New test - October 20, 2014 [363 marks]

1a. [2 marks]

The line L passes through the point and is parallel to the vector .

Write down a vector equation for line L .

Markschemeany correct equation in the form (accept any parameter for t)

where a is , and b is a scalar multiple of A2 N2

e.g.

Note: Award A1 for the form , A1 for , A0 for .

[2 marks]

Examiners reportIn part (a), the majority of candidates correctly recognized the equation that contains the position and direction vectors of a line.However, we saw a large number of candidates who continue to write their equations using " ", rather than the mathematically

correct " " or " ". r and represent vectors, whereas L is simply the name of the line. For part (b), very few

candidates recognized that a general point on the x-axis will be given by the vector . Common errors included candidates

setting their equation equal to , or , or even just the number .

(5, −4, 10)⎛⎝⎜

4−25

⎞⎠⎟

r = a + tb

⎛⎝⎜

5−410

⎞⎠⎟

⎛⎝⎜

4−25

⎞⎠⎟

r = + t , r = 5i − 4j + 10k + t(−8i + 4j − 10k)⎛⎝⎜

5−410

⎞⎠⎟

⎛⎝⎜

4−25

⎞⎠⎟

a + tb L = a + tb r = b + ta

L =

r = =⎛⎝⎜

x

y

z

⎞⎠⎟

⎛⎝⎜

x

y

z

⎞⎠⎟

⎛⎝⎜

x

00

⎞⎠⎟

⎛⎝⎜

000

⎞⎠⎟

⎛⎝⎜

100

⎞⎠⎟ 0

1b. [6 marks]The line L intersects the x-axis at the point P. Find the x-coordinate of P.

Markschemerecognizing that or at x-intercept (seen anywhere) (R1)

attempt to set up equation for x-intercept (must suggest ) (M1)

e.g. , ,

one correct equation in one variable (A1)

e.g. ,

finding A1

correct working (A1)

e.g.

(accept ) A1 N3

[6 marks]

Examiners reportIn part (a), the majority of candidates correctly recognized the equation that contains the position and direction vectors of a line.However, we saw a large number of candidates who continue to write their equations using " ", rather than the mathematically

correct " " or " ". r and represent vectors, whereas L is simply the name of the line. For part (b), very few

candidates recognized that a general point on the x-axis will be given by the vector . Common errors included candidates

setting their equation equal to , or , or even just the number .

y = 0 z = 0

x ≠ 0

L =⎛⎝⎜

x

00

⎞⎠⎟ 5 + 4t = x r =

⎛⎝⎜

100

⎞⎠⎟

−4 − 2t = 0 10 + 5t = 0

t = −2

x = 5 + (−2)(4)

x = −3 (−3, 0, 0)

L =

r = =⎛⎝⎜

x

y

z

⎞⎠⎟

⎛⎝⎜

x

y

z

⎞⎠⎟

⎛⎝⎜

x

00

⎞⎠⎟

⎛⎝⎜

000

⎞⎠⎟

⎛⎝⎜

100

⎞⎠⎟ 0

2a. [1 mark]

Let A and B be points such that and .

Show that .

=OA−→− ⎛

⎝⎜521

⎞⎠⎟ =OB

−→ ⎛⎝⎜

603

⎞⎠⎟

=AB−→ ⎛

⎝⎜1

−22

⎞⎠⎟

Markschemecorrect approach A1

e.g.

AG N0

[1 mark]

Examiners reportPart (a) was answered correctly by nearly every candidate.

+ , −AO−→−

OB−→ ⎛

⎝⎜603

⎞⎠⎟

⎛⎝⎜

521

⎞⎠⎟

=AB−→ ⎛

⎝⎜1

−22

⎞⎠⎟

2b. [4 marks]Let C and D be points such that ABCD is a rectangle.

Given that , show that .

Markschemerecognizing is perpendicular to (may be seen in sketch) (R1)

e.g. adjacent sides of rectangle are perpendicular

recognizing dot product must be zero (R1)

e.g.

correct substitution (A1)

e.g. ,

equation which clearly leads to A1

e.g. ,

AG N0

[4 marks]

Examiners reportIn part (b), the candidates who realized that the vectors must be perpendicular were successful using the scalar product to find p .Incorrect approaches included using magnitudes, or creating vector equations of lines for the sides and setting them equal to eachother. In addition, there were a good number of candidates who worked backwards, using the given value of 3 for p to find thecoordinates of point D. Candidates who work backwards on a "show that" question will earn no marks.

=AD−→− ⎛

⎝⎜4p

1

⎞⎠⎟ p = 3

AD−→−

AB−→

∙ = 0AD−→−

AB−→−

(1 × 4) + (−2 × p) + (2 × 1) 4 − 2p + 2 = 0

p = 3

6 − 2p = 0 2p = 6

p = 3

2c. [4 marks]Let C and D be points such that ABCD is a rectangle.

Find the coordinates of point C.

Markschemecorrect approach (seen anywhere including sketch) (A1)

e.g. ,

recognizing opposite sides are equal vectors (may be seen in sketch) (R1)

e.g. , , ,

coordinates of point C are (10, 3, 4) (accept ) A2 N4

Note: Award A1 for two correct values.

[4 marks]

Examiners reportPart (c) was more difficult for candidates, and was left blank by some. Some candidates found rather than , as required.Many candidates recognized that the opposite sides of the rectangle must be equal, but did not consider the directions of the vectorsfor those sides. There were also a good number of candidates who mislabelled the vertices of their rectangles, which led to themworking with a rectangle ABDC, rather than ABCD.

= +OC−→−

OB−→

BC−→

+OD−→−

DC−→−

=BC−→

AD−→−

=DC−→−

AB−→

+⎛⎝⎜

603

⎞⎠⎟

⎛⎝⎜

431

⎞⎠⎟ +

⎛⎝⎜

952

⎞⎠⎟

⎛⎝⎜

1−22

⎞⎠⎟

⎛⎝⎜

1034

⎞⎠⎟

AC−→−

OC−→−

2d. [5 marks]Let C and D be points such that ABCD is a rectangle.

Find the area of rectangle ABCD.

Markschemeattempt to find one side of the rectangle (M1)

e.g. substituting into magnitude formula

two correct magnitudes A1A1

e.g. , 3 ; ,

multiplying magnitudes (M1)

e.g.

(accept ) A1 N3

[5 marks]

Examiners reportThe majority of candidates who attempted part (d) were successful in multiplying the magnitudes of the sides. Unfortunately, therewere some who set up their solutions correctly, then had arithmetic errors in their working.

+ +(1)2 (−2)2 22− −−−−−−−−−−−−√ 16 + 9 + 1− −−−−−−−√ 26−−√

×26−−√ 9√

area = (= 3 )234−−−√ 26−−√ 3 × 26−−√

−→− −→− −→−

3a. [5 marks]

The following diagram shows the cuboid (rectangular solid) OABCDEFG, where O is the origin, and , , .

(i) Find .

(ii) Find .

(iii) Show that .

Markscheme(i) valid approach (M1)

e.g.

A1 N2

(ii) valid approach (M1)

e.g. ; ;

A1 N2

(iii) correct approach A1

e.g. ; ;

AG N0

[5 marks]

Examiners reportAlthough a large proportion of candidates managed to answer this question, their biggest challenge was the use of a proper notation torepresent the vectors and vector equations of lines.

In part (a), finding and was generally well done, although many lost the mark for (iii) due to poor working or not clearlyshowing the result.

= 4iOA−→−

= 3jOC−→−

= 2kOD−→−

OB−→

OF−→

= −4i + 3j + 2kAG−→−

OA + OB

= 4i + 3jOB−→

+ +OA−→−

AB−→

BF−→

+OB−→

BF−→

+ +OC−→−

CG−→−

GF−→

= 4i + 3j + 2kOF−→

+ +AO−→−

OC−→−

CG−→−

+ +AB−→

BF−→

FG−→

+ +AB−→

BC−→

CG−→−

= −4i + 3j + 2kAG−→−

OB−→

OF−→

3b. [4 marks]Write down a vector equation for

(i) the line OF;

(ii) the line AG.

Markscheme(i) any correct equation for (OF) in the form A2 N2

where is 0 or , and is a scalar multiple of

e.g. , ,

(ii) any correct equation for (AG) in the form A2 N2

where is or and is a scalar multiple of

e.g. , ,

[4 marks]

Examiners reportPart (b) was very poorly done. Not all the students recognized which correct position vectors they had to use to write the equations ofthe lines. It was seen that they frequently failed to present the equations in the required format, which prevented these candidates fromachieving full marks. The notations generally seen were , or .

r = a + tb

a 4i + 3j + 2k b 4i + 3j + 2k

r = t(4, 3, 2) r =⎛⎝⎜

4t

3t

2t

⎞⎠⎟ r = 4i + 3j + 2k + t(4i + 3j + 2k)

r = a + sb

a 4i 3j + 2k b −4i + 3j + 2k

r = (4, 0, 0) + s(−4, 3, 2) r =⎛⎝⎜

4 − 4s

3s

2s

⎞⎠⎟ r = 3j + 2k + s(−4i + 3j + 2k)

AG = a + bt r = 4 + t(4, 3, 2) L = a + bt

3c. [7 marks]Find the obtuse angle between the lines OF and AG.

Markschemechoosing correct direction vectors, and (A1)(A1)

scalar product (A1)

magnitudes , , (A1)(A1)

substitution into formula M1

e.g.

,

or A1 N4

[7 marks]

Examiners reportMost achieved the correct result in part (c) with many others gaining most of the marks as follow through from choosing incorrectvectors. Some students did not state which vectors had been used, another cause for losing marks. A few showed poor notation,including i, j and k in the working.

OF−→

AG−→−

= −16 + 9 + 4 (= −3)

+ +42 32 22− −−−−−−−−√ + +(−4)2 32 22− −−−−−−−−−−−√ ( , )29−−√ 29−−√

cosθ = = (− )−16+9+4

( )×+ +42 32 22√ + +(−4)2 32 22√3

29

95.93777∘ 1.67443 radians

θ = 95.9∘ 1.67

4a. [3 marks]

A line passes though points P(−1, 6, −1) and Q(0, 4, 1) .

(i) Show that .

(ii) Hence, write down an equation for in the form .

