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IB Math Standard Level – Vector Practice 0506 Alei -‐ Desert Academy

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SL Vector Practice 0506 1. The points P(−2, 4), Q (3, 1) and R (1, 6) are shown in the diagram below.

(a) Find the vector . (b) Find a vector equation for the line through R parallel to the line (PQ). .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. ..............................................................................................................................................

(Total 6 marks) 2. The position vector of point A is 2i + 3 j + k and the position vector of point B is 4i − 5 j + 21k.

(a) (i) Show that = 2i −8 j + 20k. (ii) Find the unit vector u in the direction of . (iii) Show that u is perpendicular to .

(6)

Let S be the midpoint of [AB]. The line L1 passes through S and is parallel to . (b) (i) Find the position vector of S.

(ii) Write down the equation of L1. (4)

The line L2 has equation r = (5i +10 j +10k) + s (−2i + 5 j − 3k). (c) Explain why L1 and L2 are not parallel.

(2) (d) The lines L1 and L2 intersect at the point P. Find the position vector of P.

(7) (Total 19 marks)

3. The line L passes through the points A (3, 2, 1) and B (1, 5, 3).

(a) Find the vector . (b) Write down a vector equation of the line L in the form r = a + tb. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. ..............................................................................................................................................

(Total 6 marks)

PQ

ABAB

OA

OA

AB



4. The line L passes through A (0, 3) and B (1, 0). The origin is at O. The point R (x, 3 − 3x) is on L, and (OR) is perpendicular to L.

(a) Write down the vectors and . (b) Use the scalar product to find the coordinates of R. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. ..............................................................................................................................................

(Total 6 marks)

5. In this question the vector represents a displacement of 1 km east, and the vector

represents a displacement of 1 km north. The diagram below shows the positions of towns A, B and C in relation to an airport O, which is at

the point (0, 0). An aircraft flies over the three towns at a constant speed of 250 km h–1.

Town A is 600 km west and 200 km south of the airport.

Town B is 200 km east and 400 km north of the airport. Town C is 1200 km east and 350 km south of the airport. (a) (i) Find .

(ii) Show that the vector of length one unit in the direction of is .

(4) An aircraft flies over town A at 12:00, heading towards town B at 250 km h–1.

Let be the velocity vector of the aircraft. Let t be the number of hours in flight after 12:00. The

position of the aircraft can be given by the vector equation

.

(b) (i) Show that the velocity vector is .

(ii) Find the position of the aircraft at 13:00. (iii) At what time is the aircraft flying over town B?

(6) Over town B the aircraft changes direction so it now flies towards town C. It takes five hours to

travel the 1250 km between B and C. Over town A the pilot noted that she had 17 000 litres of fuel left. The aircraft uses 1800 litres of fuel per hour when travelling at 250 km h–1. When the fuel gets below 1000 litres a warning light comes on. (c) How far from town C will the aircraft be when the warning light comes on? (7)

(Total 17 marks)

AB OR

⎟⎟⎠

⎞⎜⎜⎝

⎛01

⎟⎟⎠

⎞⎜⎜⎝

⎛10

A

B

C

x

y

O

AB

AB ⎟⎟⎠

⎞⎜⎜⎝

⎛6.08.0

⎟⎟⎠

⎞⎜⎜⎝

⎛qp

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛qp

tyx

200600

⎟⎟⎠

⎞⎜⎜⎝

⎛150200



6. A boat B moves with constant velocity along a straight line. Its velocity vector is given by

v = .

At time t = 0 it is at the point (−2, 1). (a) Find the magnitude of v. (b) Find the coordinates of B when t = 2. (c) Write down a vector equation representing the position of B, giving your answer in the form r

= a + tb.

(Total 6 marks)

7. Consider the point D with coordinates (4, 5), and the point E, with coordinates (12, 11). (a) Find .

(2)

(b) Find .

(2) (c) The point D is the centre of a circle and E is on the circumference as shown in the following

diagram. The point G is also on the circumference. is perpendicular to . Find the possible

coordinates of G. (8)

(Total 12 marks) 8. Car 1 moves in a straight line, starting at point A (0, 12). Its

position p seconds after it starts is given by = + p .

(a) Find the position vector of the car after 2 seconds. (2)

Car 2 moves in a straight line starting at point B (14, 0). Its

position q seconds after it starts is given by = + q .

Cars 1 and 2 collide at point P. (b) (i) Find the value of p and the value of q when the collision occurs.

