Vectors 2 Questions and Answers

Vectors 2 Questions

1. In the figure below E is the mid point of BC. AD:DC = 3:2 and F is the meeting point of BD and AE

 If AB = and AC =

 (a) Express the following in terms of and

 (i) (1mk)

 (ii) (2mks)

 (b) If = t and = n find the value of t and n. (5mks)

 (c) State the ratio of BD to BF. (1mk)

2. In the figure below OA = a and OB = b. Points P and T divide internally in the ration 2:3 and 1:3 respectively. Lines meet at Q.

 Find in terms of

 (i) (3mks)

 (ii) (1mk)

 (iii) (1mk)

 (iv) (5mks)

 If OQ = kOT and AQ = hAP where k and h are constants express OQ in two different ways and hence find the values of h and k. (10mks)

1. In the figure below, and DB is parallel to OA. C is on AB extended such that AB: BC = 2:1 and that OA = 3DB.
   1. Express the vector BC in terms of a and b. (1mk)
   2. Show by vector methods that the points O, D and C are collinear. (3mks)

4. In the figure below , where k and m are scalars 2PS = 3SR.

 (a) Express as simply as possible in terms of each of the following vectors.

 (i) (1mk)

 (ii) (1mk)

 (iii) (1mk)

 (b) Express in terms of a, b, k and m. (2mks)

 (c) If Q lies on produced with = 5:4, find the value of k and m. (5mks)

5. In the figure below, DE = ½ AB and BC = 2/3 BD and the co ordinates of A,B and D are (5,4),(9,6) and ( 12,0) respectively.

 Find the vectors

 (i) BD (1mk)

 (ii) BC (1mk)

 (iii) CD (1mk)

 (iv) AC (2mks)

 b) Given that AC = kCE; where k is a scalar,

Find

(i) the value of k (4mks)

(ii) the ratio in which C divide AE. (1mk)

6.

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.

(a) Express in terms of a and b in the simplest form

(i) AN

(ii) AT

(iii) AM (b) Show that O, T and M are collinear and state the ratio of OT: TM

7. 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)

8. Given that O is the origin, OA = 3i + 2j – 4k and OB = 6i + 11j + 2k. If x divides AB

in the ratio1:2, find the modulus of OX to 2d.p

9. a) Expand (2 – ¹/₅x)⁵

 b) Hence use the expansion to find the value of (1.96)⁵ correct to 3 decimal places

10. 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

 (a) Express AS and BC in terms of a and c

 (b) Given further that AT = hAS and BT = KBC where h and k are constants

(i) Express AT in two ways in terms a, c , h and k

 (c) The obtuse angle between the lines PQ

 (d) Hence find the ratio BT: BC

11.

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.

(a) Given that OP = P and OQ = q, express the following vectors in term of p and q

(i) PQ

(ii) OR

(iii) SQ

(b) You area further given that ST = m SQ and OT = n OR. Determine the values of m and n

Vectors 2 Answers

1. a) (i) AN = OA + ON

= -a + 2 b

7

= 2 b – a

7

(ii) AT = 7 AN

13

1. – a + 2 b
   
   1. 7
      

2 b – 7 a

13 13

(iii) AM = 1 AB

4

= 1 (AO + OB)

4

= 1 ( b – a )

4

(b) OT = OA + AT

= a 2 b – 7 a

13 13

= 2 3a + b

13

OM = OA + AM

= a + – 1 a + 1 b

4 4

3 a + 1 b

4 4

1 3a + b

4

OT = 2 (3a + b)

OM 13

1 ( 3a + b)

4

OT = 8 OM

13

Or OM = 13 OT

8

Since OT = 8 OM

13

Then OT : TM = 8 : 5

13 13

= 8 : 5

2. 3 5

(X,Y) T(-3,4) B(6,-5)

TB = ⁵/₈ AB

6 – -3 = 5 AB

-5 4 8

9 = 5 6 – x

-9 8 -5 y

9 = ⁵/₈ (6-x)

-9 ⁵/₈ ( -5– y)

30 – 5 X = 9

8 8

–25 – ⁵/₈ y = -9

8

30 – 5x = 72 -5x = 42

-25 – 5y = -72 -5y = -47

X = – 8.4 y = 9.4

3. OX = 2 (3i + 2j – 4k) + 1 (6i + 11j + 2k)

3 3

= 2i + 4j – 8k + 2i + 11 j + 2

3 3 3

= 4i + 5j -2k

10×1 = 16 + 25 + 4

= 6.71units

4. a) 2⁵ – 5(2⁴) (¹/₅ ) + 10 (2³) (¹/₅ x)² – 10(2²) (¹/₅ x)³ + 5(2) (¹/₅x)⁴ – (¹/₅x)⁵

32 – 16x + ¹⁶/₅x² – ⁸/₂₅x³ + ²/₁₂₅x⁴ – ¹/₃₁₂₅x⁵

 – ¹/₅x = -0.04

x = 0.2

b) 32 – 16 (0.2) + ¹⁶/₅ (0.2)^{2
– 8}/₂₅ (0.2)³ + ………

 = 32 – 3.2 + 0.128 – 0.00256

= 28.92544

 = 29.925

5. AS = AO + OS

= – a + 2 (3 c )

= 2 c – a…………

BC = BA + AC

= a – b + AC

But AC = AO + OC = -a + 3 c

= 3 c – a……….

AB + 2 OC = 2 3 c = 2 c

3 3

BA = 2 c…….

BC = -12c +3c – a = c -a.

 b) (i) AT =  AS =  (2c –a)

= 2c –a

AT = AB + BT = 2c + K ( c -a)

 = 2 c + Kc – K a

= ( 2 + k) c – K a

 (ii) 2 + K = 2 (i) K =  (ii)

2 +  = 2

2 = 2 – 

 2 = , K = 2

(c ) BT : BC

 BT = 2 BC

6. (a) (i) PQ = PO + OQ

= P + q or q – p

(ii) OR = OP + PR

= P + 2 PQ

3

= P + 2 (q – p)

3

= P + 2 q – 2 p

3 3

= 1 p + 2 q

3 3

(iii) SQ = SO + OQ

= – 3 OP + OQ

4

= – 3 p + q or q – 3 p

1. 4
   

(b) Express OT in two different ways:

Given OT = n OR

= n 1P + 2 q

1. 3
   

= n p + 2n q

1. 3
   

From ∆OST,

OT = OS + ST

= 3 OP + M SQ

4

= 3 P + M –3 P +q

4 4

= 3 – 3m p + mq

4 4

∴
n p + 2n q = 3 – 3m p + mq

3 3 4 4

Compare the coefficients of p and q

n = 3 – 3 m

3 4 4

4n = 9 – 9m

4n + 9m = 9 ………………..eq (1)

2n = m

3

m
= 2n …………….eq. (2)

3

Substitutes form in equation (1)

4n + 9 2n = 9

3

4n + 6n = 9

10n = 9

n = 9

10

Substitute for n in equation (2)

m = 2 x 9 = 3

3 10 5