O lvl E Maths: 2D Vectors

In ∆ABC, D, E and F are the midpoints of BC, AB and AC respectively. The lines AD and CE intersect at point K.

(a) Given that AB = 6p and AC = 10q, express in terms of p and q:
(i) BC
(ii) AD
(iii) CE

(b) Given that AK = h AD and by considering ∆ABK, express BK in terms of h, p and q.

(c) Given that CK = k CE and by considering ∆BCK, express BK in terms of k, p and q.

(d) Using these 2 expressions for BK, find the values of h and k.

(e) Prove that BK, when produced, will pass through F.

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Answer:

(a)
(i) BC = BA + AC = -6p + 10q

(ii) AD = AB + BD
= AB + ½BC
= 6p + ½(-6p + 10q)
= 3p + 5q

(iii) CE = CA + AE
= CA + ½AB
= -10q + ½(6p)
= 3p - 10q


(b) BK = BA + AK
= BA + hAD
= -6p + h(3p + 5q)
= (3h - 6)p + 5hq


(c) BK = BC + CK
= BC + kCE
= -6p + 10q + k(3p - 10q)
= (3k - 6)p + (10 - 10k)q


(d) Comparing coefficients of p,
3h - 6 = 3 k - 6
h = k

Comparing coefficients of q,
5h = 10 - 10k
Sub h = k
5k = 10 - 10k
15k = 10
k = h = 2/3


(e) BF = BA + AF
= BA + ½AC
= -6p + ½(10q)
= -6p + 5q

BK = [3(⅔) - 6] p + 5(⅔)q
= -4p + (10/3) q
= ⅔(-6p + 5q)
= ⅔BF

Since BK = ⅔BF, hence, BK produced will meed AC at F.

O lvl E Maths: 2D Vectors

OABC is a parallelogram whose diagonals intersect at E; the diagonal AB produced to D and OD and CD are joined.

Given that OA = a, OB = b and DB = 2(a - b), express, as simply as possible, in terms of a and b
(i) AB
(ii) ED
(iii) OD
(iv) CD

Given also that X is the point on BD such that OX = ⅓ (8b - 5a), find the value of BX/BD.

If, in addition |a| = |b| = 3 and ∡AOB = 60°, find the numerical value of |b - a|.

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Answer:

(i) AB = AO + OB
= - OA + OB
= -a + b

(ii) EB = ½ AB (since E is midpoint of AB, properties of a parallelogram)

ED = EB + BD
= ½ AB - DB
= -½ a + ½ b - 2(a - b)
= -5/2 a + 5/2 b

(iii) OD = OB + BD
= OB - DB
= b - 2(a - b)
= -2a + 3b

(iv) OC = OA + AC
= OA + OB (AC = OB)
= a + b

CD = CO + OD
= - OC + OD
= -(a + b) + (-2a + 3b)
= -3a + 2b



BD = -DB = 2(b - a)

BX = BO + OX
= -b + ⅓(8b - 5a)
= ⅓(5b - 5a)
= (5/3)(b - a)
= (5/3) ½BD
= (5/6) BD

Hence, BX/BD = 5/6


From (i), AB = b - a
Hence, |b - a| = length AB

From |a| = |b| = 3, OA = OB = 3

AB² = OA² + OB² - 2 (OA)(OB) cos 60°
AB² = 3² + 3² - 2(3)(3)(½)
AB² = 9
AB = 3

Hence, |b - a| = 3

O lvl E Maths: 2D Vectors

In the diagram, OABC is a parallelogram. X is a point on OB such that OX = 3 XB. P is on OA such that OA = ⅓ OP. It is given that OA = 2a, and OC = 4c.

(a) Express as simply as possible in terms of a and/or c
(i) OB
(ii) CX
(iii) CP

(b) Prove that CXP is a straight line.

(c) Find the ratio of CX:XP

(d) Given that Y is a point on AB such that CY = h CX and BY = k BA, write down a vector equation, relating vector CX, BA and CB. Hence solve for h and k.

(e) State the ratio of XY:YP.

