vectors worksheet a

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VECTORS C4 Worksheet A 1

G H I D E F A B C The diagram shows three sets of equally-spaced parallel lines.

Given that AC = p and that AD = q, express the following vectors in terms of p and q.

a CA b AG c AB d DF e HE f AF

g AH h DC i CG j IA k EC l IB 2 B O

C

In the quadrilateral shown, OA = u, AB = v and OC = w.

Find expressions in terms of u, v and w for

a OB b AC c CB 3 A B D C E F H G

The diagram shows a cuboid.

Given that AB = p, AD = q and AE = r, find expressions in terms of p, q and r for

a BC b AF c DE d AG e GB f BH 4 R S O T The diagram shows parallelogram ORST.

Given that OR = a + 2b and that OT = a − 2b,

a find expressions in terms of a and b for

i OS ii TR

Given also that OA = a and that OB = b,

b copy the diagram and show the positions of the points A and B.

A

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5 A

C D O B

The diagram shows triangle OAB in which OA = a and OB = b.

The points C and D are the mid-points of OA and AB respectively.

a Find and simplify expressions in terms of a and b for

i OC ii AB iii AD iv OD v CD

b Explain what your expression for CD tells you about OB and CD . 6 Given that vectors p and q are not parallel, state whether or not each of the following pairs of

vectors are parallel.

a 2p and 3p b (p + 2q) and (2p − 4q) c (3p − q) and (p − 13 q)

d (p − 2q) and (4q − 2p) e ( 34 p + q) and (6p + 8q) f (2q − 3p) and ( 3

2 q − p)

7 The points O, A, B and C are such that OA = 4m, OB = 4m + 2n and OC = 2m + 3n, where m and n are non-parallel vectors.

a Find an expression for BC in terms of m and n.

The point M is the mid-point of OC.

b Show that AM is parallel to BC.

8 The points O, A, B and C are such that OA = 6u − 4v, OB = 3u − v and OC = v − 3u, where u and v are non-parallel vectors.

The point M is the mid-point of OA and the point N is the point on AB such that AN : NB = 1 : 2

a Find OM and ON .

b Prove that C, M and N are collinear. 9 Given that vectors p and q are not parallel, find the values of the constants a and b such that

a ap + 3q = 5p + bq b (2p + aq) + (bp − 4q) = 0

c 4aq − p = bp − 2q d (2ap + bq) − (aq − 6p) = 0 10 A C

O B

The diagram shows triangle OAB in which OA = a and OB = b. The point C is the mid-point of OA and the point D is the mid-point of BC.

a Find an expression for OD in terms of a and b.

b Show that if the point E lies on AB then OE can be written in the form a + k(b − a), where k is a constant.

Given also that OD produced meets AB at E,

c find OE ,

d show that AE : EB = 2 : 1

C4 VECTORS Worksheet A continued

D

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VECTORS C4 Worksheet B 1 The points A, B and C have coordinates (6, 1), (2, 3) and (−4, 3) respectively and O is the origin.

Find, in terms of i and j, the vectors

a OA b AB c BC d CA 2 Given that p = i − 3j and q = 4i + 2j, find expressions in terms of i and j for

a 4p b q − p c 2p + 3q d 4p − 2q

3 Given that p = 34

−

and q = 12

, find

a | p | b | 2q | c | p + 2q | d | 3q − 2p | 4 Given that p = 2i + j and q = i − 3j, find, in degrees to 1 decimal place, the angle made with

the vector i by the vector

a p b q c 5p + q d p − 3q 5 Find a unit vector in the direction

a 43

b 724

−

c 11−

d 24

6 Find a vector

a of magnitude 26 in the direction 5i + 12j,

b of magnitude 15 in the direction −6i − 8j,

c of magnitude 5 in the direction 2i − 4j. 7 Given that m = 2i + λj and n = µi − 5j, find the values of λ and µ such that

a m + n = 3i − j b 2m − n = −3i + 8j 8 Given that r = 6i + cj, where c is a positive constant, find the value of c such that

a r is parallel to the vector 2i + j b r is parallel to the vector −9i − 6j

c | r | = 10 d | r | = 53 9 Given that p = i + 3j and q = 4i − 2j,

a find the values of a and b such that ap + bq = −5i + 13j,

b find the value of c such that cp + q is parallel to the vector j,

c find the value of d such that p + dq is parallel to the vector 3i − j.

10 Relative to a fixed origin O, the points A and B have position vectors 36

and 52

−

respectively.

Find

a the vector AB ,

b AB ,

c the position vector of the mid-point of AB,

d the position vector of the point C such that OABC is a parallelogram.

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11 Given the coordinates of the points A and B, find the length AB in each case.

a A (4, 0, 9), B (2, −3, 3) b A (11, −3, 5), B (7, −1, 3) 12 Find the magnitude of each vector.

a 4i + 2j − 4k b i + j + k c −8i − j + 4k d 3i − 5j + k 13 Find

a a unit vector in the direction 5i − 2j + 14k,

b a vector of magnitude 10 in the direction 2i + 11j − 10k,

c a vector of magnitude 20 in the direction −5i − 4j + 2k. 14 Given that r = λi + 12j − 4k, find the two possible values of λ such that | r | = 14.