L1

=PQ−→ ⎛

⎝⎜1

−22

⎞⎠⎟

L1 r = a + tb

Markscheme(i) evidence of correct approach A1

e.g. ,

AG N0

(ii) any correct equation in the form A2 N2

where a is either or and b is a scalar multiple of

e.g. , ,

[3 marks]

Examiners reportA pleasing number of candidates were successful on this straightforward vector and line question. Part (a) was generally wellanswered, although a few candidates still labelled their line or used a position vector for the direction vector. Follow-throughmarking allowed full recovery from the latter error.

= −PQ−→

OQ−→−

OP−→

Q − P

=PQ−→ ⎛

⎝⎜1

−22

⎞⎠⎟

r = a + tb

OP−→

OQ−→−

PQ−→

r = + t⎛⎝⎜

−16

−1

⎞⎠⎟

⎛⎝⎜

1−22

⎞⎠⎟ r =

⎛⎝⎜

t

4 − 2t

1 + 2t

⎞⎠⎟ r = 4j + k + t(i − 2j + 2k)

L =

4b. [7 marks]

A second line has equation .

Find the cosine of the angle between and .

Markschemechoosing a correct direction vector for (A1)

e.g.

finding scalar products and magnitudes (A1)(A1)(A1)

scalar product

magnitudes ,

substitution into formula M1

e.g.

A2 N5

[7 marks]

Examiners reportFew candidates wrote down their direction vector in part (b) which led to lost follow-through marks, and a common error was findingan incorrect scalar product due to difficulty multiplying by zero.

L2 r = + s⎛⎝⎜

42

−1

⎞⎠⎟

⎛⎝⎜

30

−4

⎞⎠⎟

PQ−→

L2

L2

⎛⎝⎜

30

−4

⎞⎠⎟

= 1(3) − 2(0) + 2(−4) (= −5)

= + +12 (−2)2 22− −−−−−−−−−−−√ (= 3) + +32 02 (−4)2

− −−−−−−−−−−−√ (= 5)

cosθ = −5×9√ 25√

cosθ = − 13

4c. [7 marks]The lines and intersect at the point R. Find the coordinates of R.L1 L2

Markschemeevidence of valid approach (M1)

e.g. equating lines,

EITHER

one correct equation in one variable A2

e.g.

OR

two correct equations in two variables A1A1

e.g. ,

THEN

attempt to solve (M1)

one correct parameter A1

e.g. ,

correct substitution of either parameter (A1)

e.g. ,

coordinates A1 N3

[7 marks]

Examiners reportPart (c) was generally well understood with some candidates realizing that the equation in just one variable led to the correctparameter more quickly than solving a system of two equations to find both parameters. Some candidates gave the answer as (s, t)instead of substituting those parameters, indicating a more rote understanding of the problem. Another common error was using thesame parameter for both lines.

There were an alarming number of misreads of negative signs from the question or from the candidate working.

=L1 L2

6 − 2t = 2

2t + 4s = 0 t − 3s = 5

t = 2 s = −1

r = + (−1)⎛⎝⎜

42

−1

⎞⎠⎟

⎛⎝⎜

30

−4

⎞⎠⎟ r = + (+2)

⎛⎝⎜

−16

−1

⎞⎠⎟

⎛⎝⎜

1−22

⎞⎠⎟

R(1, 2, 3)

5a. [4 marks]

The line passes through the points P(2, 4, 8) and Q(4, 5, 4) .

(i) Find .

(ii) Hence write down a vector equation for in the form .

L1

PQ−→

L1 r = a + sb

Markscheme(i) evidence of approach (M1)

e.g. ,

A1 N2

(ii) any correct equation in the form (accept any parameter for s)

where a is or , and b is a scalar multiple of A2 N2

e.g. , ,

Note: Award A1 for the form , A1 for , A0 for .

[4 marks]

Examiners reportIn part (a), nearly all the candidates correctly found the vector PQ, and the majority went onto find the correct vector equation of theline. There are still many candidates who do not write this equation in the correct form, using "r = ", and these candidates werepenalized one mark.

+PO−→

OQ−→−

P − Q

=PQ−→ ⎛

⎝⎜21

−4

⎞⎠⎟

r = a + sb

⎛⎝⎜

248

⎞⎠⎟

⎛⎝⎜

454

⎞⎠⎟

⎛⎝⎜

21

−4

⎞⎠⎟

r = + s⎛⎝⎜

248

⎞⎠⎟

⎛⎝⎜

21

−4

⎞⎠⎟ r =

⎛⎝⎜

4 + 2s

5 + 1s

4 − 4s

⎞⎠⎟ r = 2i + 4j + 8k + s(2i + 1j − 4k)

a + sb L = a + sb r = b + sa

5b. [7 marks]

The line is perpendicular to , and parallel to , where .

(i) Find the value of p .

(ii) Given that passes through , write down a vector equation for .

Markscheme(i) choosing correct direction vectors for and (A1) (A1)

e.g. ,

evidence of equating scalar product to 0 (M1)

correct calculation of scalar product A1

e.g. ,

A1 N3

(ii) any correct expression in the form (accept any parameter for t)

where a is , and b is a scalar multiple of A2 N2

e.g. , ,

Note: Award A1 for the form , A1 for (unless they have been penalised for in part (a)), A0 for .

[7 marks]

L2 L1

⎛⎝⎜

3p

2p

4

⎞⎠⎟ p ∈ Z

L2 R(10, 6, − 40) L2

L1 L2

⎛⎝⎜

21

−4

⎞⎠⎟

⎛⎝⎜

3p

2p

4

⎞⎠⎟

2 × 3p + 1 × 2p + (−4) × 4 8p − 16 = 0

p = 2

r = a + tb

⎛⎝⎜

106

−40

⎞⎠⎟

⎛⎝⎜

644

⎞⎠⎟

r = + t⎛⎝⎜

106

−40

⎞⎠⎟

⎛⎝⎜

644

⎞⎠⎟ r =

⎛⎝⎜

10 + 6s

6 + 4s

−40 + 4s

⎞⎠⎟ r = 10i + 6j − 40k + s(6i + 4j + 4k)

a + tb L = a + tb L = a + sb

r = b + ta

Examiners reportIn part (b), the majority of candidates knew to set the scalar product equal to zero for the perpendicular vectors, and were able to findthe correct value of p.

5c. [7 marks]The lines and intersect at the point A. Find the x-coordinate of A.

Markschemeappropriate approach (M1)

e.g.

any two correct equations with different parameters A1A1

e.g. , ,

attempt to solve simultaneous equations (M1)

correct working (A1)

e.g. , , ,

one correct parameter , A1

(accept (22, 14, −32)) A1 N4

[7 marks]

Examiners reportA good number of candidates used the correct method to find the intersection of the two lines, though some algebraic and arithmeticerrors kept some from finding the correct final answer.

L1 L2

+ s = + t⎛⎝⎜

248

⎞⎠⎟

⎛⎝⎜

21

−4

⎞⎠⎟

⎛⎝⎜

106

−40

⎞⎠⎟

⎛⎝⎜

644

⎞⎠⎟

2 + 2s = 10 + 6t 4 + s = 6 + 4t 8 − 4s = −40 + 4t

−6 = −2 − 2t 4 = 2t −4 + 5s = 46 5s = 50

s = 10 t = 2

x = 22

6a. [2 marks]

A line L passes through and is parallel to the line .

Write down a vector equation for L in the form .

Markschemecorrect equation in the form A2 N2

[2 marks]

Examiners reportMany candidates answered this question well. Some continue to write the vector equation in (a) using "L =", which does not earn fullmarks.

A(1, − 1, 2) r = + s⎛⎝⎜

−215

⎞⎠⎟

⎛⎝⎜

13

−2

⎞⎠⎟

r = a + tb

r = a + tb

r = + t⎛⎝⎜

1−12

⎞⎠⎟

⎛⎝⎜

13

−2

⎞⎠⎟

6b. [4 marks]

The line L passes through point P when .

Find

(i) ;

(ii) .

Markscheme(i) attempt to substitute into the equation (M1)

e.g. ,

A1 N2

(ii) correct substitution into formula for magnitude A1

e.g. ,

A1 N1

[4 marks]

Examiners reportPart (b) proved accessible for most, although small arithmetic errors were not uncommon. Some candidates substituted into the

original equation, and a few answered . A small but surprising number of candidates left this question blank,

suggesting the topic was not given adequate attention in course preparation.

t = 2

OP−→

| |OP−→

t = 2

⎛⎝⎜

26

−4

⎞⎠⎟ + 2

⎛⎝⎜

1−12

⎞⎠⎟

⎛⎝⎜

13

−2

⎞⎠⎟

=OP−→ ⎛

⎝⎜35

−2

⎞⎠⎟

+ + −32 52 22− −−−−−−−−−√ + +32 52 22− −−−−−−−−√

| | =OP−→

38−−√

t = 2

=OP−→ ⎛

⎝⎜26

−4

⎞⎠⎟

7a. [3 marks]

The following diagram shows the obtuse-angled triangle ABC such that and .

(i) Write down .

(ii) Find .

=AB−→ ⎛

⎝⎜−30

−4

⎞⎠⎟ =AC

−→− ⎛⎝⎜

−22

−6

⎞⎠⎟

BA−→

BC−→

Markscheme

(i) A1 N1

(ii) evidence of combining vectors (M1)

e.g. , ,

A1 N2

[3 marks]

Examiners reportMany candidates answered (a) correctly, although some reversed the vectors when finding , while others miscopied the vectorsfrom the question paper.

=BA−→ ⎛

⎝⎜304

⎞⎠⎟

+ =AB−→

BC−→

AC−→−

+BA−→

AC−→−

−⎛⎝⎜

−22

−6

⎞⎠⎟

⎛⎝⎜

−30

−4

⎞⎠⎟

=BC−→ ⎛

⎝⎜12

−2

⎞⎠⎟

BC−→

7b. [7 marks](i) Find .

(ii) Hence, find .