(ii) Find the coordinates of P. (6)

(Total 8 marks)

9. Find the cosine of the angle between the two vectors and .

Working: Answer:

....…………………………………….......... (Total 6 marks)

⎟⎟⎠

⎞⎜⎜⎝

⎛34

Answers:

Working:

(a) .....................................................

(b) .....................................................

(c) .....................................................

DE

DE

DE DG

⎟⎟⎠

⎞⎜⎜⎝

⎛yx

⎟⎟⎠

⎞⎜⎜⎝

⎛120

⎟⎟⎠

⎞⎜⎜⎝

⎛− 35

⎟⎟⎠

⎞⎜⎜⎝

⎛yx

⎟⎟⎠

⎞⎜⎜⎝

⎛014

⎟⎟⎠

⎞⎜⎜⎝

⎛31

⎟⎟⎠

⎞⎜⎜⎝

⎛43

⎟⎟⎠

⎞⎜⎜⎝

⎛−12



10. The following diagram shows a solid figure ABCDEFGH. Each of the six faces is a parallelogram.

The coordinates of A and B are A (7, –3, –5), B(17, 2, 5). (a) Find

(i)

(ii)

(4) The following information is given.

= , = 9, = , = 6

(b) (i) Calculate • .

(ii) Calculate • .

(iii) Calculate • . (iv) Hence, write down the size of the angle between any two intersecting edges.

(5) (c) Calculate the volume of the solid ABCDEFGH.

(2) (d) The coordinates of G are (9, 4, 12). Find the coordinates of H.

(3) (e) The lines (AG) and (HB) intersect at the point P.

Given that = , find the acute angle at P.

(5) (Total 19 marks)

AB;.AB

AD⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

366

AD AE⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−−

442

AE

AD AEAB ADAB AE

AG⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

1772

IB Math Standard Level – Vector Practice 0506 -‐ MarkScheme Alei -‐ Desert Academy 2013-‐14


SL Vector Practice 0506 MarkScheme

1. (a) = A1A1 N2

(b) Using r = a + tb

A2A1A1N4

[6]

2. (a) (M1)

A2 N3

(b) Using r = a + tb

A1A1A1N3

[6] 3. (a) (i) Evidence of subtracting all three components in the correct order M1

eg = 2i −8j + 20k AG N0

(ii) ⎜ ⎜ = (A1)

u = A1 N2

(iii) If the scalar product is zero, the vectors are perpendicular. R1 Note: Award R1 for stating the relationship between the scalar product and perpendicularity, seen anywhere in the solution.

Finding an appropriate scalar product M1

eg

A1 N0 (b) (i) EITHER

→PQ ⎟⎟⎠

⎞⎜⎜⎝

⎛− 35

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛35

61

tyx

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=−=

→→→

123

351

OAOBAB

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

→

232

AB

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

232

351

or232

123

tzyx

tzyx

( ) ( )kjikji ++−+−=−=→→→

322154OAOBAB

→AB ( ) ( )6.2111721364682082 222 ====+−+

( )kji 20824681 +−

⎟⎟⎠

⎞⎜⎜⎝

⎛+−+−= etc.,925.0370.00925.0,

46820

4688

4682 kjikji

⎟⎠⎞⎜

⎝⎛ ••

→→→OAABorOAu

1468203

46882

4682OA ×⎟⎟⎠

⎞⎜⎜⎝

⎛+×⎟⎟⎠

⎞⎜⎜⎝

⎛ −+×⎟⎟⎠

⎞⎜⎜⎝

⎛=•

→u

⎟⎟⎠

⎞⎜⎜⎝

⎛ +−=46820244

( ) 1203822OAAB ×+×−+×=•→→

0OAABor0OA =•=•→→→

u



(M1)(A1)

Therefore, = 3i − j + 11k (accept (3, −1, 11)) A1 N3 OR

(M1)

= (2i + 3j + k) + (2i + 8j + 20k) (A1)

= 3i − j + 11k A1 N3 (ii) L1 : r = (3i − j + 11k) + t (2i + 3j + 1k) A1 N1

(c) Using direction vectors (eg 2i + 3j + 1k and −2i + 5j − 3k) (M1) Valid explanation of why L1 is not parallel to L2 R1 N2 eg. Direction vectors are not scalar multiples of each other.

Angle between the direction vectors is not zero or 180. Finding the angle d1 • d2 ⎜d1⎜⎜d2⎜.