15 Given that p = 131

−

, q = 42

1

−

and r = 2

53

− −

, find as column vectors,

a p + 2q b p − r c p + q + r d 2p − 3q + r 16 Given that r = −2i + λj + µk, find the values of λ and µ such that

a r is parallel to 4i + 2j − 8k b r is parallel to −5i + 20j − 10k 17 Given that p = i − 2j + 4k, q = −i + 2j + 2k and r = 2i − 4j − 7k,

a find | 2p − q |,

b find the value of k such that p + kq is parallel to r. 18 Relative to a fixed origin O, the points A, B and C have position vectors (−2i + 7j + 4k),

(−4i + j + 8k) and (6i − 5j) respectively.

a Find the position vector of the mid-point of AB.

b Find the position vector of the point D on AC such that AD : DC = 3 : 1 19 Given that r = λi − 2λj + µk, and that r is parallel to the vector (2i − 4j − 3k),

a show that 3λ + 2µ = 0.

Given also that | r | = 292 and that µ > 0,

b find the values of λ and µ.

20 Relative to a fixed origin O, the points A, B and C have position vectors 624

− −

, 12

74

− −

and 618

−

respectively.

a Find the position vector of the point M, the mid-point of BC.

b Show that O, A and M are collinear. 21 The position vector of a model aircraft at time t seconds is (9 − t)i + (1 + 2t)j + (5 − t)k, relative

to a fixed origin O. One unit on each coordinate axis represents 1 metre.

a Find an expression for d 2 in terms of t, where d metres is the distance of the aircraft from O.

b Find the value of t when the aircraft is closest to O and hence, the least distance of the aircraft from O.

C4 VECTORS Worksheet B continued

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VECTORS C4 Worksheet C 1 Sketch each line on a separate diagram given its vector equation.

a r = 2i + sj b r = s(i + j) c r = i + 4j + s(i + 2j) d r = 3j + s(3i − j) e r = −4i + 2j + s(2i − j) f r = (2s + 1)i + (3s − 2)j 2 Write down a vector equation of the straight line

a parallel to the vector (3i − 2j) which passes through the point with position vector (−i + j),

b parallel to the x-axis which passes through the point with coordinates (0, 4),

c parallel to the line r = 2i + t(i + 5j) which passes through the point with coordinates (3, −1). 3 Find a vector equation of the straight line which passes through the points with position vectors

a 10

and 31

b 34

−

and 11−

c 22

−

and 23

−

4 Find the value of the constant c such that line with vector equation r = 3i − j + λ(ci + 2j)

a passes through the point (0, 5),

b is parallel to the line r = −2i + 4j + µ(6i + 3j). 5 Find a vector equation for each line given its cartesian equation.

a x = −1 b y = 2x c y = 3x + 1

d y = 34 x − 2 e y = 5 − 1

2 x f x − 4y + 8 = 0 6 A line has the vector equation r = 2i + j + λ(3i + 2j).

a Write down parametric equations for the line.

b Hence find the cartesian equation of the line in the form ax + by + c = 0, where a, b and c are integers.

7 Find a cartesian equation for each line in the form ax + by + c = 0, where a, b and c are integers.

a r = 3i + λ(i + 2j) b r = i + 4j + λ(3i + j) c r = 2j + λ(4i − j) d r = −2i + j + λ(5i + 2j) e r = 2i − 3j + λ(−3i + 4j) f r = (λ + 3)i + (−2λ − 1)j 8 For each pair of lines, determine with reasons whether they are identical, parallel but not identical

or not parallel.

a r = 12

+ s 31

−

b r = 12−

+ s 14

c r = 25

−

+ s 24

r = 23

−

+ t 62

−

r = 24

−

+ t 41

r = 11−

+ t 36

9 Find the position vector of the point of intersection of each pair of lines.

a r = i + 2j + λi b r = 4i + j + λ(−i + j) c r = j + λ(2i − j) r = 2i + j + µ(3i + j) r = 5i − 2j + µ(2i − 3j) r = 2i + 10j + µ(−i + 3j) d r = −i + 5j + λ(−4i + 6j) e r = −2i + 11j + λ(−3i + 4j) f r = i + 2j + λ(3i + 2j) r = 2i − 2j + µ(−i + 2j) r = −3i − 7j + µ(5i + 3j) r = 3i + 5j + µ(i + 4j)

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10 Write down a vector equation of the straight line

a parallel to the vector (i + 3j − 2k) which passes through the point with position vector (4i + k),

b perpendicular to the xy-plane which passes through the point with coordinates (2, 1, 0),

c parallel to the line r = 3i − j + t(2i − 3j + 5k) which passes through the point with coordinates (−1, 4, 2).

11 The points A and B have position vectors (5i + j − 2k) and (6i − 3j + k) respectively.

a Find AB in terms of i, j and k.

b Write down a vector equation of the straight line l which passes through A and B.

c Show that l passes through the point with coordinates (3, 9, −8). 12 Find a vector equation of the straight line which passes through the points with position vectors

a (i + 3j + 4k) and (5i + 4j + 6k) b (3i − 2k) and (i + 5j + 2k)

c 0 and (6i − j + 2k) d (−i − 2j + 3k) and (4i − 7j + k) 13 Find the value of the constants a and b such that line r = 3i − 5j + k + λ(2i + aj + bk)

a passes through the point (9, −2, −8),

b is parallel to the line r = 4j − 2k + µ(8i − 4j + 2k). 14 Find cartesian equations for each of the following lines.