Markscheme(i) METHOD 1

finding , ,

e.g. , ,

substituting into formula for M1

e.g. ,

A1 N3

METHOD 2

finding , , (A1)(A1)(A1)

e.g. , ,

substituting into cosine rule M1

e.g. ,

A1 N3

(ii) evidence of using Pythagoras (M1)

e.g. right-angled triangle with values,

A1 N2

[7 marks]

Examiners reportStudents had no difficulty finding the scalar product and magnitudes of the vectors used in finding the cosine. However, fewrecognized that is the vector to apply in the formula to find the cosine value. Most used to obtain a positive cosine, whichneglects that the angle is obtuse and thus has a negative cosine. Surprisingly few students could then take a value for cosine and use itto find a value for sine. Most left (bii) blank entirely.

cosA CB

sinA CB

∙BA−→

BC−→

| |BA−→

| |BC−→

∙ = 3 × 1 + 0 + 4 × −2BA−→

BC−→

| =|BA−→−

+32 42− −−−−√ | = 3|BC−→−

cosθ

3×1+0+4×−2

3 +0+32 42√

−55×3

cosA C = −B 515

(= − )13

| |AC−→−

| |BA−→

| |BC−→

| | =AC−→−

+ +22 22 62− −−−−−−−−√ | | =BA−→

+32 42− −−−−√ | | = 3BC−→

+ −52 32 ( )44√ 2

2×5×325+9−44

30

cosA C = −B 1030

(= − )13

x + x = 1sin2 cos2

sin A C =B 8√3

(= )2 2√3

BA−→

AB−→

7c. [6 marks]The point D is such that , where .

(i) Given that , show that .

(ii) Hence, show that is perpendicular to .

Markscheme(i) attempt to find an expression for (M1)

e.g. ,

correct equation A1

e.g. ,

A1

AG N0

(ii) evidence of scalar product (M1)

e.g. ,

correct substitution

e.g. , A1

A1

is perpendicular to AG N0

[6 marks]

Examiners reportPart (c) proved accessible for many candidates. Some created an expression for and then substituted the given to obtain

, which does not satisfy the "show that" instruction. Many students recognized that the scalar product must be zero for vectors tobe perpendicular, and most provided the supporting calculations.

=CD−→− ⎛

⎝⎜−45p

⎞⎠⎟ p > 0

=|CD|− →−

50−−√ p = 3

CD−→−

BC−→

| |CD−→−

+ +(−4)2 52 p2− −−−−−−−−−−−√ | = + +CD

−→−|2 42 52 p2

=+ +(−4)2 52 p2− −−−−−−−−−−−√ 50−−√ + + = 5042 52 p2

= 9p2

p = 3

∙⎛⎝⎜

−453

⎞⎠⎟

⎛⎝⎜

12

−2

⎞⎠⎟ ∙CD

−→−BC−→

−4 × 1 + 5 × 2 + 3 × −2 −4 + 10 − 6

∙ = 0CD−→−

BC−→

CD−→−

BC−→

| |CD−→−

p = 350−−√

8a. [2 marks]

The following diagram shows quadrilateral ABCD, with , , and .

Find .

=AD−→−

BC−→

= ( )AB−→ 3

1= ( )AC

−→− 44

BC−→

Markschemeevidence of appropriate approach (M1)

e.g. ,

A1 N2

[2 marks]

Examiners reportThis question on two-dimensional vectors was generally very well done.

−AC−→−

AB−→ ( )4 − 3

4 − 1

= ( )BC−→ 1

3

8b. [2 marks]Show that .

MarkschemeMETHOD 1

(A1)

correct approach A1

e.g. ,

AG N0

METHOD 2

recognizing (A1)

correct approach A1

e.g. ,

AG N0

[2 marks]

Examiners reportThis question on two-dimensional vectors was generally very well done. A very small number of candidates had trouble with the"show that" in part (b) of the question.

= ( )BD−→ −2

2

= ( )AD−→− 1

3

−AD−→−

AB−→ ( )1 − 3

3 − 1

= ( )BD−→ −2

2

=CD−→−

BA−→

+BC−→

CD−→− ( )1 − 3

3 − 1

= ( )BD−→ −2

2

8c. [3 marks]Show that vectors and are perpendicular.BD−→

AC−→−

MarkschemeMETHOD 1

evidence of scalar product (M1)

e.g. ,

correct substitution A1

e.g. ,

A1

therefore vectors and are perpendicular AG N0

METHOD 2

attempt to find angle between two vectors (M1)

e.g.

correct substitution A1

e.g. ,

A1

therefore vectors and are perpendicular AG N0

[3 marks]

Examiners reportNearly all candidates knew to use the scalar product in part (c) to show that the vectors are perpendicular.

∙BD−→

AC−→− ( ) ∙ ( )−2

244

(−2)(4) + (2)(4) −8 + 8

∙ = 0BD−→

AC−→−

BD−→

AC−→−

a∙b

ab

(−2)(4)+(2)(4)

8√ 32√cosθ = 0

θ = 90∘

BD−→

AC−→−

9. [6 marks]

Consider the vectors and .

(a) Find

(i) ;

(ii) .

Let , where is the zero vector.

(b) Find .

a = ( )2−3

b = ( )14

2a + b

|2a + b|

2a + b + c = 0 0

c

Markscheme(a) (i) (A1)

correct expression for A1 N2

eg , ,

(ii) correct substitution into length formula (A1)

eg ,

A1 N2

[4 marks]

(b) valid approach (M1)

eg , ,

A1 N2

[2 marks]

Examiners reportMost candidates comfortably applied algebraic techniques to find new vectors. However, a significant number of candidatesanswered part (b) as the absolute numerical value of the vector components, which suggests a misunderstanding of the modulusnotation. Those who understood the notation easily made the calculation.

2a = ( )4−6

2a + b

( )5−2

(5,−2) 5i − 2j

+52 22− −−−−√ + −52 22− −−−−−−√

|2a + b| = 29−−√

c = −(2a + b) 5 + x = 0 −2 + y = 0

c = ( )−52

10a. [2 marks]

Consider points A( , , ) , B( , , ) and C( , , ) . The line passes through C and is parallel to .

Find .

Markschemevalid approach (M1)

eg , ,

A1 N2

[2 marks]

Examiners reportWhile many candidates can find a vector given two points, few could write down a fully correct vector equation of a line.

1 −2 −1 7 −4 3 1 −2 3 L1 AB−→

AB−→

−⎛⎝⎜

7−43

⎞⎠⎟

⎛⎝⎜

1−2−1

⎞⎠⎟ A − B = +AB

−→AO−→−

OB−→

=AB−→ ⎛

⎝⎜6

−24

⎞⎠⎟

10b. [2 marks]Hence, write down a vector equation for .L1

Markschemeany correct equation in the form (accept any parameter for )

where and is a scalar multiple of A2 N2

eg , ,

Note: Award A1 for , A1 for , A0 for .

[2 marks]

Examiners reportWhile many candidates can find a vector given two points, few could write down a fully correct vector equation of a line. Mostcandidates wrote their equation as “ ”, which misrepresents that the resulting equation must still be a vector.

r = a + tb t

a =⎛⎝⎜

1−23

⎞⎠⎟ b AB

−→

r = + t⎛⎝⎜

1−23

⎞⎠⎟

⎛⎝⎜

6−24

⎞⎠⎟ (x,y,z) = (1,−2,3) + t(3,−1,2) r =

⎛⎝⎜

1 + 6t

−2 − 2t

3 + 4t

⎞⎠⎟

a + tb = a + tbL1 r = b + ta

=L1

10c. [3 marks]

A second line, , is given by .

Given that is perpendicular to , show that .

Markschemerecognizing that scalar product (seen anywhere) R1

correct calculation of scalar product (A1)

eg ,

correct working A1

eg ,

AG N0

[3 marks]

Examiners reportThose who recognized that vector perpendicularity means the scalar product is zero found little difficulty answering part (b).Occasionally a candidate would use the given to show the scalar product is zero. However, working backward from the givenanswer earns no marks in a question that requires candidates to show that this value is achieved.

L2 r = + s⎛⎝⎜

−12

15

⎞⎠⎟

⎛⎝⎜

3−3p

⎞⎠⎟

L1 L2 p = −6

= 0

6(3) − 2(−3) + 4p 18 + 6 + 4p

24 + 4p = 0 4p = −24

p = −6

p = 6

10d. [7 marks]The line intersects the line at point Q. Find the -coordinate of Q.L1 L2 x

Markschemesetting lines equal (M1)

eg ,

any two correct equations with different parameters A1A1

eg , ,

attempt to solve their simultaneous equations (M1)

one correct parameter A1

eg ,

attempt to substitute parameter into vector equation (M1)

eg ,

(accept (4, -3, 5), ignore incorrect values for and ) A1 N3

[7 marks]

Examiners reportWhile many candidates knew to set the lines equal to find an intersection point, a surprising number could not carry the process tocorrect completion. Some could not solve a simultaneous pair of equations, and for those who did, some did not know what to dowith the parameter value. Another common error was to set the vector equations equal using the same parameter, from which thecandidates did not recognize a system to solve. Furthermore, it is interesting to note that while only one parameter value is needed toanswer the question, most candidates find or attempt to find both, presumably out of habit in the algorithm.

=L1 L2 + t = + s⎛⎝⎜

1−23

⎞⎠⎟

⎛⎝⎜

6−24

⎞⎠⎟

⎛⎝⎜

−12

15

⎞⎠⎟

⎛⎝⎜

3−3−6

⎞⎠⎟

1 + 6t = 1 + 3s −2 − 2t = 2 − 3s 3 + 4t = 15 − 6s

t = 12

s = 53

+⎛⎝⎜

1−23

⎞⎠⎟ 1

2

⎛⎝⎜

6−24

⎞⎠⎟ 1 + × 61

2

x = 4 y z

11. [7 marks]Line has equation and line has equation .

Lines and intersect at point A. Find the coordinates of A.

Markschemeappropriate approach (M1)

eg ,

any two correct equations A1A1

eg , ,

attempt to solve (M1)

eg substituting one equation into another

one correct parameter A1

eg ,

correct substitution (A1)

eg , ,

A ( , , ) (accept column vector) A1 N4

[7 marks]

L1 = + sr1

⎛⎝⎜

106

−1

⎞⎠⎟

⎛⎝⎜

2−5−2

⎞⎠⎟ L2 = + tr2

⎛⎝⎜

21

−3

⎞⎠⎟

⎛⎝⎜

352

⎞⎠⎟

L1 L2

+ s = + t⎛⎝⎜

106

−1

⎞⎠⎟

⎛⎝⎜

2−5−2

⎞⎠⎟

⎛⎝⎜

21

−3

⎞⎠⎟

⎛⎝⎜

352

⎞⎠⎟ =L1 L2

10 + 2s = 2 + 3t 6 − 5s = 1 + 5t −1 − 2s = −3 + 2t

s = −1 t = 2

2 + 3(2) 1 + 5(2) −3 + 2(2)

= 8 11 1

Examiners reportMost students were able to set up one or more equations, but few chose to use their GDCs to solve the resulting system. Algebraicerrors prevented many of these candidates from obtaining the final three marks. Some candidates stopped after finding the value of sand/or .t

12a. [3 marks]

Consider the points A( , , ) , B( , , ) , and C( , , ) , .

