vectors exam questions (soln)

7/28/2019 Vectors Exam Questions (Soln)

1/96


2/96


3/96


4/96


5/96


6/96


7/96


8/96


9/96


10/96


11/96


12/96


13/96


14/96


15/96


16/96


17/96


18/96


19/96


20/96


21/96


22/96


23/96


24/96


25/96


26/96


27/96


28/96


29/96


30/96


31/96


32/96


33/96


34/96


35/96


36/96


37/96


38/96


39/96


40/96


41/96


42/96


43/96


44/96


45/96


46/96


47/96


48/96


49/96


50/96


51/96


52/96


53/96


54/96


55/96


56/96


57/96


58/96


59/96


60/96


61/96


62/96


63/96


64/96


65/96


66/96


67/96


68/96


69/96


70/96


71/96


72/96


73/96


74/96


75/96


76/96


77/96


78/96


79/96


80/96


81/96


82/96


83/96

Vectors (GDC OPTIONAL) IB Questionbank Maths SL 9

A1. evidence of appropriate approach (M1)

e.g.

+

=

+

15

3

2

2

9

2

3

5

1

3

2

t s

two correct equations A1A1e.g. 2 + 5 s = 9 3 t , 3 3 s = 2 + 5 t , 1 + 2 s = 2 t attempting to solve the equations (M1) one correct parameter s = 2, t = 1 A1

P is (12, 3, 3)

3

3-

12

accept A1 N3

[6]


84/96


A2. (a) Attempting to find unit vector ( eb) in the direction of b (M1)

Correct values =

++ 0

4

3

043

1

222A1

=

0

8.06.0

A1

Finding direction vector for b, vb = 18 eb (M1)

b =

0

4.14

8.10

A1

Using vector representation b = b0 + t vb (M1)

=

+

0

4.14

8.10

5

0

0

t AG N0

(b) (i) t = 0 (49, 32, 0) A1 N1

(ii) Finding magnitude of velocity vector (M1)

Substituting correctly vh =222 6)24()48( ++ A1

= 54 (km h 1) A1 N2

(c) (i) At R,

=

t

t

t

t

t

6

2432

4849

5

4.14

8.10

A1

t =6

5(= 0.833)(hours) A1 N1

(ii) For substituting t =6

5into expression for b or h (M1)

(9, 12, 5) A2 N3

[15]


85/96


A3. (a) Finding correct vectors, AB =

3

4 AC =

1

3A1A1

Substituting correctly in the scalar productACAB = 4(3) + 3(1) A1

= 9 AG 3

(b) | AB | = 5 | AC | = 10 (A1)(A1)

Attempting to use scalar product formula cos BAC =105

9 M1

= 0.569 (3 s.f) AG 3[6]


86/96


A4. (a) Attempting to find unit vector ( eb) in the direction of b (M1)

Correct values =

++ 0

4

3

043

1222

A1

=

0

8.0

6.0

A1

Finding direction vector for b, vb = 18 eb (M1)

b =

0

4.14

8.10

A1

Using vector representation b = b0 + tvb (M1)

=

+

0

4.14

8.10

5

0

0

t AG 6

(b) (i) t = 0 (49, 32, 0) A1 1

(ii) Finding magnitude of velocity vector (M1)

Substituting correctly vh =222 6)24()48( ++ A1

= 54(km h 1) A1 3

(c) (i) At R,

=

t

t

t

t

t

6

2432

4849

5

4.14

8.10

A1

t =6

5(= 0.833) (hours) A1 2

(ii) For substituting t =6

5into expression for b or h M1

(9,12,5) A2 3[15]


87/96


A5.

T

S

U

V

y

x

(a) ST = t s (M1)

=

2

2

7

7

=

9

9(A1)

VU = ST (M1)

u v =

9

9

v = u

9

9

=

=

6

4

9

9

15

5 (A1)

V(4, 6) (A1) 5

(b) Equation of (UV): direction is =

1

1or

9

9k (A1)

r =

+

9

9

15

5 or

+

1

1

15

5 (A1)

OR

r =

+

9

9

6

4 or

+

1

1

6

4 (A1) 2

(c)

11

1is on the line because it gives the same value of , for both the x

and y coordinates. (R1)

For example, 1 = 5 + 9 = 9

4

11 = 15 + 9 = 94 (A1) 2


88/96


A5. continued

(d) (i)

=

11

1

17EW

a

(M1)

=

6

1 a(A1)

EW = 2 13 ( ) 361 2 +a = 2 13 (or ( a 1)2 + 36 = 52)(M1)

a 2 2 a + 1 +36 = 52a 2 2 a 15 = 0 (A1)a = 5 or a = 3 (A1)(AG)

(ii) For a = 3

EW =

6

4 ET = t e =

4

6 (A1)(A1)

cos TEW =ETEW

ETEW (M1)

=5252

24 24 (A1)

= 13

12

Therefore, TEW = 157 (3 sf) (A1) 10[19]


89/96


A6. (i) a =22 512 + = 13 (A1)

(ii) b = 22 86 + = 10 (A1)

=> unit vector in direction of b =10

1(6 i + 8 j ) (A1)

= 0.6 i + 0.8 j

(iii) a . b = ab cos (M1)

=> cos =( ) ( )

( )101385612 +

(A1)

=65

56

130

112= (A1) 6

[6]


90/96


A7. (a) At t = 2,

=

+

2

4.3

1

7.02

0

2(M1)

Distance from (0, 0) = 22 24.3 + = 3.94 m (A1) 2

(b) 22 17.01

7.0+=

(M1)