a r = 230

+ λ352

b r = 41

3

−

+ λ163

c r = 1

52

− −

+ λ421

− −

15 Find a vector equation for each line given its cartesian equations.

a 13

x − = 42

y + = z − 5 b 4x = 1

2y −−

= 73

z + c 54

x +−

= y + 3 = z 16 Show that the lines with vector equations r = 4i + 3k + s(i − 2j + 2k) and r = 7i + 2j − 5k + t(−3i + 2j + k) intersect, and find the coordinates of their point

of intersection. 17 Show that the lines with vector equations r = 2i − j + 4k + λ(i + j + 3k) and

r = i + 4j + 3k + µ(i − 2j + k) are skew. 18 For each pair of lines, find the position vector of their point of intersection or, if they do not

intersect, state whether they are parallel or skew.

a r = 315

+ λ411

−

and r = 324

−

+ µ102

b r = 031

+ λ213

− −

and r = 621

− −

+ µ4

26

−

c r = 824

−

+ λ132

−

and r = 2

28

−

+ µ434

− −

d r = 152

+ λ142

−

and r = 765

− −

+ µ213

−

e r = 41

3

−

+ λ253

−

and r = 32

1

−

+ µ534

− −

f r = 072

−

+ λ64

8

−

and r = 121

11

− −

+ µ523

−

C4 VECTORS Worksheet C continued

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VECTORS C4 Worksheet D 1 Calculate

a (i + 2j).(3i + j) b (4i − j).(3i + 5j) c (i − 2j).(−5i − 2j) 2 Show that the vectors (i + 4j) and (8i − 2j) are perpendicular. 3 Find in each case the value of the constant c for which the vectors u and v are perpendicular.

a u = 31

−

, v = 3c

b u = 21

, v = 3c

c u = 25

−

, v = 4

c −

4 Find, in degrees to 1 decimal place, the angle between the vectors

a (4i − 3j) and (8i + 6j) b (7i + j) and (2i + 6j) c (4i + 2j) and (−5i + 2j) 5 Relative to a fixed origin O, the points A, B and C have position vectors (9i + j), (3i − j)

and (5i − 2j) respectively. Show that ∠ABC = 45°. 6 Calculate

a (i + 2j + 4k).(3i + j + 2k) b (6i − 2j + 2k).(i − 3j − k)

c (−5i + 2k).(i + 4j − 3k) d (3i + 2j − 8k).(−i + 11j − 4k)

e (3i − 7j + k).(9i + 4j − k) f (7i − 3j).(−3j + 6k) 7 Given that p = 2i + j − 3k, q = i + 5j − k and r = 6i − 2j − 3k,

a find the value of p.q,

b find the value of p.r,

c verify that p.(q + r) = p.q + p.r 8 Simplify

a p.(q + r) + p.(q − r) b p.(q + r) + q.(r − p) 9 Show that the vectors (5i − 3j + 2k) and (3i + j − 6k) are perpendicular. 10 Relative to a fixed origin O, the points A, B and C have position vectors (3i + 4j − 6k),

(i + 5j − 2k) and (8i + 3j + 2k) respectively. Show that ∠ABC = 90°. 11 Find in each case the value or values of the constant c for which the vectors u and v are

perpendicular.

a u = (2i + 3j + k), v = (ci − 3j + k) b u = (−5i + 3j + 2k), v = (ci − j + 3ck)

c u = (ci − 2j + 8k), v = (ci + cj − 3k) d u = (3ci + 2j + ck), v = (5i − 4j + 2ck) 12 Find the exact value of the cosine of the angle between the vectors

a 122

−

and 814

−

b 412

−

and 2

36

− −

c 121

−

and 17

2

−

d 53

4

−

and 341

− −

13 Find, in degrees to 1 decimal place, the angle between the vectors

a (3i − 4k) and (7i − 4j + 4k) b (2i − 6j + 3k) and (i − 3j − k)

c (6i − 2j − 9k) and (3i + j + 4k) d (i + 5j − 3k) and (−3i − 4j + 2k)

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14 The points A (7, 2, −2), B (−1, 6, −3) and C (−3, 1, 2) are the vertices of a triangle.

a Find BA and BC in terms of i, j and k.

b Show that ∠ABC = 82.2° to 1 decimal place.

c Find the area of triangle ABC to 3 significant figures. 15 Relative to a fixed origin, the points A, B and C have position vectors (3i − 2j − k),

(4i + 3j − 2k) and (2i − j) respectively.

a Find the exact value of the cosine of angle BAC.

b Hence show that the area of triangle ABC is 3 2 . 16 Find, in degrees to 1 decimal place, the acute angle between each pair of lines.

a r = 131

−

+ λ44

2

−

and r = 52

1

−

+ µ806

−

b r = 03

7

−

+ λ61

18

− −

and r = 463

−

+ µ4123

−

c r = 715

+ λ11

3

−

and r = 2

63

− −

+ µ25

3

−

d r = 239

− −

+ λ46

7

− −

and r = 1112

−

+ µ518

− −

17 Relative to a fixed origin, the points A and B have position vectors (5i + 8j − k) and

(6i + 5j + k) respectively.

a Find a vector equation of the straight line l1 which passes through A and B.

The line l2 has the equation r = 4i − 3j + 5k + µ(−5i + j − 2k).

b Show that lines l1 and l2 intersect and find the position vector of their point of intersection.

c Find, in degrees, the acute angle between lines l1 and l2. 18 Find, in degrees to 1 decimal place, the acute angle between the lines with cartesian equations

23

x − = 2y = 5

6z +−

and 44

x −−

= 17

y + = 34

z −−

. 19 The line l has the equation r = 7i − 2k + λ(2i − j + 2k) and the line m has the equation r = −4i + 7j − 6k + µ(5i − 4j − 2k).

a Find the coordinates of the point A where lines l and m intersect.

b Find, in degrees, the acute angle between lines l and m.