Find

(i) ;

(ii) .

Markscheme(i) appropriate approach (M1)

eg ,

A1 N2

(ii) A1 N1

[3 marks]

Examiners reportThe majority of candidates successfully found the vectors between the given points in part (a).

5 2 1 6 5 3 7 6 a + 1 a ∈ R

AB−→

AC−→−

+AO−→−

OB−→

B − A

=AB−→ ⎛

⎝⎜132

⎞⎠⎟

=AC−→− ⎛

⎝⎜24a

⎞⎠⎟

12b. [4 marks]

Let be the angle between and .

Find the value of for which .

Markschemevalid reasoning (seen anywhere) R1

eg scalar product is zero,

correct scalar product of their and (may be seen in part (c)) (A1)

eg

correct working for their and (A1)

eg ,

A1 N3

[4 marks]

q AB−→

AC−→−

a q = π

2

cos =π

2u∙v

|u||v|

AB−→

AC−→−

1(2) + 3(4) + 2(a)

AB−→

AC−→−

2a + 14 2a = −14

a = −7

Examiners reportIn part (b), while most candidates correctly found the value of , many unnecessarily worked with the magnitudes of the vectors,sometimes leading to algebra errors.

a

12c. [4 marks]Show that .

Markschemecorrect magnitudes (may be seen in (b)) (A1)(A1)

,

substitution into formula (M1)

eg ,

simplification leading to required answer A1

eg

AG N0

[4 marks]

Examiners reportSome candidates showed a minimum of working in part (c)(i); in a “show that” question, candidates need to ensure that their workingclearly leads to the answer given. A common error was simplifying the magnitude of vector AC to instead of .

cosq = 2a+14

14 +280a2√

(= )+ +12 32 22− −−−−−−−−√ 14−−√ (= )+ +22 42 a2− −−−−−−−−√ 20 + a2

− −−−−−√

cosθ 1×2+3×4+2×a

+ +12 32 22√ + +22 42 a2√

14+2a

14√ 4+16+a2√

cosθ = 14+2a

14√ 20+a2√

cosθ = 2a+14

14 +280a2√

20a2− −−−√ 20 + a2− −−−−−

√

13a. [5 marks]

The diagram shows quadrilateral ABCD with vertices A(1, 0), B(1, 5), C(5, 2) and D(4, −1) .

(i) Show that .

(ii) Find .

(iii) Show that is perpendicular to .

= ( )AC−→− 4

2

BD−→

AC−→−

BD−→

Markscheme(i) correct approach A1

e.g. ,

AG N0

(ii) appropriate approach (M1)

e.g. , , move 3 to the right and 6 down

A1 N2

(iii) finding the scalar product A1

e.g. ,

valid reasoning R1

e.g. , scalar product is zero

is perpendicular to AG N0

[5 marks]

Examiners reportThe majority of candidates were successful on part (a), finding vectors between two points and using the scalar product to show twovectors to be perpendicular.

−OC−→−

OA−→− ( ) − ( )5

210

= ( )AC−→− 4

2

D − B ( ) − ( )4−1

15

= ( )BD−→ 3

−6

4(3) + 2(−6) 12 − 12

4(3) + 2(−6) = 0

AC−→−

BD−→

13b. [4 marks]The line (AC) has equation .

(i) Write down vector u and vector v .

(ii) Find a vector equation for the line (BD).

Markscheme(i) correct “position” vector for u; “direction” vector for v A1A1 N2

e.g. , ; ,

accept in equation e.g.

(ii) any correct equation in the form , where

, A2 N2

[4 marks]

Examiners reportAlthough a large number of candidates answered part (b) correctly, there were many who had trouble with the vector equation of aline. Most notably, there were those who confused the position vector with the direction vector, and those who wrote their equation inan incorrect form.

r = u + sv

u = ( )52

u = ( )10

v = ( )42

v = ( )−2−1

( ) + t ( )52

−4−2

r = a + tb b = BD−→

r = ( ) + t ( )15

3−6

( ) = ( ) + t ( )x

y

4−1

−12

13c. [3 marks]The lines (AC) and (BD) intersect at the point .

Show that .

P(3, k)

k = 1

MarkschemeMETHOD 1

substitute (3, k) into equation for (AC) or (BD) (M1)

e.g. ,

value of t or s A1

e.g. , , ,

substituting A1

e.g.

AG N0

METHOD 2

setting up two equations (M1)

e.g. , ; setting vector equations of lines equal

value of t or s A1

e.g. , , ,

substituting A1

e.g.

AG N0

[3 marks]

Examiners reportIn part (c), most candidates seemed to know what was required, though there were many who made algebraic errors when solving forthe parameters. A few candidates worked backward, using , which is not allowed on a "show that" question.

3 = 1 + 4s 3 = 1 + 3t

s = 12

− 12

t = 23

− 13

k = 0 + (2)12

k = 1

1 + 4s = 4 + 3t 2s = −1 − 6t

s = 12

− 12

t = 23

− 13

r = ( ) − ( )4−1

13

3−6

k = 1

k = 1

13d. [5 marks]The lines (AC) and (BD) intersect at the point .

Hence find the area of triangle ACD.

Markscheme (A1)

(A1)

(A1)

area ( ) M1

A1 N4

[5 marks]

Examiners reportIn part (d), candidates attempted many different geometric and vector methods to find the area of the triangle. As the question said"hence", it was required that candidates should use answers from their previous working - i.e. and . Somegeometric approaches, while leading to the correct answer, did not use "hence" or lacked the required justification.

P(3, k)

= ( )PD−→ 1

−2

| | =PD−→

+22 12− −−−−√ (= )5√

| | =AC−→−

+42 22− −−−−√ (= )20−−√

= × | | × | |12

AC−→−

PD−→

= × ×12

20−−√ 5√

= 5

AC⊥BD P(3, 1)

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

14a. [2 marks]

The line is represented by the vector equation .

A second line is parallel to and passes through the point B( , , ) .

Write down a vector equation for in the form .

Markschemeany correct equation in the form (accept any parameter) A2 N2

e.g.

Note: Award A1 for , A1 for , A0 for .

[2 marks]

Examiners reportMany candidates gave a correct vector equation for the line.

L1 r = + p⎛⎝⎜

−3−1

−25

⎞⎠⎟

⎛⎝⎜

21

−8

⎞⎠⎟

L2 L1 −8 −5 25

L2 r = a + tb

r = a + tb

r = + t⎛⎝⎜

−8−525

⎞⎠⎟

⎛⎝⎜

21

−8

⎞⎠⎟

a + tb L = a + tb r = b + ta

14b. [5 marks]A third line is perpendicular to and is represented by .

Show that .

Markschemerecognizing scalar product must be zero (seen anywhere) R1

e.g.

evidence of choosing direction vectors (A1)(A1)

correct calculation of scalar product (A1)

e.g.

simplification that clearly leads to solution A1

e.g. ,

AG N0

[5 marks]

Examiners reportA common error was to misplace the initial position and direction vectors. Those who set the scalar product of the direction vectors tozero typically solved for k successfully. Those who substituted earned fewer marks for working backwards in a "show that"question.

L3 L1 r = + q⎛⎝⎜

503

⎞⎠⎟

⎛⎝⎜

−7−2k

⎞⎠⎟

k = −2

a ∙ b = 0

,⎛⎝⎜

21

−8

⎞⎠⎟

⎛⎝⎜

−7−2k

⎞⎠⎟

2(−7) + 1(−2) − 8k

−16 − 8k −16 − 8k = 0

k = −2

k = −2

14c. [6 marks]The lines and intersect at the point A.

Find the coordinates of A.

L1 L3

Markschemeevidence of equating vectors (M1)

e.g. ,

any two correct equations A1A1

e.g. , ,

attempting to solve equations (M1)

finding one correct parameter ( , ) A1

the coordinates of A are A1 N3

[6 marks]

Examiners reportMany went on to find the coordinates of point A, however some used the same letter, say p, for each parameter and thus could notsolve the system.

=L1 L3 + p = + q⎛⎝⎜

−3−1

−25

⎞⎠⎟

⎛⎝⎜

21

−8

⎞⎠⎟

⎛⎝⎜

503

⎞⎠⎟

⎛⎝⎜

−7−2−2

⎞⎠⎟

−3 + 2p = 5 − 7q −1 + p = −2q −25 − 8p = 3 − 2q

p = −3 q = 2

(−9,−4,−1)

14d. [5 marks]The lines and intersect at point C where .

(i) Find .

(ii) Hence, find .

Markscheme(i) evidence of appropriate approach (M1)

e.g. ,

A1 N2

(ii) finding A1

evidence of finding magnitude (M1)

e.g.

A1 N3

[5 marks]

Examiners reportPart (d) proved challenging as many candidates did not consider that . Rather, many attempted to find thecoordinates of point C, which became a more arduous and error-prone task.

L2 L3 =BC−→ ⎛

⎝⎜63

−24

⎞⎠⎟

AB−→

| |AC−→−

+ =OA−→−

AB−→

OB−→

= −AB−→ ⎛

⎝⎜−8−525

⎞⎠⎟

⎛⎝⎜

−9−4−1

⎞⎠⎟

=AB−→ ⎛

⎝⎜1

−126

⎞⎠⎟

=AC−→− ⎛

⎝⎜722

⎞⎠⎟

| | =AC−→−

+ +72 22 22− −−−−−−−−√

| | =AC−→−

57−−√

+ =AB−→

BC−→

AC−→−

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

15a. [2 marks]

Let and .

Find .

Markschemeevidence of appropriate approach (M1)

e.g. ,

A1 N2

[2 marks]

Examiners reportPart (a) was generally done well with candidates employing different correct methods to find the vector . Some candidatessubtracted the given vectors in the wrong order and others simply added them. Calculation errors were seen with some frequency.

