Vectors 2 Questions and Answers - Form 3 Topical Mathematics

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Questions

  1.  
    vectors 2 1q
    In the figure alongside OA = a , OB = b. T lies on AN such that AN : TN = 13:6. M lies on AB such that AM:MB=1:3 and N lies on OB such that OB:BN = 7:-5.
    1. Express in terms of a and b in the simplest form.
      1. AN
      2. AT
      3. AM
    2. Show that O, T and M are collinear and state the ratio of OT: TM
  2. A point (-3, 4) divides AB internally in the ratio 3:5. Find the coordinates of point A given that point B is (6, -5)
  3. Given that O is the origin, OA = 3i + 2j – 4k and OB = 6i + 11j + 2k. If x divides AB in the ratio 1:2, find the modulus of OX to 2d.p
  4. In the figure OABC is a trapezium in which 3 AB = 2OC. S divides OC in the ratio 2:1
    and AS produced meets BC produced at T.
    vectors 2 4q
    Given that OC = 3c and OA = a
    1. Express AS and BC in terms of a and c
    2. Given further that AT = hAS and BT = kBC where h and k are constants
      1. Express AT in two ways in terms a, c , h and k
    3. The obtuse angle between the lines PQ
    4. Hence find the ratio BT: BC

  5. vectors 2 5q
    In the figure above, OPQ is a triangle in which OS = ¾ OP and PR: RQ = 2 : 1. Lines OR and SQ meet at T.
    1. Given that OP = P and OQ = q, express the following vectors in term of p and q
      1. PQ
      2. OR
      3. SQ
    2. You area further given that ST = mSQ and OT = nOR. Determine the values of m and n.

Answers

  1.  
    1.  
      1. AN = OA + ON
        = -a + 2/7b
        = 2/7b - a
      2. AT = 7/13AN
        7/13 (-a + 2/7b)
        2/13b - 7/13a
      3. AM = 1/4AB
        = 1/4 (AO + OB)
        = 1/4 (b - a)
    2. OT = OA + AT
      = a + (2/13b - 7/13a)
      = 2/13(3a + b)
      OM = OA + AM
      = a + (-1/4a + 1/4b)
      =3/4a + 1/4 b
      =1/4(3a + b)
      OT = 2/13 (3a + b)
      OM     1/4 ( 3a + b)
      OT = 8/13 OM
      Or OM = 13/8OT
      Since OT = 8/13OM
      Then OT : TM = 8/13 : 5/13
      = 8 : 5

  2. vectors 2 2a
  3. OX = 2/3(3i + 2j – 4k) + 1/3(6i + 11j + 2k)
    = 2i + 4j8/3k + 2i + 11/3j + 2/3
    = 4i + 5j -2k
    10x1 = √(16 + 25 + 4)
    = 6.71units
    1. AS = AO + OS
      = -a + 2(3c)
      = 2 ca…………
      BC = BA + AC
      = a - b + AC
      But AC = AO + OC = -a + 3c
      = 3ca……….
      AB + 2/3OC = 2/3 3 c = 2 c
      BA = 2 c…….
      BC = -12c +3ca = c -a.
    2.  
      1. AT = hAS = h(2ca)
        = 2hc -ha
        AT
        = AB + BT = 2c + K ( c -a)
        = 2c + Kc – Ka
        = ( 2 + k)c – Ka
      2. 2 + K = 2h
        (i) K = h
        (ii) 2 + h = 2h
        2 = 2h - h
        2 = h, K = 2
    3. BT : BC
      BT = 2 BC
  4.  
    1.  
      1. PQ = PO + OQ
        = P + q or qp
      2. OR = OP + PR
        = P + 2/3PQ
        = P + 2/3(qp)
        = P + 2/3q - 2p
        = 1/3p + 2/3q
      3. SQ = SO + OQ
        = -3/4OP + OQ
        = -3/4p + q or q3/4p
    2. Express OT in two different ways:
      Given OT = nOR
      = n (1/3P + 2/3q)
      = n/3p + 2n/3q
      From ΔOST,
      OT = OS + ST
      = 3/4OP + MSQ
      = 3/4P + M (-3/4P +q)
      = (3/4 - 3/4m) p + mq
      n/3p + 2nq = (3/4 - 3/4m) p + mq
      Compare the coefficients of p and q
      n/3 = 3/4 - 3/4m
      4n = 9 – 9m
      4n + 9m = 9 ………………..eq (1)
      2/3n = m
      m = 2/3n …………….eq. (2)
      Substitutes form in equation (1)
      4n + 9 (2/3n) = 9
      4n + 6n = 9
      10n = 9
      n = 9/10
      Substitute for n in equation (2)
      m = 2/3 x 9/10 = 3/5

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