The point B has coordinates (5, 1, −4).

c Show that B lies on the line l.

d Find the distance of B from m. 20 Relative to a fixed origin O, the points A and B have position vectors (9i + 6j) and (11i + 5j + k)

respectively.

a Show that for all values of λ, the point C with position vector (9 + 2λ)i + (6 − λ)j + λk lies on the straight line l which passes through A and B.

b Find the value of λ for which OC is perpendicular to l.

c Hence, find the position vector of the foot of the perpendicular from O to l. 21 Find the coordinates of the point on each line which is closest to the origin.

a r = −4i + 2j + 7k + λ(i + 3j − 4k) b r = 7i + 11j − 9k + λ(6i − 9j + 3k)

C4 VECTORS Worksheet D continued

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VECTORS C4 Worksheet E 1 Relative to a fixed origin, the line l has vector equation

r = i − 4j + pk + λ(2i + qj − 3k),

where λ is a scalar parameter.

Given that l passes through the point with position vector (7i − j − k),

a find the values of the constants p and q, (3)

b find, in degrees, the acute angle l makes with the line with equation

r = 3i + 4j − 3k + µ(−4i + 5j − 2k). (4)

2 The points A and B have position vectors 164

and 506

−

respectively, relative to a

fixed origin.

a Find, in vector form, an equation of the line l which passes through A and B. (2)

The line m has equation

r = 55

3

−

+ t14

2

−

.

Given that lines l and m intersect at the point C,

b find the position vector of C, (5)

c show that C is the mid-point of AB. (2) 3 Relative to a fixed origin, the points P and Q have position vectors (5i − 2j + 2k) and

(3i + j) respectively.

a Find, in vector form, an equation of the line L1 which passes through P and Q. (2)

The line L2 has equation

r = 4i + 6j − k + µ(5i − j + 3k).

b Show that lines L1 and L2 intersect and find the position vector of their point of intersection. (6)

c Find, in degrees to 1 decimal place, the acute angle between lines L1 and L2. (4) 4 Relative to a fixed origin, the lines l1 and l2 have vector equations as follows:

l1 : r = 5i + k + λ(2i − j + 2k),

l2 : r = 7i − 3j + 7k + µ(−i + j − 2k),

where λ and µ are scalar parameters.

a Show that lines l1 and l2 intersect and find the position vector of their point of intersection. (6)

The points A and C lie on l1 and the points B and D lie on l2.

Given that ABCD is a parallelogram and that A has position vector (9i − 2j + 5k),

b find the position vector of C. (3)

Given also that the area of parallelogram ABCD is 54,

c find the distance of the point B from the line l1. (4)

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5 Relative to a fixed origin, the points A and B have position vectors (4i + 2j − 4k) and

(2i − j + 2k) respectively.

a Find, in vector form, an equation of the line l1 which passes through A and B. (2)

The line l2 passes through the point C with position vector (4i − 7j − k) and is parallel to the vector (6j − 2k).

b Write down, in vector form, an equation of the line l2. (1)

c Show that A lies on l2. (2)

d Find, in degrees, the acute angle between lines l1 and l2. (4)


10

− −

and 418

−


fixed origin O.


The line l intersects the y-axis at the point C.

b Find the coordinates of C. (2)

The point D on the line l is such that OD is perpendicular to l.

c Find the coordinates of D. (5)

d Find the area of triangle OCD, giving your answer in the form 5k . (3) 7 Relative to a fixed origin, the line l1 has the equation

r = 162

− −

+ s041

−

.

a Show that the point P with coordinates (1, 6, −5) lies on l1. (1)

The line l2 has the equation

r = 441

− −

+ t32

2

−

,

and intersects l1 at the point Q.

b Find the position vector of Q. (3)

The point R lies on l2 such that PQ = QR.

c Find the two possible position vectors of the point R. (5) 8 Relative to a fixed origin, the points A and B have position vectors (4i + 5j + 6k) and

(4i + 6j + 2k) respectively.

a Find, in vector form, an equation of the line l1 which passes through A and B. (2)

The line l2 has equation

r = i + 5j − 3k + µ(i + j − k).

b Show that l1 and l2 intersect and find the position vector of their point of intersection. (4)

c Find the acute angle between lines l1 and l2. (3)

d Show that the point on l2 closest to A has position vector (−i + 3j − k). (5)

C4 VECTORS Worksheet E continued

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VECTORS C4 Worksheet F


− −

and 034

−


fixed origin.


The line m has equation

r = 65

1

−

+ µ 31

a −

,

where a is a constant.

Given that lines l and m intersect,

b find the value of a and the coordinates of the point where l and m intersect. (6) 2 Relative to a fixed origin, the points A, B and C have position vectors (−4i + 2j − k),

(2i + 5j − 7k) and (6i + 4j + k) respectively.

a Show that cos (∠ABC) = 13 . (3)

The point M is the mid-point of AC.

b Find the position vector of M. (2)

c Show that BM is perpendicular to AC. (3)

d Find the size of angle ACB in degrees. (3)

3 Relative to a fixed origin O, the points A and B have position vectors 953

−

and 1173

−

respectively.

a Find, in vector form, an equation of the line L which passes through A and B. (2)

The point C lies on L such that OC is perpendicular to L.