=AB−→ ⎛

⎝⎜6

−23

⎞⎠⎟ =AC

−→− ⎛⎝⎜

−2−32

⎞⎠⎟

BC−→

= +BC−→

BA−→

AC−→−

−⎛⎝⎜

−2−32

⎞⎠⎟

⎛⎝⎜

6−23

⎞⎠⎟

=BC−→ ⎛

⎝⎜−8−1−1

⎞⎠⎟

BC−→

15b. [3 marks]Find a unit vector in the direction of .

Markschemeattempt to find the length of (M1)

(A1)

unit vector is A1 N2

[3 marks]

Examiners reportMany candidates did not appear to know how to find a unit vector in part (b). Some tried to write down the vector equation of a line,indicating no familiarity with the concept of unit vectors while others gave the vector (1, 1, 1) or wrote the same vector as a linearcombination of i, j and k. A number of candidates correctly found the magnitude but did not continue on to write the unit vector.

AB−→

AB−→

| =|AB−→−

+ +62 (−2)2 32− −−−−−−−−−−−√ (= = = 7)36 + 4 + 9− −−−−−−−√ 49−−√

17

⎛⎝⎜

6−23

⎞⎠⎟ =

⎛⎝⎜⎜⎜

⎛⎝⎜⎜⎜

67

− 27

37

⎞⎠⎟⎟⎟

⎞⎠⎟⎟⎟

AB−→

15c. [3 marks]Show that is perpendicular to .AB−→

AC−→−

Markschemerecognizing that the dot product or being 0 implies perpendicular (M1)

correct substitution in a scalar product formula A1

e.g. ,

correct calculation A1

e.g. ,

therefore, they are perpendicular AG N0

[3 marks]

Examiners reportCandidates were generally successful in showing that the vectors in part (c) were perpendicular. Many used the efficient approach ofshowing that the scalar product equalled zero, while others worked a little harder than necessary and used the cosine rule to find theangle between the two vectors.

cosθ

(6) × (−2) + (−2) × (−3) + (3) × (2) cosθ = −12+6+67× 17√

∙ = 0AB−→

AC−→−

cosθ = 0

16a. [3 marks]Let u and w . Given that u is perpendicular to w , find the value of p .

Markschemeevidence of equating scalar product to 0 (M1)

( , ) A1

A1 N2

[3 marks]

Examiners reportMost candidates knew to set the scalar product equal to zero.

=⎛⎝⎜

23

−1

⎞⎠⎟ =

⎛⎝⎜

3−1p

⎞⎠⎟

2 × 3 + 3 × (−1) + (−1) × p = 0 6 − 3 − p = 0 3 − p = 0

p = 3

16b. [3 marks]Let . Given that , find the possible values of .

Markschemeevidence of substituting into magnitude formula (M1)

e.g. ,

setting up a correct equation A1

e.g. , ,

A1 N2

[3 marks]

Examiners reportMost candidates knew to set the scalar product equal to zero. A pleasing number found both answers for , although some oftenneglected to provide both solutions.

v =⎛⎝⎜

1q

5

⎞⎠⎟ |v| = 42−−√ q

1 + + 25q2− −−−−−−−−√ 1 + + 25q2

=1 + + 25q2− −−−−−−−−√ 42−−√ 1 + + 25 = 42q2 = 16q2

q = ±4

q

17a. [1 mark]

Consider the points P(2, −1, 5) and Q(3, − 3, 8). Let be the line through P and Q.

Show that .

Markschemeevidence of correct approach A1

e.g. ,

AG N0

[1 mark]

Examiners reportMost candidates answered part (a) easily.

L1

=PQ−→ ⎛

⎝⎜1

−23

⎞⎠⎟

= −PQ−→

OQ−→−

OP−→

−⎛⎝⎜

3−38

⎞⎠⎟

⎛⎝⎜

2−15

⎞⎠⎟

=PQ−→ ⎛

⎝⎜1

−23

⎞⎠⎟

17b. [3 marks]The line may be represented by .

(i) What information does the vector give about ?

(ii) Write down another vector representation for using .

Markscheme(i) correct description R1 N1

e.g. reference to being the position vector of a point on the line, a vector to the line, a point on the line.

(ii) any correct expression in the form A2 N2

where is , and is a scalar multiple of

e.g. ,

[3 marks]

Examiners reportFor part (b), a number of candidates stated that the vector was a "starting point," which misses the idea that it is a position vector tosome point on the line.

L1 r = + s⎛⎝⎜

3−38

⎞⎠⎟

⎛⎝⎜

1−23

⎞⎠⎟

⎛⎝⎜

3−38

⎞⎠⎟ L1

L1

⎛⎝⎜

3−38

⎞⎠⎟

⎛⎝⎜

3−38

⎞⎠⎟

r = a + tb

a⎛⎝⎜

3−38

⎞⎠⎟ b

⎛⎝⎜

1−23

⎞⎠⎟

r = + t⎛⎝⎜

3−38

⎞⎠⎟

⎛⎝⎜

−12

−3

⎞⎠⎟ r =

⎛⎝⎜

3 + 2s

−3 − 4s

8 + 6s

⎞⎠⎟

17c. [3 marks]The point lies on .

Find the value of .

T(−1, 5, p) L1

p

Markschemeone correct equation (A1)

e.g. ,

A1

A1 N2

[3 marks]

Examiners reportParts (c) and (d) proved accessible to many.

3 + s = −1 −3 − 2s = 5

s = −4

p = −4

17d. [3 marks]The point T also lies on with equation .

Show that .

Markschemeone correct equation A1

e.g. ,

A1

substituting

e.g. , A1

AG N0

[3 marks]

Examiners reportParts (c) and (d) proved accessible to many.

L2 = + t⎛⎝⎜

x

y

z

⎞⎠⎟

⎛⎝⎜

−392

⎞⎠⎟

⎛⎝⎜

1−2q

⎞⎠⎟

q = −3

−3 + t = −1 9 − 2t = 5

t = 2

t = 2

2 + 2q = −4 2q = −6

q = −3

17e. [7 marks]Let be the obtuse angle between and . Calculate the size of .

Markscheme

choosing correct direction vectors and (A1)(A1)

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

scalar product

magnitudes ,

evidence of substituting into scalar product M1

e.g.

radians (or ) A1 N4

[7 marks]

θ L1 L2 θ

⎛⎝⎜

1−23

⎞⎠⎟

⎛⎝⎜

1−2−3

⎞⎠⎟

(1)(1) + (−2)(−2) + (−3)(3) (= −4)

+ +12 (−2)2 32− −−−−−−−−−−−√ = 14−−√ + +12 (−2)2 (−3)2

− −−−−−−−−−−−−−√ = 14−−√

cosθ = −43.741…×3.741…

θ = 1.86 107∘

Examiners reportFor part (e), a surprising number of candidates chose incorrect vectors. Few candidates seemed to have a good conceptualunderstanding of the vector equation of a line.

18a. [3 marks]

The vertices of the triangle PQR are defined by the position vectors

, and .

Find

(i) ;

(ii) .

Markscheme(i) evidence of approach (M1)

e.g. ,

A1 N2

(ii) A1 N1

[3 marks]

Examiners reportCombining the vectors in (a) was generally well done, although some candidates reversed the subtraction, while others calculated themagnitudes.

=OP−→ ⎛

⎝⎜4

−31

⎞⎠⎟ =OQ

−→− ⎛⎝⎜

3−12

⎞⎠⎟ =OR

−→− ⎛⎝⎜

6−15

⎞⎠⎟

PQ−→

PR−→

= +PQ−→

PO−→

OQ−→−

Q − P

=PQ−→ ⎛

⎝⎜−121

⎞⎠⎟

=PR−→ ⎛

⎝⎜224

⎞⎠⎟

18b. [7 marks]Show that .cosR Q =P 12

MarkschemeMETHOD 1

choosing correct vectors and (A1)(A1)

finding , , (A1) (A1)(A1)

,

substituting into formula for angle between two vectors M1

e.g.

simplifying to expression clearly leading to A1

e.g. , ,

AG N0

METHOD 2

evidence of choosing cosine rule (seen anywhere) (M1)

A1

, and (A1)(A1)(A1)

A1

A1

AG N0

[7 marks]

Examiners reportMany candidates successfully used scalar product and magnitude calculations to complete part (b). Alternatively, some used thecosine rule, and often achieved correct results. Some assumed the triangle was a right-angled triangle and thus did not earn full marks.Although PQR is indeed right-angled, in a “show that” question this attribute must be directly established.

PQ−→

PR−→

∙PQ−→

PR−→ ∣

∣PQ−→∣

∣∣∣PR−→∣

∣

∙ = −2 + 4 + 4(= 6)PQ−→

PR−→

=∣∣PQ−→∣

∣ + +(−1)2 22 12− −−−−−−−−−−−√ (= )6√ =∣

∣PR−→∣

∣ + +22 22 42− −−−−−−−−√ (= )24−−√

cosR Q =P 6×6√ 24√

12

6×26√ 6√

6144√

612

cosR Q =P 12

=QR−→− ⎛

⎝⎜303

⎞⎠⎟

=∣∣QR−→− ∣

∣ 18−−√ =∣∣PQ−→∣

∣ 6√ =∣∣PR−→∣

∣ 24−−√

cosR Q =P+ −( )6√ 2 ( )24√ 2 ( )18√ 2

2 ×6√ 24√

cosR Q =P 6+24−1824

(= )1224

cosR Q =P 12

19. [7 marks]Let and , for . The angle between v and w is .

Find the value of .

v =⎛⎝⎜

2−36

⎞⎠⎟ w =

⎛⎝⎜

k

−24

⎞⎠⎟ k > 0 π

3

k

Markschemecorrect substitutions for ; ; (A1)(A1)(A1)

e.g. , ; , ; ,

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

e.g.

correct substitution A1

e.g.

A2 N5

[7 marks]

Examiners reportFor the most part, this question was well done and candidates had little difficulty finding the scalar product, the appropriatemagnitudes and then correctly substituting into the formula for the angle between vectors. However, few candidates were able tosolve the resulting equation using their GDCs to obtain the correct answer. Problems arose when candidates attempted to solve theresulting equation analytically.

v ∙ w |v| |w|

2k + (−3) × (−2) + 6 × 4 2k + 30 + +22 (−3)2 62− −−−−−−−−−−−√ 49−−√ + +k2 (−2)2 42

− −−−−−−−−−−−√ + 20k2− −−−−−√

2k+30

7× +20k2√

cos =π

32k+30

7× +20k2√

k = 18.8

20a. [4 marks]

In this question, distance is in metres.