Note: Award R0 for “direction vectors are not equal”. (d) Setting up any two of the three equations (M1)

For each correct equation A1A1 eg 3 + 2t = 5 − 2s, −1 + 3t = 10 + 5s, 11 + t = 10 − 3s Attempt to solve these equations (M1) Finding one correct parameter (s = −1, t = 2) (A1) P has position vector 7i + 5j + 13k A2 N4

Notes: Award (M1)A2 if the same parameter is used for both lines in the initial correct equations. Award no further marks.

[19]

4. (a) A1A1 N2

(b) A1

M1

R is A1A1 N2

[6]

5. (a) (i) (A1)

(A1)(N2)

(ii) (must be seen) (M1)

unit vector (A1)

(AG)(N0) 4

Note: A reverse method is not acceptable in “show that” questions.

⎟⎠⎞⎜

⎝⎛ +−+

2211,

253,

242S

→OS

→→→+= AB21OAOS

21

→OS

≠

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

=→→

xx33

OR,31

AB

( )xx 333ORAB −−=•→→

( )09100ORAB =−=•→→

x

⎟⎠⎞⎜

⎝⎛

103,

109

200 600AB

400 200

→ −⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠800600

⎛ ⎞= ⎜ ⎟⎝ ⎠

2 2AB 800 600 1000→

⎜ ⎟ = + =80016001000

⎛ ⎞= ⎜ ⎟

⎝ ⎠0.80.6

⎛ ⎞= ⎜ ⎟⎝ ⎠



(b) (i) (M1)

(AG)(N0)

Note: A correct alternative method is using the given vector equation with t = 4. (ii) at 13:00, t = 1

(M1)

(A1)(N1)

(iii)

Time (M1)(A1)

over town B at 16:00 (4 pm, 4:00 pm) (Do not accept 16 or 4:00 or 4) (A1)(N3) 6

(c) Note: There are a variety of approaches. The table shows some of them, with the mark allocation. Use discretion, following this allocation as closely as possible.

Time for A to B to C = 9 hours

Distance from A to B to C

= 2250 km

Fuel used from A to B = litres (A1)

Light goes on after 16000 litres

Light goes on after 16000 litres

Fuel remaining = 9800 litres (A1)

Time for 16 000 litres

Time remaining is

= hour

Distance on 16000 litres

km

Hours before light

Time remaining is

hour

(A1)(A1) (A1)

Distance

= 27.8 km

Distance to C = 2250 – 2222.22

= 27.8 km

Distance

= 27.8 km

(A2)

(N4) 7

[17]

6. (a) = = 5 (M1)(A1) (C2)

(b) (so B is (6, 7) ) (M1)(A1) (C2)

(c) r = (not unique) (A2)(C2)

Note: Award (A1) if “ r = ” is omitted, ie not an equation.

[6]

7. (a) = = (M1)(A1) (N2)

0.8250

0.6⎛ ⎞

= ⎜ ⎟⎝ ⎠

v

200150⎛ ⎞

= ⎜ ⎟⎝ ⎠

600 2001

200 150xy

−⎛ ⎞ ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠40050

−⎛ ⎞= ⎜ ⎟−⎝ ⎠

AB 1000→

⎜ ⎟ =1000 4 (hours)250

= =

1800 4 7200× =

)889.8(988

180016000

==

=

)111.0(91 =

250180016000 ×=

)22.2222(922222 ==

88001800

( )84 4.8899

= =

( )1 0.1119

= =

1 2509

= × 1 2509

= ×

916+ 25

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛−76

34

212

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛−34

12

t

→DE ⎟⎟⎠

⎞⎜⎜⎝

⎛−−511412

⎟⎟⎠

⎞⎜⎜⎝

⎛68



(b) = (M1) = 10 (A1)(N2)

(c) Vector geometry approach Using DG = 10 (M1) (x − 4)2 + (y − 5)2 = 100 (A1) Using (DG) perpendicular to (DE) (M1)

Leading to = , = (A1)(A1)

Using = (O is the origin) (M1) G (−2, 13), G (10, −3) (accept position vectors) (A1)(A1) Algebraic approach

gradient of DE = (A1)

gradient of DG = (A1)

equation of line DG is y − 5 = (A1)

Using DG = 10 (M1) (x − 4)2 + (y − 5)2 = 100 (A1) Solving simultaneous equation (M1) G (− 2, 13), G (10, −3) (accept position vectors) (A1)(A1)

Note: Award full marks for an appropriately labelled diagram (eg showing that DG =10 , displacements of 6 and 8), or an accurate diagram leading to the correct answers.