= 1.22 m s 1 (A1) 2

(c) x = 2 + 0.7 t and y = t (M1) x 0.7 y = 2 (A1) 2

(d) y = 0.6 x + 2 and x 0.7 y = 2 (M1)

x = 5.86 and y = 5.52

==

29

160and

29

170or

y x (A1)(A1) 3

(e) The time of the collision may be found by solving

+

=

1

7.0

0

2

52.5

86.5t for t (M1)

t = 5.52 s (A1)[ie collision occurred 5.52 seconds after the vehicles set out].Distance d travelled by the motorcycle is given by

d = 22 )52.3()86.5(2

0

52.5

86.5+=

(M1)

= 73.46 = 6.84 m (A1)

Speed of the motorcycle =52.5

84.6=

t

d

= 1.24 m s 1 (A1) 5[14]


91/96


A8. (a)

+

51

3

2 x

x

x

= 0 (M1)(M1)

2 x( x + 1) + ( x 3)(5) = 0 (A1) 2 x2 + 7 x 15 = 0 (C3)

(b) METHOD 1

2 x2 + 7 x 15 = (2 x 3)( x + 5) = 0

x =2

3or x = 5 (A1) (C1)

METHOD 2

x =)2(2

)15)(2(477 2

x =2

3or x = 5 (A1) (C1)

[4]


92/96

Vectors (GDC) IB Questionbank Maths SL 5

A1. (a) appropriate approach (M1)

e.g. A B , + OBAD

=1

1

1

AB A1 N2 2

(b) any correct equation in the form r = a + t b A2 N2 2

where b is a scalar multiple of

1

1

1

e.g. ( )k j i k j i r r r +++=

+

+

=

+

= t

t

t

t

t 522,

5

2

2

,

1

1

1

4

1

1

(c) choosing correct direction vectors

1

1

1

,

3

1

2

(A1)(A1)

finding scalar product and magnitudes (A1)(A1)(A1)

scalar product =1 2 + 1 1+1 3 (= 4)

magnitudes ( ) ( ) ( ) 74.3914,.73.1111 222 =++=++

substitution into

==

vu

vu.

vu

v .u

vu

vu. sinnot but,accept M1

e.g.( ) 42

4cos,

312111

311121cos

222222=

++++

++=

= 0.906 (51.9) A1 N5 7


93/96


(d) METHOD 1

+

=

1

1

1

4

1

1

from t r

appropriate approach (M1)

e.g. p = r , st

+

=

+

3

1

2

7

4

2

1

1

1

4

1

1

,

two correct equations A1A1

e.g. 1 + t = 2 + 2 s, 1 t = 4 + s, 4 + t = 7 + 3 s

attempt to solve (M1)

one correct parameter A1

e.g. t = 3, s = 2

C is (2, 2, 1) A1 N3 6

METHOD 2

+

=

1

1

1

5

2

2

from t r 1

appropriate approach (M1)

e.g. p = r ,

3

1

2

7

4

2

1

1

1

5

2

2

st

+

=

+

two correct equations A1A1

e.g. 2 + t = 2 + 2 s, 2 t = 4 + s, 5 + t = 7 + 3 s

attempt to solve (M1)

one correct parameter A1

e.g. t = 4, s = 2

C is (2, 2, 1) A1 N3 6[17]


94/96


A2. (a) (i) (3, 4, 0) A1 N1

(ii) choosing velocity vector

1

3

2

(M1)

finding magnitude of velocity vector (A1)e.g. 194,13)2( 222 ++++

speed = 3.74 ( 14 ) A1 N2

(b) (i) substituting p = 7 (M1)B = (11, 17, 7) A1 N2

(ii) METHOD 1

appropriate method to find BAor AB (M1)

e.g. OBAO + , A B

=

=

7

21

14

BAor

7

21

14

AB (A1)

distance = 26.2 )147( A1 N3

METHOD 2

evidence of applying distance is speed time (M2)e.g. 3.74 7distance = 26.2 )147( A1 N3

METHOD 3

attempt to find AB 2, AB (M1)

e.g. (3 (11)) 2 + (4 17) 2 + (0 7) 2, 222 )70()174())11(3( ++

AB 2 = 686, AB = 686 (A1)

distance AB = 26.2 ( 147 ) A1 N3

(c) correct direction vectors

a

2

1

and13

2

(A1)(A1)

+=

a

a

a

2

1

1

3

2

,52

12 = a + 8 (A1)(A1)

substituting M1

e.g. cos 40 =514

8

2+

+

a

a

a = 3.21, a = 0.990 A1A1 N3[16]


95/96


A3. correct substitutions for v w; v; w (A1)(A1)(A1)

e.g. 2k + ( 3) (2) + 6 4, 2 k + 30; 20,4)2(;49,6)3(2 2222222 +++++ k k

evidence of substituting into the formula for scalar product (M1)

e.g. 207

3022

+

+

k k

correct substitution A1

e.g. 207

302

3

cos

2+

+=

k

k

k = 18.8 A2 N5[7]


96/96

A4. (a) Finding correct vectors,

=

=

1

3AC

3

4AB A1A1

Substituting correctly in the scalar product ACAB = 4(3) + 3(1) A1= 9 AG N0

(b) 10AC5AB == (A1)(A1)

Evidence of using scalar product formula M1

e.g. 105

9CABcos

= = 0.569 (3 s.f.)

CAB = 2.47 (radians), 125 A1 N3[7]

vectors exam questions (soln)

Documents