Vectors Questions and Answers

Vectors Questions

1.  Given that and find

1. (i) (3 mks)
   1. || (3 mks)
2. Show that A (1, -1), B (3, 5) and C (5, 11) are collinear (4 mks)

2.  Given the column vectors and that

1.  (i) Express p as a column vector  (2mks)
2.  (ii) Determine the magnitude of p   (1mk)

3.  Given the points P(-6, -3), Q(-2, -1) and R(6, 3) express PQ and QR as column vectors. Hence show that the points P, Q and R are collinear. (3mks)

4.  The position vectors of points x and y are and respectively. Find x y as a column vector   (2 mks)

5.  Given that (3mks)

6.  The position vectors of A and B are 2 and 8 respectively. Find the coordinates of M

5 -7

which divides AB in the ratio 1:2. (3 marks)

7.  The diagram shows the graph of vectors and .

 Find the column vectors;

 (a) (1mk)

 (b) || (2mks)  

8.  . Find (2mks)

9.  Show that P (4, 0 -4), Q (8, 2, -1) and R (24, 10, 11) are collinear. (3 mks)

10.  Given that  
= 2i – j + k and q = i + j +2k, determine

a. │p + q│ (1 mk)

(b) │ ½ p – 2q │  (2 mks)

11.  Express in surds form and rationalize the denominator.  

  1

Sin 60^o Sin 45^o – Sin 45^o

12.  If OA = 12i + 8j and OB = 16i + 4j. Find the coordinates of the point which divides AB

internally in the ratio1:3

13.  Find scalars m and n such that

 m 4 + n -3 = 5

3 2 8

14.  In a triangle OAB, M and N are points on OA and OB respectively, such that OM: MA = 2:3

 and ON: NB = 2:1. AN and BM intersect at X. Given that OA = a and OB = b

 (a) Express in terms of a and b

  (i) BM

  (ii) AN

 (b) By taking BX = t and AX = h
AN, where t and h are scalars, express OX in two

different ways  

 (c) Find the values of the scalars t and h  

 (d) Determine the ratios in which X divides :-

  (i) BM

  (ii) AN    

15.  OABC is a parallelogram, M is the mid-point of OA and AX = ²/₇ AC, OA=a and OC = c

 (a) Express the following in terms of a and c

(i) MA

(ii) AB

(iii) AC

(iv) AX

 (b) Using triangle MAX, express MX in terms of a and c

 (c)The co-ordinates of A and B are (1, 6, 8) and (3, 0, 4) respectively. If O is the origin and P

  the midpoint of AB. Find;

  (i) Length of OP  

  (ii) How far are the midpoints of OA and OB?

16.  a) If A, B & C are the points (2, – 4), (4, 0) and (1, 6) respectively, use the vector method

to find the coordinates of point D given that ABCD is a parallelogram.

b) The position vectors of points P and Q are p and q respectively. R is another point with

position vector r = ³/₂ q – ½ p. Express in terms of P and q

(i) PR

(ii) PQ, hence show that P, Q & R are collinear.

(iii) Determine the ratio PQ : QR

17.  The figure shows a triangle of vectors in which OS: SP = 1:3, PR:RQ = 2:1 and T is the

midpoint of OR

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

i) OR

ii) QT

 b) Express TS in terms of p and q and hence show that the points Q, T and S are collinear

 c) M is a point on OQ such that OM = KOQ and PTM is a straight line. Given that

PT: TM = 5:1, find the value of k

18.  Given that a = , b = and c = and that p = 3q – ½ b +¹/₁₀c

  Express p as a column vector and hence calculate its magnitude /P/ correct to two decimal places

19.  In a triangle OAB, M and N are points on OA and OB respectively, such that OM:MA= 2:3 and

 ON:NB= 2:1. AN and BM intersect at X. Given that OA = a and OB = b

 (a) Express in terms of a and b:-

  (i) BM

(ii) AN

  (b) Taking BX = kBM and AX =hAN where k and h are constants express OX in terms of

(i) a, b and k only

 (ii) a, b, and h only

 (c) Use the expressions in (b) above to find values of k and h  

.  