b Find the position vector of C. (5)

c Find, to 3 significant figures, the area of triangle OAC. (3)

d Find the exact ratio of the area of triangle OAB to the area of triangle OAC. (2) 4 Relative to a fixed origin O, the points A and B have position vectors (7i − 5j − k) and

(4i − 5j + 3k) respectively.

a Find cos (∠AOB), giving your answer in the form 6k , where k is an exact fraction. (4)

b Show that AB is perpendicular to OB. (3)

The point C is such that OC = 32 OB .

c Show that AC is perpendicular to OA. (3)

d Find the size of ∠ACO in degrees to 1 decimal place. (3)

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VECTORS C4 Answers - Worksheet A

1 a) −p b) 2q c) 1

2 p d) p e) −q f) p + q

g) 12 p + 2q h) p − q i) 2q − p j) −p − 2q k) 1

2 p − q l) 12− p − 2q

2 a) u + v b) w − u c) u + v − w

3 a) q b) p + r c) r − q d) p + q + r e) −q − r f) q + r − p

4 a i = (a + 2b) + (a − 2b) = 2a ii = (a + 2b) − (a − 2b) = 4b b OA = 1

2 OS , OB = 14 TR

∴ R S A

B

O T

5 a i = 1

2 a ii = b − a iii = 1

2 (b − a) iv = a + 1

2 (b − a) = 12 (a + b)

v = 12 (a + b) − 1

2 a = 12 b

b they are parallel (and the magnitude of CD is half that of OB )

6 a parallel, 3p = 3

2 (2p) b not parallel c parallel, (p − 1

3 q) = 13 (3p − q)

d parallel, (4q − 2p) = −2(p − 2q) e parallel, (6p + 8q) = 8( 3

4 p + q) f not parallel

7 a = (2m + 3n) − (4m + 2n) = n − 2m b OM = 1

2 OC = m + 32 n

AM = (m + 32 n) − 4m = 3

2 n − 3m

∴ AM = 32 BC

∴ AM is parallel to BC

C4 VECTORS Answers - Worksheet A page 2

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8 a OM = 1

2 OA = 3u − 2v

AB = (3u − v) − (6u − 4v) = 3v − 3u ON = OA + 1

3 AB = (6u − 4v) + 13 (3v − 3u) = 5u − 3v

b CM = (3u − 2v) − (v − 3u) = 6u − 3v CN = (5u − 3v) − (v − 3u) = 8u − 4v ∴ CN = 4

3 CM

∴ CN and CM are parallel common point C ∴ C, M and N are collinear

9 a a = 5, b = 3 b 2 + b = 0 and a − 4 = 0 ∴ a = 4, b = −2

c −1 = b and 4a = −2 d 2a + 6 = 0 and b − a = 0 ∴ a = 1

2− , b = −1 ∴ a = −3, b = −3

10 a OC = 1

2 a

CB = b − 12 a

OD = 12 a + 1

2 ( b − 12 a) = 1

4 a + 12 b

b AB = b − a OE = OA + k AB ∴ OE = a + k(b − a) c OE = l OD ∴ a + k(b − a) = l( 1

4 a + 12 b)

∴ 1 − k = 14 l

and k = 12 l

adding 1 = 34 l

l = 43

∴ OE = 43 ( 1

4 a + 12 b) = 1

3 a + 23 b

d k = 12 l = 2

3

∴ AE = 23 AB

∴ AE : EB = 2 : 1

Solomon Press

VECTORS C4 Answers - Worksheet B

1 a 6i + j b −4i + 2j c −6i d 10i − 2j 2 a = 4(i − 3j) b = (4i + 2j) − (i − 3j) = 4i − 12j = 3i + 5j

c = 2(i − 3j) + 3(4i + 2j) d = 4(i − 3j) − 2(4i + 2j) = 14i = −4i − 16j 3 a = 9 16+ = 5 b = 2 1 4+ = 2 5

c p + 2q = 34

−

+ 2 12

= 50

d 3q − 2p = 3 12

− 2 34

−

= 314−

| p + 2q | = 5 | 3q − 2p | = 9 196+ = 205 = 14.3 (3sf) 4 a = tan−1 1

2 = 26.6° b = tan−1 3 = 71.6°

c 5p + q = 5(2i + j) + (i − 3j) = 11i + 2j d p − 3q = (2i + j) − 3(i − 3j) = −i + 10j angle = tan−1 2

11 = 10.3° angle = 180° − tan−1 10 = 95.7°

5 a 43

= 16 9+ = 5 b 724

−

= 49 576+ = 25

∴ 15

43

∴ 125

724

−

c 11−

= 1 1+ = 2 d 24

= 4 16+ = 20 = 2 5

∴ 12

11−

= 12 2 1

1−

∴ 12 5

24

= 15 5 1

2

6 a | 5i + 12j | = 25 144+ = 13 ∴ 26

13 (5i + 12j) = 10i + 24j

b | −6i − 8j | = 36 64+ = 10 ∴ 15

10 (−6i − 8j) = −9i − 12j

c | 2i − 4j | = 4 16+ = 20 = 2 5 ∴ 5

2 5(2i − 4j) = 5 (i − 2j)

7 a (2i + λj) + (µi − 5j) = 3i − j b 2(2i + λj) − (µi − 5j) = −3i + 8j 2 + µ = 3 and λ − 5 = −1 4 − µ = −3 and 2λ + 5 = 8 ∴ λ = 4, µ = 1 ∴ λ = 3