Toy airplanes fly in a straight line at a constant speed. Airplane 1 passes through a point A.

Its position, p seconds after it has passed through A, is given by .

(i) Write down the coordinates of A.

(ii) Find the speed of the airplane in .

Markscheme(i) (3, , 0) A1 N1

(ii) choosing velocity vector (M1)

finding magnitude of velocity vector (A1)

e.g. ,

speed A1 N2

[4 marks]

Examiners reportMany candidates demonstrated a good understanding of the vector equation of a line and its application to a kinematics problem bycorrectly answering the first two parts of this question.

= + p⎛⎝⎜

x

y

z

⎞⎠⎟

⎛⎝⎜

3−40

⎞⎠⎟

⎛⎝⎜

−231

⎞⎠⎟

ms−1

−4

⎛⎝⎜

−231

⎞⎠⎟

+ +(−2)2 32 12− −−−−−−−−−−−√ 4 + 9 + 1− −−−−−−√

= 3.74 ( )14−−√

20b. [5 marks]After seven seconds the airplane passes through a point B.

(i) Find the coordinates of B.

(ii) Find the distance the airplane has travelled during the seven seconds.

Markscheme(i) substituting (M1)

A1 N2

(ii) METHOD 1

appropriate method to find or (M1)

e.g. ,

or (A1)

distance A1 N3

METHOD 2

evidence of applying distance is speed × time (M2)

e.g.

distance A1 N3

METHOD 3

attempt to find AB , AB (M1)

e.g. ,

AB , AB (A1)

distance AB A1 N3

[5 marks]

Examiners reportMany candidates demonstrated a good understanding of the vector equation of a line and its application to a kinematics problem bycorrectly answering the first two parts of this question.

Some knew that speed and distance were magnitudes of vectors but chose the wrong vectors to calculate magnitudes.

p = 7

B = (−11, 17, 7)

AB−→

BA−→

+AO−→−

OB−→

A − B

=AB−→ ⎛

⎝⎜−14217

⎞⎠⎟ =BA

−→ ⎛⎝⎜

14−21−7

⎞⎠⎟

= 26.2 (7 )14−−√

3.74 × 7

= 26.2 (7 )14−−√

2

(3 − (−11) + (−4 − 17 + (0 − 7))2 )2 )2 + + (0 − 7)(3 − (−11))2 (−4 − 17)2 )2− −−−−−−−−−−−−−−−−−−−−−−−−−−−√

2 = 686 = 686−−−√

= 26.2 (7 )14−−√

20c. [7 marks]Airplane 2 passes through a point C. Its position q seconds after it passes through C is given by

.

The angle between the flight paths of Airplane 1 and Airplane 2 is . Find the two values of a.

Markscheme

correct direction vectors and (A1)(A1)

, (A1)(A1)

substituting M1

e.g.

, A1A1 N3

[7 marks]

= + q ,a ∈ R⎛⎝⎜

x

y

z

⎞⎠⎟

⎛⎝⎜

2−58

⎞⎠⎟

⎛⎝⎜

−12a

⎞⎠⎟

40∘

⎛⎝⎜

−231

⎞⎠⎟

⎛⎝⎜

−12a

⎞⎠⎟

=∣

∣

∣∣

−12a

∣

∣

∣∣ + 5a2

− −−−−√ ∙ = a + 8

⎛⎝⎜

−231

⎞⎠⎟

⎛⎝⎜

−12a

⎞⎠⎟

cos =40∘ a+8

14√ +5a2√

a = 3.21 a = −0.990

Examiners reportVery few candidates were able to get the two correct answers in (c) even if they set up the equation correctly. Much contorted algebrawas seen and little evidence of using the GDC to solve the equation. Many made simple algebraic errors by combining unlike termsin working with the scalar product (often writing rather than ) or the magnitude (often writing rather than ).8a 8 + a 5a2 5 + a2

21a. [2 marks]

Line passes through points and .

Find .

Markschemeappropriate approach (M1)

e.g. ,

A1 N2

[2 marks]

Examiners reportFinding was generally well done, although some candidates reversed the subtraction.

L1 A(1, − 1, 4) B(2, − 2, 5)

AB−→

+AO−→−

OB−→

B − A

=AB−→ ⎛

⎝⎜1

−11

⎞⎠⎟

AB−→

21b. [2 marks]Find an equation for in the form .

Markschemeany correct equation in the form A2 N2

where is a scalar multiple of

e.g. , ,

[2 marks]

Examiners reportIn part (b) not all the candidates recognized that was the direction vector of the line, as some used the position vector of point Bas the direction vector.

L1 r = a + tb

r = a + tb

b⎛⎝⎜

1−11

⎞⎠⎟

r = + t⎛⎝⎜

1−14

⎞⎠⎟

⎛⎝⎜

1−11

⎞⎠⎟ r =

⎛⎝⎜

2 + t

−2 − t

5 + t

⎞⎠⎟ r = 2i − 2j + 5k + t(i − j + k)

AB−→

21c. [7 marks]

Line has equation .

Find the angle between and .

L2 r = + s⎛⎝⎜

247

⎞⎠⎟

⎛⎝⎜

213

⎞⎠⎟

L1 L2

Markscheme

choosing correct direction vectors , (A1)(A1)

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

scalar product

magnitudes ,

substitution into (accept , but not ) M1

e.g. ,

A1 N5

[7 marks]

Examiners reportMany candidates successfully used scalar product and magnitudes in part (c), although a large number did choose vectors other thanthe direction vectors and many did not state clearly which vectors they were using.

⎛⎝⎜

213

⎞⎠⎟

⎛⎝⎜

1−11

⎞⎠⎟

= 1 × 2 + −1 × 1 + 1 × 3 (= 4)

(= 1.73…)+ +12 (−1)2 12− −−−−−−−−−−−√ (= 3.74…)4 + 1 + 9− −−−−−−√

u∙v

|u||v|θ = u∙v

|u||v|sin θ = u∙v

|u||v|

cosθ = 1×2+−1×1+1×3

+ +12 (−1)2 12√ + +22 12 32√cosθ = 4

42√

θ = 0.906 ( )51.9∘

21d. [6 marks]The lines and intersect at point C. Find the coordinates of C.L1 L2

Markscheme

METHOD 1 (from )

appropriate approach (M1)

e.g. ,

two correct equations A1A1

e.g. , ,

attempt to solve (M1)

one correct parameter A1

e.g. ,

C is A1 N3

METHOD 2 (from )

appropriate approach (M1)

e.g. ,

two correct equations A1A1

e.g. , ,

attempt to solve (M1)

one correct parameter A1

e.g. ,

C is A1 N3

[6 marks]

Examiners reportCandidates who were comfortable on the first three parts often had little difficulty with the final part. While the resulting systems wereeasily solved algebraically, a surprising number of candidates did not check their solutions either manually or with technology. Anoccasionally seen error in the final part was using a midpoint to find C. Some candidates found the point of intersection in part (c)rather than in part (d), indicating a familiarity with the type of question but a lack of understanding of the concepts involved.

r = + t⎛⎝⎜

1−14

⎞⎠⎟

⎛⎝⎜

1−11

⎞⎠⎟

p = r + t = + s⎛⎝⎜

1−14

⎞⎠⎟

⎛⎝⎜

1−11

⎞⎠⎟

⎛⎝⎜

247

⎞⎠⎟

⎛⎝⎜

213

⎞⎠⎟

1 + t = 2 + 2s −1 − t = 4 + s 4 + t = 7 + 3s

t = −3 s = −2

(−2, 2, 1)

r = + t⎛⎝⎜

2−25

⎞⎠⎟

⎛⎝⎜

1−11

⎞⎠⎟

p = r + t = + s⎛⎝⎜

2−25

⎞⎠⎟

⎛⎝⎜

1−11

⎞⎠⎟

⎛⎝⎜

247

⎞⎠⎟

⎛⎝⎜

213

⎞⎠⎟

2 + t = 2 + 2s −2 − t = 4 + s 5 + t = 7 + 3s

t = −4 s = −2

(−2, 2, 1)

22a. [4 marks]

A particle is moving with a constant velocity along line L . Its initial position is A(6 , −2 , 10) . After one second the particle has moved to B(9, −6 , 15) .

(i) Find the velocity vector, .

(ii) Find the speed of the particle.

AB−→

Markscheme(i) evidence of approach (M1)

e.g. , ,

(accept ) A1 N2

(ii) evidence of finding the magnitude of the velocity vector M1

e.g.

A1 N1

[4 marks]

Examiners reportThis question was quite well done. Marks were lost when candidates found the vector instead of in part (a) and for notwriting their vector equation as an equation.

+AO−→−

OB−→

B − A⎛⎝⎜

9 − 6−6 + 215 − 10

⎞⎠⎟

=AB−→ ⎛

⎝⎜3

−45

⎞⎠⎟ (3,−4,5)

speed = + +32 42 52− −−−−−−−−√

speed = 50−−√ (= 5 )2√

BA−→

AB−→

22b. [2 marks]Write down an equation of the line L .

Markschemecorrect equation (accept Cartesian and parametric forms) A2 N2

e.g. ,

[2 marks]

Examiners reportIn part (b), a few candidates switched the position and velocity vectors or used the vectors and to incorrectly write the vectorequation.

r = + t⎛⎝⎜

6−210

⎞⎠⎟

⎛⎝⎜

3−45

⎞⎠⎟ r = + t

⎛⎝⎜

9−615

⎞⎠⎟

⎛⎝⎜

3−45

⎞⎠⎟

OA−→−

OB−→−

23a. [5 marks]

The diagram shows a parallelogram ABCD.

The coordinates of A, B and D are A(1, 2, 3) , B(6, 4,4 ) and D(2, 5, 5) .

(i) Show that .

(ii) Find .

(iii) Hence show that .

Markscheme(i) evidence of approach M1

e.g. , ,

AG N0

(ii) evidence of approach (M1)

e.g. , ,

A1 N2

(iii) evidence of approach (M1)

e.g.

correct substitution A1

e.g.

AG N0

[5 marks]

Examiners reportCandidates performed very well in this question, showing a strong ability to work with the algebra and geometry of vectors.

