d = b − a −−→ −−→ Now AB.AB = (b − a).(b − a) d2 = b2 − b.a − a.b + a2 = a2 + b2 − 2ab cos ∠AOB 2ab cos ∠AOB = a2 + b2 − d2 cos ∠AOB = a2 +b2 −d2 2ab Q4 Relative to the origin O, the position vectors of points A and B are given by 3  3    −→   OA =  0   −4  and  6    −−→   OB =  −3   2  (i) Find the cosine of ∠AOB The position vector of C is given by   k    −−→   OC =  −2k     2k − 3 . (ii) Given that AB and OC have the same length, find the possible values of k. Solution p √ −→ |OA| = 32 + 02 + (−4)2 = 9 + 0 + 16 = 5 p √ −−→ |OB| = 62 + (−3)2 + 22 = 36 + 9 + 4 = 7 −→ −−→ OA.OB = 3 × 6 + 0 × (−3) + (−4) × 2 −→ −−→ OA.OB = 18 − 8 = 10 Putting in the formula cos ∠AOB = a1 b1 +a2 b2 +a3 b3 |OA||OB| 4 cos ∠AOB = 10 5×7 = 2 7 alternate method −→ |OA| = a = 5 −−→ |OB| = b = 7 −−→ AB = d = b − a  3     −−→  AB = d =  −3   6  p √ √ −−→ |AB| = d = 32 + (−3)2 + 62 = 9 + 9 + 36 = 3 6 Putting in the formula cos ∠AOB = a2 +b2 −d2 2ab 2 2 +7 −54 20 cos ∠AOB = 5 2×5×7 = 2×5×7 = 27 √ −−→ (ii) |AB| = |d| = 3 6| p √ −−→ |OC| = | k 2 + (−2k)2 + (2k − 3)2 | = | k 2 + 4k 2 + +4k 2 + 9 − 12k| p √ −−→ |OC| = | k 2 + (−2k)2 + (2k − 3)2 | = | 9k 2 − 12k + 9| According to the condition √ √ 9k 2 − 12k + 9 = 3 6 9k 2 − 12k + 9 = 54 9k 2 − 12k + 9 − 54 = 0 9k 2 − 12k − 45 = 0 3k 2 − 4k − 15 = 0 3k 2 − 9k + 5k − 15 = 0 3k(k − 3) + 5(k − 3) = 0 (3k + 5)(k − 3) = 0 k = 0. k = − 35 5