[12]

8. (a) p = 2 ⇒ (A1)

= (accept any other vector notation, including (10, 6) ) (A1)(N2)

(b) METHOD 1 (i) equating components (M1)

0 + 5p = 14 + q , 12 − 3p = 0 + 3q (A1) p = 3, q =1 (A1)(A1)(N1)(N1)

(ii) The coordinates of P are (15, 3) (accept x = 15, y = 3 ) (A1)(A1)(N1)(N1) METHOD 2 (i) Setting up Cartesian equations (M1)

x = 5p x = 14 + q y =12 − 3p y = 3q giving 3x + 5y = 60 3x − y = 42 (A1) Solving simultaneously gives x = 15, y = 3 Substituting to find p and q

p = 3 q = 1 (A1)(A1)(N1)(N1) (ii) From above, P is (15, 3) (accept x = 15, y = 3 seen above) (A1)(A1)(N1)(N1)

[8] 9. METHOD 1

⎟⎟→DE 22 68 + ( )3664+=

→DG ⎟⎟⎠

⎞⎜⎜⎝

⎛−86 →

DG ⎟⎟⎠

⎞⎜⎜⎝

⎛− 86

→DG

→→+OGDO

86

68−

4)(34 −− x

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛35

2120

⎟⎟⎠

⎞⎜⎜⎝

⎛610

,31

014

315

,35

120

315

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛qp



Using a ⋅ b = ab cos θ (may be implied) (M1)

(A1)

Correct value of scalar product (A1)

Correct magnitudes (A1)(A1)

(A1)(C6)

METHOD 2

(A1)

(A1)

(A1)

Using cosine rule (M1) (A1)

(A1)(C6)

[6]

10. (a) (i) (A1)

=

= A1 N2

(ii) (M1) = 15 A1 N2

(b) Evidence of correct calculation of scalar product (may be in (i), (ii) or (iii)) A1

(i) ((−6)(−2) + 6(−4) + 3(4)) A1 N1

(ii) ((10)(−6) + 5(6) + 10(3)) A1 N1

(iii) ((10)(−2) + 5(−4) + 10(4)) A1 N1

(iv) 90° A1 N1

(c) Volume = ⎜ ⎜ × ⎜ ⎜ × ⎜ ⎜ (A1) = 15 × 9 × 6 = 810 (cubic units) A1 N2

θcos12–

43

12–

43

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛•⎟⎟⎠

⎞⎜⎜⎝

⎛

( ) ( ) 2–142–312–

43

=×+×=⎟⎟⎠

⎞⎜⎜⎝

⎛•⎟⎟⎠

⎞⎜⎜⎝

⎛

( )3 225 5 , 5

4 1−⎛ ⎞ ⎛ ⎞

= = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2cos125

θ −=

325

4⎛ ⎞

=⎜ ⎟⎝ ⎠2

51−⎛ ⎞

=⎜ ⎟⎝ ⎠5

343⎛ ⎞

=⎜ ⎟⎝ ⎠

34 25 5 25 5cosθ= + −2cos125

θ = −

→→→−= OAOBAB

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

537

5217

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

10510

222 10510AB ++=→

0AEAB =⋅→→

0ADAB =⋅→→

0AEAB =⋅→→

⎟⎠⎞⎜

⎝⎛ π

2or

→AB

→AD

→AE



(d) Setting up a valid equation involving H. There are many possibilities.

eg (M1)

Using equal vectors (M1)

eg

coordinates of H are (− 1, −1, 2) A1 N3

(e) A1

Attempting to use formula cos (M1)

= A1

= 0.31578... (A1) (= 1.25 radians) A1 N3

[19]

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

++=+=→→→→→→→

10510

1249

,EHAEOAOH,GHOGOHzyx

→→→→=−= ADEH,ABGH

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−−

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−−

+⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−−

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

→→

211

366

442

537

OH,211

10510

1249

OH

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

→

3318

HB

→→

→→⋅=HBAG

HBAGP̂

⎟⎟⎠

⎞⎜⎜⎝

⎛=

++++

×+×+×342342

108

33181772

31737182222222

°= 6.71P̂

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IBMath#Standard#Level#–#Vector#Practice#0506# …users.desertacademy.org/balei/Math/SL/SL.VectorPractice0506.pdf · IBMath#Standard#Level#–#Vector#Practice#0506# Alei#:#Desert#Academy#