=AB−→ ⎛

⎝⎜521

⎞⎠⎟

AD−→−

=AC−→− ⎛

⎝⎜653

⎞⎠⎟

B − A +AO−→−

OB−→

−⎛⎝⎜

644

⎞⎠⎟

⎛⎝⎜

123

⎞⎠⎟

=AB−→ ⎛

⎝⎜521

⎞⎠⎟

D − A +AO−→−

OD−→−

−⎛⎝⎜

255

⎞⎠⎟

⎛⎝⎜

123

⎞⎠⎟

=AD−→− ⎛

⎝⎜132

⎞⎠⎟

= +AC−→−

AB−→

AD−→−

= +AC−→− ⎛

⎝⎜521

⎞⎠⎟

⎛⎝⎜

132

⎞⎠⎟

=AC−→− ⎛

⎝⎜653

⎞⎠⎟

23b. [3 marks]Find the coordinates of point C.

Markschemeevidence of combining vectors (there are at least 5 ways) (M1)

e.g. , ,

correct substitution A1

e.g. coordinates of C are A1 N1

[3 marks]

Examiners reportCandidates performed very well in this question, showing a strong ability to work with the algebra and geometry of vectors.

= +OC−→−

OA−→−

AC−→−

= +OC−→−

OB−→

AD−→−

= −AB−→

OC−→−

OD−→−

= + =OC−→− ⎛

⎝⎜123

⎞⎠⎟

⎛⎝⎜

653

⎞⎠⎟

⎛⎝⎜

⎛⎝⎜

776

⎞⎠⎟

⎞⎠⎟

(7, 7, 6)

23c. [7 marks](i) Find .

(ii) Hence find angle A.

Markscheme(i) evidence of using scalar product on and (M1)

e.g.

A1 N2

(ii) , (A1)(A1)

evidence of using (M1)

correct substitution A1

e.g.

A1 N3

[7 marks]

Examiners reportSome candidates were unable to find the scalar product in part (c), yet still managed to find the correct angle, able to use the formulain the information booklet without knowing that the scalar product is a part of that formula.

∙AB−→

AD−→−

AB−→

AD−→−

∙ = 5(1) + 2(3) + 1(2)AB−→

AD−→−

∙ = 13AB−→

AD−→−

= 5.477…∣∣AB−→∣

∣ = 3.741…∣∣AD−→− ∣

∣

cosA = ∙AB→

AD→

∣

∣∣∣AB→∣

∣∣∣∣

∣∣∣AD→∣

∣∣∣

cosA = 1320.493

= 0.884A ( )50.6∘

24. [7 marks]Let and . The vector is perpendicular to w. Find the value of p.v = 3i + 4j + k w = i + 2j − 3k v + pw

Markscheme (seen anywhere) (A1)

attempt to find (M1)

e.g.

collecting terms A1

attempt to find the dot product (M1)

e.g.

setting their dot product equal to 0 (M1)

e.g.

simplifying A1

e.g. ,

A1 N3

[7 marks]

Examiners reportThis question was very poorly done with many leaving it blank. Of those that did attempt it, most were able to find but reallydid not know how to proceed from there. They tried many approaches, such as, finding magnitudes, using negative reciprocals, orcalculating the angle between two vectors. A few had the idea that the scalar product should equal zero but had trouble trying to set itup.

pw = pi + 2pj − 3pk

v + pw

3i + 4j + k + p(i + 2j − 3k)

(3 + p)i + (4 + 2p)j + (1 − 3p)k

1(3 + p) + 2(4 + 2p) − 3(1 − 3p)

1(3 + p) + 2(4 + 2p) − 3(1 − 3p) = 0

3 + p + 8 + 4p − 3 + 9p = 0 14p + 8 = 0

p = −0.571 (− )814

v + pw

25a. [8 marks]

The point O has coordinates (0 , 0 , 0) , point A has coordinates (1 , – 2 , 3) and point B has coordinates (– 3 , 4 , 2) .

(i) Show that .

(ii) Find .

Markscheme(i) evidence of approach M1

e.g.

AG N0

(ii) for choosing correct vectors, ( with or with ) (A1)(A1)

Note: Using with will lead to . If they then say , this is a correct solution.

calculating , , (A1)(A1)(A1)

e.g.

,

evidence of using the formula to find the angle M1

e.g. , ,

radians (accept ) A1 N3

[8 marks]

=AB−→ ⎛

⎝⎜−46

−1

⎞⎠⎟

B OA

+ =AO−→−

OB−→

AB−→

=AB−→ ⎛

⎝⎜−46

−1

⎞⎠⎟

AO−→−

AB−→

OA−→−

BA−→

AO−→−

BA−→

π − 0.799 B O = 0.799A

∙AO−→−

AB−→ ∣

∣AO−→− ∣

∣∣∣AB−→∣

∣

∙ = (−1)(−4) + (2)(6) + (−3)(−1)(= 19)d1 d2

| | = (= )d1 + +(−1)2 22 (−3)2− −−−−−−−−−−−−−√ 14−−√ | | = (= )d2 + +(−4)2 62 (−1)2

− −−−−−−−−−−−−−√ 53−−√

cosθ = (−1)(−4)+(2)(6)+(−3)(−1)

+ +(−1)2 22 (−3)2√ + +(−4)2 62 (−1)2√19

14√ 53√0.69751…

B O = 0.799A 45.8∘

Examiners reportPart (ai) was done well by most students. Most knew how to approach finding the angle in part (aii). The problems occurred when theincorrect vectors were chosen. If the vectors being used were stated, then follow through marks could be given.

25b. [2 marks]The line has equation .

Write down the coordinates of two points on .

Markschemetwo correct answers A1A1

e.g. (1, , 3) , ( , 4, 2) , ( , 10, 1), ( , 16, 0) N2

[2 marks]

Examiners reportPart (b) was well done.

L1 = + s⎛⎝⎜

x

y

z

⎞⎠⎟

⎛⎝⎜

−342

⎞⎠⎟

⎛⎝⎜

−46

−1

⎞⎠⎟

L1

−2 −3 −7 −11

25c. [6 marks]The line passes through A and is parallel to .

(i) Find a vector equation for , giving your answer in the form .

(ii) Point is on . Find the coordinates of C.

Markscheme

(i) A2 N2

(ii) C on , so (M1)

evidence of equating components (A1)

e.g. , ,

one correct value , (seen anywhere) (A1)

coordinates of C are A1 N3

[6 marks]

Examiners reportIn part (ci), the error that occurred most often was the incorrect choice for the direction vector.

L2 OB−→

L2 r = a + tb

C(k,−k,5) L2

r = + t⎛⎝⎜

1−23

⎞⎠⎟

⎛⎝⎜

−342

⎞⎠⎟

L2 = + t⎛⎝⎜

k

−k

5

⎞⎠⎟

⎛⎝⎜

1−23

⎞⎠⎟

⎛⎝⎜

−342

⎞⎠⎟

1 − 3t = k −2 + 4t = −k 5 = 3 + 2t

t = 1 k = −2

(−2, 2, 5)

25d. [2 marks]The line has equation and passes through the point C.

Find the value of p at C.

L3 = + p⎛⎝⎜

x

y

z

⎞⎠⎟

⎛⎝⎜

3−80

⎞⎠⎟

⎛⎝⎜

1−2−1

⎞⎠⎟

Markscheme

for setting up one (or more) correct equation using (M1)

e.g. , ,

A1 N2

[2 marks]

Examiners reportThose that were able to find the coordinates in part (cii) were also able to be successful in part (d).

= + p⎛⎝⎜

−225

⎞⎠⎟

⎛⎝⎜

3−80

⎞⎠⎟

⎛⎝⎜

1−2−1

⎞⎠⎟

3 + p = −2 −8 − 2p = 2 −p = 5

p = −5

26a. [3 marks]

Consider the points A (1 , 5 , 4) , B (3 , 1 , 2) and D (3 , k , 2) , with (AD) perpendicular to (AB) .

Find

(i) ;

(ii) giving your answer in terms of k .

[3 marks]

Markscheme(i) evidence of combining vectors (M1)

e.g. (or in part (ii))

A1 N2

(ii) A1 N1

[3 marks]

Examiners reportThis question was well done by many candidates. Most found and correctly.

AB−→

AD−→−

= −AB−→

OB−→

OA−→−

= +AD−→−

AO−→−

OD−→−

=AB−→ ⎛

⎝⎜2

−4−2

⎞⎠⎟

=AD−→− ⎛

⎝⎜2

k − 5−2

⎞⎠⎟

AB−→

AD−→−

26b. [3 marks]Show that .

Markschemeevidence of using perpendicularity scalar product = 0 (M1)

e.g.

A1

(accept any correct equation clearly leading to ) A1

AG N0

[3 marks]

k = 7

⇒

∙ = 0⎛⎝⎜

2−4−2

⎞⎠⎟

⎛⎝⎜

2k − 5−2

⎞⎠⎟

4 − 4(k − 5) + 4 = 0

−4k + 28 = 0 k = 7

k = 7

Examiners reportThe majority of candidates correctly used the scalar product to show .k = 7

26c. [4 marks]The point C is such that .

Find the position vector of C.

Markscheme

(A1)

A1

evidence of correct approach (M1)

e.g. , ,

A1 N3

[4 marks]

Examiners reportSome confusion arose in substituting into , but otherwise part (c) was well done, though finding the position vector of Cpresented greater difficulty.

=BC−→ 1

2AD−→−

=AD−→− ⎛

⎝⎜22

−2

⎞⎠⎟

=BC−→ ⎛

⎝⎜11

−1

⎞⎠⎟

= +OC−→−

OB−→

BC−→

+⎛⎝⎜

312

⎞⎠⎟

⎛⎝⎜

11

−1

⎞⎠⎟ =

⎛⎝⎜

x − 3y − 1z − 2

⎞⎠⎟

⎛⎝⎜

11

−1

⎞⎠⎟

=OC−→− ⎛

⎝⎜421

⎞⎠⎟

k = 7 AD−→−

26d. [3 marks]Find .

MarkschemeMETHOD 1

choosing appropriate vectors, , (A1)

finding the scalar product M1

e.g. ,

A1 N1

METHOD 2

parallel to (may show this on a diagram with points labelled) R1

(may show this on a diagram with points labelled) R1

A1 N1

[3 marks]

cosA CB

BA−→

BC−→

−2(1) + 4(1) + 2(−1) 2(1) + (−4)(1) + (−2)(−1)

cosA C = 0B

BC−→

AD−→−

⊥BC−→

AB−→

A C =B 90∘

cosA C = 0B

Examiners reportOwing to and being perpendicular, no problems were created by using these two vectors to find , and themajority of candidates answering part (d) did exactly that.