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Answer:

(a)
(i) AB = OC
Hence, OB = OA + AB = 2a + 4c

(ii) OX = 3XB
Hence, OX = ¾ OB

CX = CO + OX
= -4c + ¾ (2a + 4c)
= ½ (3a - 2c)

(iii) OA = ⅓OP
OP = 3 OA = 6a

CP = CO + OP
= -4c + 6a


(b) CP = 2 (3a - 2c)
Thus, CP = 4 CX

Hence, CXP is a straight line.


(c) CP = 4 CX
Thus, CX : XP = 1 : 3


(d) CB = CY + YB
CB = CY - BY
CB = h CX - k BA

Since CB = OA and BA = CO,
2a = h(½ (3a - 2c)) - k (-4c)
2a = (3h/2) a + (4k - h) c

Comparing coefficients of a,
3h/2 = 2
h = 4/3

Comparing coefficients of c,
4k - h = 0
k = 1/3


(e) CY = 4/3 CX
CX : XY = 3 : 1

Also, CX : XP = 1 : 3 = 3 : 9

Thus, XY : XP = 1 : 9
XY : YP = 1 : 8

O lvl E Maths: 2D Vectors

In the diagram, OA = 2p, OB = 3q and OC = 4p + 9q.

(i) Given that the point P is such that AP = 2 PB, express the position vector of P in terms of p and q.

(ii) Given that point Q is such that OQ = 3 OP, express OQ in terms of p and q. Show that Q lies on BC and write down the numerical value of BQ/QC.

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Answer:

(i) AP = 2 PB
AO + OP = 2 (PO + OB)
OP - OA = -2 OP + 2 OB
3 OP = OA + 2 OB
3 OP = 2p + 6q
OP = ⅔p + 2q

(ii) OQ = 3 OP = 2p + 6q


BC = BO + OC
= -3q + 4p + 9q
= 4p + 6q

BQ = BO + OQ
= -3q + 2p + 6q
= 2p + 3q


Since
BQ = 2p + 3q
= ½ (4p + 6q)
= ½ BC

BQ and BC are collinear.

Hence, Q lies on BC


BQ/BC = ½
Thus, BQ / QC = 1

O lvl E Maths: 2D Vectors

OABC is a rectangle in which OC = 2x, CB = 2y, and M is the midpoint of AB. OA is produced to D such that 3OA = 2OD and OM is produced to meet BD at N.

(a) Express the following vectors in terms of x and y, giving each answer in its simplest form.
(i) OB,
(ii) OM,
(iii) BD

(b) If BN = α BD and ON = β OM, form an equation connecting α, β, x and y and find the values of α and β.

(c) Hence, find the value of
(i) OM/ON
(ii) area of ΔMDN / area of ΔMDO
(iii) area of ΔOBD / area of trapezium OCBD

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Answer:

(a)
(i) OB = OC + CB = 2x + 2y

(ii) AB = OC = 2x
OA = CB = 2y

OM = OA + ½AB
= 2y + x

(iii) 3OA = 2 OD
OD = 3/2 OA = 3y

BD = BO + OD
= OD - OB
= 3y - (2x + 2y)
= y - 2x



(b) BN = α BD
BO + ON = α( y - 2x)
-(OB) + β OM = α( y - 2x)
-(2x + 2y) + β(2y + x) = α( y - 2x)

Comparing coefficients of x,
-2 + β = -2α ----------------------(1)

Comparing coefficients of y,
-2 + 2β = α -----------------------(2)

Sub (2) into (1)
-2 + β = -2(-2 + 2β)
-2 + β = 4 - 4β
5β = 6
β = 6/5

α = -2 + 2β
α = -2 + 2(6/5)
α= 2/5


(c)
(i) ON = 6/5 OM
OM / ON = 5/6

(ii) area of ΔMDN / area of ΔMDO
= MN / OM (since they have the same common height)
= (ON - OM) / OM
= ON / OM - OM / OM
= 6/5 - 1
= 1/5

(iii) Area of ΔOBC / Area of ΔOBD
= Area of ΔOAB / Area of ΔOBD (since area of ΔOBC = area of ΔOAB)
= OA / OD (since they have the same common height)
= 2/3

area of trapezium OCBD / area of ΔOBD (find opposite first)
= (area of ΔOBD + area of ΔOBC) / area of ΔOBD
= 1 + Area of ΔOBC / Area of ΔOBD
= 1 + 2/3
= 5/3

Thus,
area of ΔOBD / area of trapezium OCBD = 3/5 (flip over)