2 , µ = 7 8 a 6i + cj = 3(2i + j) b 6i + cj = 2

3− (−9i − 6j) ∴ c = 3 ∴ c = 4

c 36 + c2 = 102 = 100 d 36 + c2 = ( 53 )2 = 45 ∴ c2 = 64 ∴ c2 = 9 c > 0 ∴ c = 8 c > 0 ∴ c = 3

C4 VECTORS Answers - Worksheet B page 2

Solomon Press

9 a a(i + 3j) + b(4i − 2j) = −5i + 13j ∴ a + 4b = −5 (1) and 3a − 2b = 13 (2) (1) + 2×(2) ⇒ 7a = 21 ∴ a = 3, b = −2 b c(i + 3j) + (4i − 2j) = kj ∴ c + 4 = 0 ∴ c = −4 c (i + 3j) + d(4i − 2j) = k(3i − j) ∴ 1 + 4d = 3k and 3 − 2d = −k (1) + 2×(2) ⇒ 7 = k ∴ d = 5

10 a AB = 52

−

− 36

= 84

− −

b AB = 64 16+ = 80 = 4 5 c = OA + 1

2 AB

= 36

+ 12

84

− −

= 14−

d OC = AB

∴ pos. vector = 84

− −

11 a = 2 2 2(2 4) ( 3 0) (3 9)− + − − + − b = 2 2 2(7 11) ( 1 3) (3 5)− + − + + −

= 4 9 36+ + = 16 4 4+ + = 7 = 24 = 2 6 = 4.90 (3sf) 12 a = 16 4 16+ + b = 1 1 1+ + c = 64 1 16+ + d = 9 25 1+ + = 6 = 3 = 1.73 (3sf) = 9 = 35 = 5.92 (3sf) 13 a | 5i − 2j + 14k | = 25 4 196+ + = 15 ∴ 1

15 (5i − 2j + 14k)

b | 2i + 11j − 10k | = 4 121 100+ + = 15 ∴ 10

15 (2i + 11j − 10k) = 23 (2i + 11j − 10k)

c | −5i − 4j + 2k | = 25 16 4+ + = 45 = 3 5 ∴ 20

3 5(−5i − 4j + 2k) = 4

3 5 (−5i − 4j + 2k)

14 λ2 + 144 + 16 = 142 = 196 λ2 = 36 λ = ± 6

C4 VECTORS Answers - Worksheet B page 3

Solomon Press

15 a = 131

−

+ 242

1

−

= 91

1

−

b = 131

−

− 2

53

− −

= 32

2

−

c = 131

−

+ 42

1

−

+ 2

53

− −

= 363

−

d = 2131

−

− 342

1

−

+ 2

53

− −

= 12

178

− −

16 a −2i + λj + µk = 12− (4i + 2j − 8k) b −2i + λj + µk = 2

5 (−5i + 20j − 10k) ∴ λ = −1, µ = 4 ∴ λ = 8, µ = −4 17 a 2p − q = 2(i − 2j + 4k) − (−i + 2j + 2k) 18 a AB = (−4i + j + 8k) − (−2i + 7j + 4k) = 3i − 6j + 6k = −2i − 6j + 4k ∴ | 2p − q | = 9 36 36+ + = 9 pos. vec of mid-point = OA + 1

2 AB b (i − 2j + 4k) + k(−i + 2j + 2k) = (−2i + 7j + 4k) + 1

2 (−2i − 6j + 4k) = l(2i − 4j − 7k) = −3i + 4j + 6k ∴ 1 − k = 2l (1) b AC = (6i − 5j) − (−2i + 7j + 4k) −2 + 2k = −4l (2) = 8i − 12j − 4k 4 + 2k = −7l (3) AD = OA + 3

4 AC [(1) and (2) are the same equation] = (−2i + 7j + 4k) + 3

4 (8i − 12j − 4k) (2) − (3) ⇒ −6 = 3l = 4i − 2j + k ∴ l = −2 ∴ k = 5

19 a (λi − 2λj + µk) = k(2i − 4j − 3k) 20 a BC = 618

−

− 12

74

− −

= 6

84

− −

∴ λ = 2k (1) OM = OB + 12 BC

−2λ = −4k (2) = 12

74

− −

+ 12

684

− −

= 936

− −

µ = −3k (3) b OM = 32 OA

[(1) and (2) are the same equation] ∴ OM and OA are parallel 3×(1) + 2×(3) ⇒ 3λ + 2µ = 0 common point O b λ2 + (−2λ)2 + µ2 = ( 292 )2 ∴ O, A and M are collinear 5λ2 + µ2 = 116 µ = 3

2− λ ⇒ 5λ2 + 94 λ2 = 116

λ2 = 16 λ = ± 4 µ = 3

2− λ and µ > 0 ∴ λ = −4, µ = 6 21 a d2 = (9 − t)2 + (1 + 2t)2 + (5 − t)2 = 81 − 18t + t2 + 1 + 4t + 4t2 + 25 − 10t + t2 = 6t2 − 24t + 107 b d2 = 6(t2 − 4t) + 107 = 6[(t − 2)2 − 4] + 107 = 6(t − 2)2 + 83 ∴ closest when t = 2 min. d = 83 = 9.11 m (3sf)

Solomon Press

VECTORS C4 Answers - Worksheet C

Note: For this worksheet especially, there may be alternative correct answers

1 a y b y c y 2

O 2 x

O x −1 O x

d y e y f r = i − 2j + s(2i + 3j)

