Questions

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. Express in terms of a and b in the simplest form.
 AN
 AT
 AM
 Show that O, T and M are collinear and state the ratio of OT: TM
 Express in terms of a and b in the simplest form.
 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)
 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
 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.
Given that OC = 3c and OA = a Express AS and BC in terms of a and c
 Given further that AT = hAS and BT = kBC where h and k are constants
 Express AT in two ways in terms a, c , h and k
 The obtuse angle between the lines PQ
 Hence find the ratio BT: BC
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. Given that OP = P and OQ = q, express the following vectors in term of p and q
 PQ
 OR
 SQ
 You area further given that ST = mSQ and OT = nOR. Determine the values of m and n.
 Given that OP = P and OQ = q, express the following vectors in term of p and q
Answers


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

 OX = ^{2}/_{3}(3i + 2j – 4k) + ^{1}/_{3}(6i + 11j + 2k)
= 2i + 4j – ^{8}/_{3}k + 2i + ^{11}/_{3}j + ^{2}/_{3}
= 4i + 5j 2k
10x1 = √(16 + 25 + 4)
= 6.71units  AS = AO + OS
= a + 2(3c)
= 2 c – a…………
BC = BA + AC
= a  b + AC
But AC = AO + OC = a + 3c
= 3c – a……….
AB + ^{2}/_{3}OC = ^{2}/_{3} 3 c = 2 c
BA = 2 c…….
BC = 12c +3c – a = c a. 
 AT = hAS = h(2c –a)
= 2hc ha
AT = AB + BT = 2c + K ( c a)
= 2c + Kc – Ka
= ( 2 + k)c – Ka  2 + K = 2h
(i) K = h
(ii) 2 + h = 2h
2 = 2h  h
2 = h, K = 2
 AT = hAS = h(2c –a)
 BT : BC
BT = 2 BC
 AS = AO + OS


 PQ = PO + OQ
= P + q or q – p  OR = OP + PR
= P + ^{2}/_{3}PQ
= P + ^{2}/_{3}(q – p)
= P + ^{2}/_{3}q  2p
= ^{1}/_{3}p + ^{2}/_{3}q  SQ = SO + OQ
= ^{3}/_{4}OP + OQ
= ^{3}/_{4}p + q or q – ^{3}/_{4}p
 PQ = PO + OQ
 Express OT in two different ways:
Given OT = nOR
= n (^{1}/_{3}P + ^{2}/_{3}q)
= ^{n}/_{3}p + ^{2n}/_{3}q
From ΔOST,
OT = OS + ST
= ^{3}/_{4}OP + MSQ
= ^{3}/_{4}P + M (^{3}/_{4}P +q)
= (^{3}/_{4}  ^{3}/_{4}m) p + mq
∴ ^{n}/_{3}p + 2nq = (^{3}/_{4}  ^{3}/_{4}m) p + mq
Compare the coefficients of p and q
^{n}/_{3} = ^{3}/_{4}  ^{3}/_{4}m
4n = 9 – 9m
4n + 9m = 9 ………………..eq (1)
^{2}/_{3}n = m
m = ^{2}/_{3}n …………….eq. (2)
Substitutes form in equation (1)
4n + 9 (^{2}/_{3}n) = 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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