20.  In the figure below OAB is a triangle in which M divides OA in the ratio 2:3 and N

divides OB in the ratio 4:1. AN and BM
intersects at X

(a) Given that OA = a and OB = b, express in terms of a and b

(i) AN  

(ii) BM  

(iii) AB  

(b) If AX = sAN and BX = tBM, where s and t are constants, write two expressions

for OX in terms of a, b, s and t. Find the value of s and t hence write OX in terms

of a and b

21.  A student traveling abroad for further studies sets a side Kshs. 115800 to be converted into US

dollars through a bank at the rate of 76.84 per dollar. The bank charges a commission of 2 ½ %

of the amount exchanged. If he plans to purchase text books and stationery worth US$270, how

much money, to the nearest dollar, will he be left with?

22.  Given that:- r = 5i – 2j and m = -2i + 6j – k are the position vectors for R and M respectively.

 Find the length of vector RM

23.  OABC is a trapezium in which OA = a and AB = b. AB is parallel to OC with 2AB = OC.

T is a point on OC produced so that OC: CT = 2:1. At and BC intersect at X so that BX = hBC and AX = KAT

(a) Express the following in terms of a and b:-

(i) OB

(ii) BC

(b) Express CX in terms of a, b and h

(c) Express CX in terms of a, b and k

(d) Hence calculate the values of h and k  

24.  Given that a = 2i + j – 2k and b = -3i + 4j – k find :-

| a + b|.

25.  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 = b and AC = c;

 (i) Express BD and AE in terms of b and c  

 (ii) If BF =tBD and AF =nAE, find the values of t ad n

 (iii) State the ratios in which F divides BD and AE

26.  The coordinates of point O, A, B and C are (0, 0) (3, 4) (11, 6) and (8, 2) respectively.

A point P is such that the vector OP, BA, BC satisfy the vector equation OP = BA + ½ BC

 Find the coordinates of P

27. A point Q divides AB in the ratio 7:2. Given that A is (-3, 4) and B(2, -1).

Find the co-ordinates of Q

Vectors Answers

1.  

Sin 60 = 
³/₂ 1

Sin 45 = ¹/₂
3
1 – 1

2 2 2

= 1

3 – 1

22 2

= 6 –  2

1. 2
   

= 6 – 22

4

2.  OP = OA + ¼ AB

= OA + ¼ (OB – OA)

= OA + ¼ OB – ¼ OA

= ¾ OA + ¼ OB

= ¾ OA + ¼ OB

= ¾ 12 + ¼ 16 = 3 + 4 = 7

  8 4 6 1 7

3. m 4 + n -3 = 5  

3 2 8

4m – 3n = 5 ……….. (i) x 2

3m + 2n = 8 ………..(ii) x 2

8m – 6n = 10

9m + 6n = 24

17m = 34

m= 2

4 x 2 – 3n = 5

-3n = -3

n = 1

m = 2, n = 1

4.  (a) (i) BM = 2a – b = 1(2a – 5b)

  5 5

(ii) AN = 2b – a = 1 (2b – 3a)

3 3

(b) BX = t (2a – 5b)

5

AX = h(2b – 3a)

    3

OX₁ = OB + BX = b + t (2a – 5b)

5

= (-t)b + 2 + a

5

OX = OA + AX = a + h (2b – 3a)

= (1-h)a + 2hb

  3

(c) OX₁ = OX₂

2 + a + (1 – t)b= (1-h)a + 2hb

5 3

2t = 1-h ….(i)

5

(1 – t) = ¾ h ….(ii) t = 5 – 5h

2

1 – (5 -5h) = 2h = 11h = 9

    2 3

h = 9

11

t = 5 – 5
9 = 5

  2 11 11

(i) BX : XM = 1:10

(ii) AX: XN = 3:8

5.  a) i) MA = ½ a

ii) AB = a

iii) AC = a + c

iv) AX = ²/₇ AC = ²/₇ (-a + c)

b) MA = ½ a

  AX = ²/₇ c – ²/₇ a

 MX = ½ a + ²/₇ – ²/₇ a

 = ³/₁₄ a + ²/₇c

Co-ordinates of P = (1 + 3, 6 + 0, 8 + 4)