3 y

O 9 x O x O 7

3 x

72−

2 a r = −i + j + s(3i − 2j)

b r = 4j + si

c r = 3i − j + s(i + 5j) 3 a dirn = 3

1

− 10

= 21

b dirn = 11−

− 34

−

= 23

−

c dirn = 23

−

− 22

−

= 45

−

∴ r = 10

+ s 21

∴ r = 34

−

+ s 23

−

∴ r = 22

−

+ s 45

−

4 a −1 + 2λ = 5 ∴ λ = 3 3 + cλ = 3 + 3c = 0 ∴ c = −1

b ci + 2j = k(6i + 3j) ∴ k = 2

3 ∴ c = 4 5 a r = −i + sj b r = s(i + 2j) c r = j + s(i + 3j)

d r = −2j + s(4i + 3j) e r = 5j + s(2i − j) f y = 14 x + 2

∴ r = 2j + s(4i + j) 6 a x = 2 + 3λ, y = 1 + 2λ

b λ = 23

x − = 12

y −

2(x − 2) = 3(y − 1) 2x − 4 = 3y − 3 2x − 3y − 1 = 0

C4 VECTORS Answers - Worksheet C page 2

Solomon Press

7 a x = 3 + λ, y = 2λ b x = 1 + 3λ, y = 4 + λ c x = 4λ, y = 2 − λ λ = x − 3 =

2y λ = 1

3x − = y − 4 λ =

4x = 2 − y

2(x − 3) = y x − 1 = 3(y − 4) x = 4(2 − y) 2x − y − 6 = 0 x − 3y + 11 = 0 x + 4y − 8 = 0 d x = −2 + 5λ, y = 1 + 2λ e x = 2 − 3λ, y = −3 + 4λ f x = λ + 3, y = −2λ − 1 λ = 2

5x + = 1

2y − λ = 2

3x −−

= 34

y + λ = x − 3 = 12

y +−

2(x + 2) = 5(y − 1) 4(x − 2) = −3(y + 3) −2(x − 3) = y + 1 2x − 5y + 9 = 0 4x + 3y + 1 = 0 2x + y − 5 = 0 8 a 3

1 −

= 12− 6

2−

b 14

≠ k 41

c 24

= 23

36

∴ parallel ∴ not parallel ∴ parallel (1, 2) lies on first line (2, −5) lies on first line when x = 1 on second line when x = 2 on second line −2 − 6t = 1 ⇒ t = 1

2− −1 + 3t = 2 ⇒ t = 1 ⇒ y = 3 + 2( 1

2− ) = 2 ⇒ y = 1 + 6(1) = 7 parallel and common point ∴ (2, −5) not on second line ∴ identical ∴ parallel but not identical 9 a 1 + λ = 2 + 3µ (1) b 4 − λ = 5 + 2µ (1) c 2λ = 2 − µ (1) 2 = 1 + µ (2) 1 + λ = −2 − 3µ (2) 1 − λ = 10 + 3µ (2) (2) ⇒ µ = 1 (1) + (2) ⇒ 5 = 3 − µ (1) + 2×(2) ⇒ 2 = 22 + 5µ ∴ 5i + 2j µ = −2 µ = −4 ∴ i + 4j ∴ 6i − 2j

d −1 − 4λ = 2 − µ (1) e −2 − 3λ = −3 + 5µ (1) f 1 + 3λ = 3 + µ (1) 5 + 6λ = −2 + 2µ (2) 11 + 4λ = −7 + 3µ (2) 2 + 2λ = 5 + 4µ (2) 2×(1) + (2) ⇒ 3 − 2λ = 2 4×(1) + 3×(2) 2×(1) − 3×(2) λ = 1

2 ⇒ 25 = −33 + 29µ ⇒ −4 = −9 − 10µ ∴ −3i + 8j µ = 2 µ = 1

2− ∴ 7i − j ∴ 5

2 i + 3j

10 a r = 4i + k + s(i + 3j − 2k)

b r = 2i + j + sk

c r = −i + 4j + 2k + s(2i − 3j + 5k) 11 a AB = (6i − 3j + k) − (5i + j − 2k) = i − 4j + 3k

b r = (5i + j − 2k) + s(i − 4j + 3k)

c 5 + s = 3 ⇒ s = −2 when s = −2, r = (5i + j − 2k) − 2(i − 4j + 3k) = 3i + 9j − 8k ∴ l passes through (3, 9, −8)


Solomon Press

12 a direction = (5i + 4j + 6k) − (i + 3j + 4k) b direction = (i + 5j + 2k) − (3i − 2k) = 4i + j + 2k = −2i + 5j + 4k ∴ r = i + 3j + 4k + s(4i + j + 2k) ∴ r = 3i − 2k + s(−2i + 5j + 4k)

c r = s(6i − j + 2k) d direction = (4i − 7j + k) − (−i − 2j + 3k) = 5i − 5j − 2k ∴ r = −i − 2j + 3k + s(5i − 5j − 2k) 13 a 3 + 2λ = 9 ∴ λ = 3 −5 + aλ = −5 + 3a = −2 ∴ a = 1 1 + bλ = 1 + 3b = −8 ∴ b = −3

b 2i + aj + bk= k(8i − 4j + 2k) ∴ k = 1

4 ∴ a = −1, b = 1

2

14 a x = 2 + 3λ, b x = 4 + λ, c x = −1 + 4λ, y = 3 + 5λ, y = −1 + 6λ, y = 5 − 2λ, z = 2λ, z = 3 + 3λ, z = −2 − λ, (λ = ) 2