AB−→

BC−→

cosA C = 0B

27. [6 marks]The line L is represented by and the line L by .

The lines L and L intersect at point T. Find the coordinates of T.

Markschemeevidence of equating vectors (M1)

e.g.

for any two correct equations A1A1

e.g. , ,

attempting to solve the equations (M1)

finding one correct parameter A1

the coordinates of T are A1 N3

[6 marks]

Examiners reportThose candidates prepared in this topic area answered the question particularly well, often only making some calculation error whensolving the resulting system of equations. Curiously, a few candidates found correct values for s and t, but when substituting back intoone of the vector equations, neglected to find the z-coordinate of T.

1 = + sr1

⎛⎝⎜

253

⎞⎠⎟

⎛⎝⎜

123

⎞⎠⎟ 2 = + tr2

⎛⎝⎜

3−38

⎞⎠⎟

⎛⎝⎜

−13

−4

⎞⎠⎟

1 2

=L1 L2

2 + s = 3 − t 5 + 2s = −3 + 3t 3 + 3s = 8 − 4t

(2 = −1, t = 2)

(1, 3, 0)

28. [6 marks]Two lines with equations and intersect at the point P. Find the coordinates

of P.

Markschemeevidence of appropriate approach (M1)

e.g.

two correct equations A1A1

e.g. , ,

attempting to solve the equations (M1)

one correct parameter , A1

P is (accept ) A1 N3

[6 marks]

= + sr1

⎛⎝⎜

23

−1

⎞⎠⎟

⎛⎝⎜

5−32

⎞⎠⎟ = + tr2

⎛⎝⎜

922

⎞⎠⎟

⎛⎝⎜

−35

−1

⎞⎠⎟

+ s = + t⎛⎝⎜

23

−1

⎞⎠⎟

⎛⎝⎜

5−32

⎞⎠⎟

⎛⎝⎜

922

⎞⎠⎟

⎛⎝⎜

−35

−1

⎞⎠⎟

2 + 5s = 9 − 3t 3 − 3s = 2 + 5t −1 + 2s = 2 − t

s = 2 t = −1

(12,−3,3)⎛⎝⎜

12−33

⎞⎠⎟

Examiners reportIf this topic had been taught well then the candidates scored highly. The question was either well answered or not at all. Manycandidates did not understand what was needed and tried to find the length of vectors or mid-points of lines. The other most commonmistake was to use the values of the parameters to write the coordinates as . P(2, − 1)

29. [6 marks]Find the cosine of the angle between the two vectors and .

Markschemefinding scalar product and magnitudes (A1)(A1)(A1)

scalar product ( )

magnitudes , ,

substitution into formula M1

e.g.

A2 N4

[6 marks]

Examiners reportMany candidates performed well in finding the magnitudes and scalar product to use the formula for angle between vectors. Someexperienced trouble with the arithmetic to obtain the required result. A significant number of candidates isolated the theta answering

with .

3i + 4j + 5k 4i − 5j − 3k

= 12 − 20 − 15 = −23

= + +32 42 52− −−−−−−−−√ = + +42 (−5)2 (−3)2− −−−−−−−−−−−−−√ ( , )50−−√ 50−−√

cosθ = 12−20−15

( )×( )+ +32 42 52√ + +42 (−5)2 (−3)2√

cosθ = − 2350

(= −0.46)

arccos ( )−2350

30a. [2 marks]

The line is parallel to the z-axis. The point P has position vector and lies on .

Write down the equation of in the form .

Markscheme

A2 N2

[2 marks]

Examiners reportVery few candidates gave a correct direction vector parallel to the z-axis. Provided they wrote down an equation here they were ableto earn most of subsequent marks on follow through.

L1

⎛⎝⎜

810

⎞⎠⎟ L1

L1 r = a + tb

: r = + tL1

⎛⎝⎜

810

⎞⎠⎟

⎛⎝⎜

001

⎞⎠⎟

30b. [4 marks]The line has equation . The point A has position vector .

Show that A lies on .

L2 r = + s⎛⎝⎜

24

−1

⎞⎠⎟

⎛⎝⎜

2−15

⎞⎠⎟

⎛⎝⎜

629

⎞⎠⎟

L2

Markschemeevidence of equating and (M1)

e.g. ,

one correct equation A1

e.g. , ,

A1

evidence of confirming for other two equations A1

e.g. , ,

so A lies on AG N0

[4 marks]

Examiners reportFor (b), many found the correct parameter but neglected to confirm it in the other two equations.

r OA−→−

= + s⎛⎝⎜

629

⎞⎠⎟

⎛⎝⎜

24

−1

⎞⎠⎟

⎛⎝⎜

2−15

⎞⎠⎟ A = r

6 = 2 + 2s 2 = 4 − s 9 = −1 + 5s

s = 2

6 = 2 + 4 2 = 4 − 2 9 = −1 + 10

L2

30c. [7 marks]Let B be the point of intersection of lines and .

(i) Show that .

(ii) Find .

Markscheme(i) evidence of approach M1

e.g.

one correct equation A1

e.g. , ,

attempt to solve (M1)

finding A1

substituting M1

e.g.

AG N0

(ii) evidence of appropriate approach (M1)

e.g. ,

A1 N2

[7 marks]

L1 L2

=OB−→ ⎛

⎝⎜81

14

⎞⎠⎟

AB−→

+ s = + t⎛⎝⎜

24

−1

⎞⎠⎟

⎛⎝⎜

2−15

⎞⎠⎟

⎛⎝⎜

810

⎞⎠⎟

⎛⎝⎜

001

⎞⎠⎟ =L1 L2

2 + 2s = 8 4 − s = 1 −1 + 5s = t

s = 3

= + 3OB−→ ⎛

⎝⎜24

−1

⎞⎠⎟

⎛⎝⎜

2−15

⎞⎠⎟

=OB−→ ⎛

⎝⎜81

14

⎞⎠⎟

= +AB−→

AO−→−

OB−→

= −AB−→

OB−→

OA−→−

=AB−→ ⎛

⎝⎜2

−15

⎞⎠⎟

Examiners reportIn (c) some performed a trial and error approach to obtaining an integer parameter and thus did not "show" the mathematical origin ofthe result. Finding vector proved accessible.AB

−→

30d. [3 marks]The point C is at (2, 1, − 4). Let D be the point such that ABCD is a parallelogram.

Find .

Markschemeevidence of appropriate approach (M1)

e.g.

correct values A1

e.g. , ,

A1 N2

[3 marks]

Examiners reportA good number of candidates had an appropriate approach to (d), although surprisingly many subtracted from in finding

.

OD−→−

=AB−→

DC−→−

+ =OD−→− ⎛

⎝⎜2

−15

⎞⎠⎟

⎛⎝⎜

21

−4

⎞⎠⎟ + =

⎛⎝⎜

x

y

z

⎞⎠⎟

⎛⎝⎜

2−15

⎞⎠⎟

⎛⎝⎜

21

−4

⎞⎠⎟ =

⎛⎝⎜

2−15

⎞⎠⎟

⎛⎝⎜

2 − x

1 − y

−4 − z

⎞⎠⎟

=OD−→− ⎛

⎝⎜02

−9

⎞⎠⎟

OC−→−

AB−→

OD−→−

31a. [3 marks]

In the following diagram, and .

The midpoint of is E and .

Express each of the following vectors in terms of and .

Markscheme A1

attempt to find M1

e.g. ,

A1 N2

[3 marks]

u = AB−→

v = BD−→

AD−→−

=BDDC

13

u v

AE−→

=AE−→ 1

2AD−→−

AD−→−

+AB−→

BD−→

u + v

= (u + v)AE−→ 1

2(= u + v)1

212

Examiners report[N/A]

31b. [4 marks]

Markscheme A1

A1

attempt to find \overrightarrow {{\rm{EC}}} M1

e.g. \overrightarrow {{\rm{ED}}} + \overrightarrow {{\rm{DC}}} , \frac{1}{2}({\boldsymbol{u}} + {\boldsymbol{v}}) +3{\boldsymbol{v}}

\overrightarrow {{\rm{EC}}} = \frac{1}{2}{\boldsymbol{u}} + \frac{7}{2}{\boldsymbol{v}} \left( { =\frac{1}{2}({\boldsymbol{u}} + 7{\boldsymbol{v}})} \right) A1 N2

[4 marks]

Examiners report[N/A]

EC−→

= = (u + v)EC−→

AE−→ 1

2

= 3vDC−→−

32a. [1 mark]

Consider the lines {L_1} , {L_2} , {L_2} , and {L_4} , with respective equations.

{L_1} : \left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1\\ 2\\ 3 \end{array}} \right) + t\left({\begin{array}{*{20}{c}} 3\\ { - 2}\\ 1 \end{array}} \right)

{L_2} : \left( \begin{array}{l} x\\ y\\ z \end{array} \right) = \left( \begin{array}{l} 1\\ 2\\ 3 \end{array} \right) + p\left( \begin{array}{l} 3\\2\\ 1 \end{array} \right)

{L_3} : \left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0\\ 1\\ 0 \end{array}} \right) + s\left({\begin{array}{*{20}{c}} { - 1}\\ 2\\ { - a} \end{array}} \right)

{L_4} : \left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = q\left( {\begin{array}{*{20}{c}} { - 6}\\ 4\\ { - 2} \end{array}}\right)

Write down the line that is parallel to {L_4} .

Markscheme{L_1} A1 N1

[1 mark]

Examiners report[N/A]

32b. [1 mark]Write down the position vector of the point of intersection of {L_1} and {L_2} .

Markscheme\left( \begin{array}{l} 1\\ 2\\ 3 \end{array} \right) A1 N1

[1 mark]

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© International Baccalaureate Organization 2014 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Examiners report[N/A]

32c. [5 marks]Given that {L_1} is perpendicular to {L_3} , find the value of a .

Markschemechoosing correct direction vectors A1A1

e.g. \left( {\begin{array}{*{20}{c}} 3\\ { - 2}\\ 1 \end{array}} \right) , \left( {\begin{array}{*{20}{c}} { - 1}\\ 2\\ { - a}\end{array}} \right)

recognizing that {\boldsymbol{a}} \bullet {\boldsymbol{b}} = 0 M1

correct substitution A1

e.g. - 3 - 4 - a = 0

a = - 7 A1 N3

[5 marks]

Examiners report[N/A]

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