  2 2 2

  = (2, 3, 6)

/OP/ = √ 2² + 3² + 6²

 = √ 4 + 9 + 36

= √49 = 7 units

c)  Co-ordinates of O (0,0,0)

Co-ordinates of A (1, 6, 8)

Mid points of AO = (1 + 0, 6 + 0, 8 + 0)

2 2 2

= (0.5, 3, 4)

6.   a) AB = DC ⇒1 – x = 2 ⇒ x = -1

6 – y = 4⇒ y = 2

 D = (-1, 2)

  b) (i) RQ = Q – R = q – ³/₂q – ½ p

= -½ q – p = ½ p – q √

  (ii) PR = ³/₂ q – ½ p – P √

= ³/₂ q – p

³/₂ q = – ½ Also –³/₂ p = ½ kp

⇒k = -3 ⇒ k = -3

 Hence P, Q, R, Q Collinear.

  (iii) PQ = q – p , QR = ½ (q – P)

PQ : QR = 2 : 1

7.  (a) PQ = PO + OQ = -p + q

  Or = OP + PR = P + 2/3 PQ

= P + 2/3 (-p+q)

= ¹/₃p + ²/₃q

QT = QO + OT = -q + ½ OR since OT = TR

= -q + ½ ( ¹/₃p – ²/₃q)

= ¹/₆p – ²//₃q OR ¹/₆ (p-4q)

(b) TS = TO + OS = – ½ OR + ¼ OP

   = – ½ (¹/₃p + ²/₃q) + ¼ p = ^-1/₆p – ¹/₃q + ¼ p

    = ¹/₁₂p – ¹/₃q or ¹/₁₂(p-4q)

QT: TS = ¹/6(p-4q): ¹/12(p-4q) = ¹/6:¹/12 = 2:1

QT = 2TS OT//TS but T is a common point hence Q, T, S are collinear

(c) Vector OT can be expressed in 2 ways

1^st OT = ½ OR given

= ½ (¹/₃ P + ²/₃q) = ¹/₆q + 1/₃q………..(i)

2^nd using OPT

OT = OP + PT = P + ⁵/₆PM

But PM = PO + OM = -P +KOQ = -P +Kq

OT = P + ⁵/₆ (-P +kq)

= P – ⁵/₆kq

= ¹/₆p + n⁵/₅kq…………………(ii)

Aqn (i) and (ii) represent the same vector OT

¹/₆p + ¹/₃q = ¹/₆p + ⁵/₆kq……………..(iii)

Comparing coefficients of q in eqn (iii) have⁵/₆k = ¹/3

15k = 6

8.  3a = 3(-3) = (-9)

2 6

½ b = ½ (4) = (2)

  -6 -3

¹/₁₀c = ¹/₁₀ (5) = (0.5)

-10 -1

P = (-9) – (2) + 0.5)

  6 -3 -1

  = (-10.5)

8

/P/ = √ (-10.5)² + 8²

= √110.25 = 64

= √174.25

= 13.20037878

  = 13.20 (2 d.p)  

9.  (i)BM = BO + OM

= ²/₅a – b

(ii) AN = AO + ON

= 2 b – a

3

(b) OX = OB + BX

  = b + k (2 a – b)

  = 2 ka + b(1- k)

5

OX = OA + AX

  = a + h ( 2b –a)

3

= a (1-h) + 2 hb

= a(10h) 2hb

(c) ²/₅ a = a (1-h) also b(1-k) = 2hb

   2k = 1-h 1-k = 2h

k = 5 – 5h

  2 2

1 – 5 +5h = 2h

2 2 3

5h – 2h = 5 -1

2 3 2

1 5h = 3

6 2

h = 3 x 6 = 9

  2 11

k = 5 – 5
9

  2 2 11

= 5 – 45

  2 22

= 5

11

10.  (i)  AN = AO + ON

= -a + 4 b

5

(ii) BM = BO + OM

  = -b + ²/₅ a

(iii) AB = AO + OB

 = -a + b

AX = sAN

BX = tBM

OX= OB + BX  

  = b + tBM

  = b + t (-b + ²/₅a)

  = b – tb + ²/₅ ta

  = b (1-t) + ²/₅ ta )

OX = OA + AX

  = a + sAN

  = a + s(-a + ⁴/₅ b)

  = a – Sa + ⁴/₅ sb

  a(1-s) + ⁴/₅ sb

b(1-t) + ²/₅ta = a (1-s) ⁴/₅sb

b(1-t) = ⁴/₅ sb

 1-t = ⁴/₅s————-(i)  

a (1-s) = ²/₅ta

1-s =²/₅ta

s=1-²/₅t————-(ii)

1-t = ⁴/₅(1- ₂/₅t)

1 – t = ⁴/₅ – 8/₂₅t

–¹⁷/₂₅ = – 1/₅

t = ⁵/₁₇

s = ¹⁵/₁₇

11.  115800 x 97.5

76.84 100

  = 1469.35 √

= 1469.35 – 270

  = 1199.35 √

= 1199 dollars

12.  

 RM = – =

RM = (-3)² + 82(-1)²

74 = 8.602 units

13.  (a) (i) OB = a + b

  (ii) BC = BA + AO +OC

  = -b + -a + 2b

  = b – a

(b) CX = CO + OA + AB + BX

= -2b + a + b + hBC

= a – b + h( b – a)

= a – b + hb – ha

= (1 – h)a + (h – 1 ) b

(c) CX = CO + OA + AX

= 2b + a + KAT

but AT = AO + OT

  = -a + 3b

CX = 2b + a + K(3b – a)

= a – Ka + 3Kb + 2b

= (1- K) a + 3 (K + 2) b

(d) I – h = 1 – k ……..(i)

  h – 1 = 3k + 2…….(ii)

from (i) h = k

sub in (ii) h-1 = 3h + 2

h = ^-3/₂

   K = ^-3/₂

14.  a + b = (2 – 3)i + (1 + 4)j + (-2-1)k

= -i + 5j – 3k

 ∣a + b∣ (-1)² + (5)² + (-3)²

  = 35

  = 5.916

15.  i) BD = BA + AD

  = -b + ³/₅c

 AE = AB + BE

 = b + ½ BC = b + ½ (c – b)

  = ½ b + ½ c

ii) BF = t (³/₅c – b)

 AF = n (½ b + ½ c) = ⁿ/₂ (b+ c)

 AF = AB + BF

 = b + t(³/₅c – b) = b + ³/₅tc + tb

 = (1 – t) b + ³/5tc

 (1 – t)b + ³/₅tc = ⁿ/₂b + ⁿ/₂c

  1 – t = ⁿ/₂  2 – 2t = n ………… (i)

 ³/₅t = ⁿ/₂ ; 6t – 5n = 0 ………… (ii)

Sub from équation (ii)

6t – 5(2 – 2t) = 0

6t – 10 + 10t = 0

16t = 10

t = ¹⁰/₁₆ = ⁵/₈

n= ¾

iii) BF = ⁵/₈ BD

 F divides BD in the ratio 5 :3

 AF = ¾ AE

 F divides AE in the ratio 3 :1

16.  BA = -8

-2

½ BC = ½ -3 = – 1 ½)

-4 -2

OP = -8 + -1 ½ = -9 ½

-2 -2 -4

Co-ordinates of P ( -9 ½ , -4)

17.  OB = 5OQ + 2OA

  7 5

OQ = 7OB – 2OA

  5 5

OQ = 7 2 – 2 -3

5 -1 5 4

= ¹⁴/₅ – ^-6/₅ = ²⁰/₅ = 4

^-7/₅
⁸/₅
^-15/₅ -3

Q= (4, -3)