3x − = 3

5y − =

2z (λ = ) x − 4 = 1

6y + = 3

3z − (λ = ) 1

4x + = 5

2y −−

= 21

z +−

15 a s = 1

3x − = 4

2y + = z − 5 b s =

4x = 1

2y −−

= 73

z + c s = 54

x +−

= y + 3 = z

x = 1 + 3s, x = 4s, x = −5 − 4s, y = −4 + 2s, y = 1 − 2s, y = −3 + s, z = 5 + s, z = −7 + 3s, z = s, r = i − 4j + 5k + s(3i + 2j + k) r = j − 7k + s(4i − 2j + 3k) r = −5i − 3j + s(−4i + j + k) 16 4 + s = 7 − 3t (1) −2s = 2 + 2t (2) 3 + 2s = −5 + t (3) (2) + (3) ⇒ 3 = −3 + 3t t = 2, s = −3 check (1) 4 + (−3) = 7 − 3(2) true ∴ intersect point of intersection: (1, 6, −3) 17 2 + λ = 1 + µ (1) −1 + λ = 4 − 2µ (2) 4 + 3λ = 3 + µ (3) (1) − (2) ⇒ 3 = −3 + 3µ µ = 2, λ = 1 check (3) 4 + 3(1) = 3 + (2) false ∴ do not intersect (i + j + 3k) ≠ k(i − 2j + k) ∴ not parallel ∴ skew


Solomon Press

18 a 3 + 4λ = 3 + µ (1) b 213

− −

= 12−

426

−

1 + λ = 2 (2) ∴ parallel 5 − λ = −4 + 2µ (3) (2) ⇒ λ = 1 sub. (1) µ = 4 check (3) 5 − (1) = −4 + 2(4) true ∴ intersect

position vector of intersection: 724

c 8 + λ = −2 + 4µ (1) d 1 + λ = 7 + 2µ (1) 2 + 3λ = 2 − 3µ (2) 5 + 4λ = −6 + µ (2) −4 − 2λ = 8 − 4µ (3) 2 − 2λ = −5 − 3µ (3) (1) + (3) ⇒ 4 − λ = 6 2×(1) + (3) ⇒ 4 = 9 + µ λ = −2, µ = 2 µ = −5, λ = −4 check (2) 2 + 3(−2) = 2 − 3(2) check (2) 5 + 4(−4) = −6 + (−5) true ∴ intersect true ∴ intersect


0

−


10

− −

e 4 + 2λ = 3 + 5µ (1) f 6λ = −12 + 5µ (1) −1 + 5λ = −2 − 3µ (2) 7 − 4λ = −1 + 2µ (2) 3 − 3λ = 1 − 4µ (3) −2 + 8λ = 11 − 3µ (3) 3×(1) + 2×(3) ⇒ 18 = 11 + 7µ 2×(2) + (3) ⇒ 12 = 9 + µ µ = 1, λ = 2 µ = 3, λ = 1

2 check (2) −1 + 5(2) = −2 − 3(1) check (1) 6( 1

2 ) = −12 + 5(3) false ∴ do not intersect true ∴ intersect

253

−

≠ k534

− −


∴ skew

Solomon Press

VECTORS C4 Answers - Worksheet D

1 a = 3 + 2 = 5 b = 12 − 5 = 7 c = −5 + 4 = −1 2 (i + 4j).(8i − 2j) = 8 − 8 = 0 ∴ perpendicular 3 a 3

1 −

.3c

= 0 b 21

. 3c

= 0 c 25

−

.4

c −

= 0

3c − 3 = 0 6 + c = 0 2c + 20 = 0 c = 1 c = −6 c = −10 4 a 4i − 3j = 16 9+ = 5, 8i + 6j = 64 36+ = 10 (4i − 3j).(8i + 6j) = 32 − 18 = 14 ∴ angle = cos−1 14

5 10× = cos−1 725 = 73.7°

b 7i + j = 49 1+ = 5 2 , 2i + 6j = 4 36+ = 2 10 (7i + j).(2i + 6j) = 14 + 6 = 20 ∴ angle = cos−1 20

5 2 2 10× = cos−1 1

5 = 63.4°

c 4i + 2j = 16 4+ = 2 5 , −5i + 2j = 25 4+ = 29 (4i + 2j).(−5i + 2j) = −20 + 4 = −16 ∴ angle = cos−1 16

2 5 29−×

= cos−1 (− 85 29

) = 131.6°

5 BA = (9i + j) − (3i − j) = 6i + 2j BC = (5i − 2j) − (3i − j) = 2i − j BA = 36 4+ = 2 10 , BC = 4 1+ = 5 BA . BC = (6i + 2j).(2i − j) = 12 − 2 = 10 ∴ ∠ABC = cos−1 10

2 10 5× = cos−1 1

2 = 45°

6 a = 3 + 2 + 8 = 13 b = 6 + 6 − 2 = 10

c = −5 + 0 − 6 = −11 d = −3 + 22 + 32 = 51

e = 27 − 28 − 1 = −2 f = 0 + 9 + 0 = 9 7 a = (2i + j − 3k).(i + 5j − k) = 2 + 5 + 3 = 10

b = (2i + j − 3k).(6i − 2j − 3k) = 12 − 2 + 9 = 19

c q + r = (i + 5j − k) + (6i − 2j − 3k) = 7i + 3j − 4k p.(q + r) = (2i + j − 3k).(7i + 3j − 4k) = 14 + 3 + 12 = 29 p.q + p.r = 10 + 19 = 29 ∴ p.(q + r) = p.q + p.r 8 a = p.q + p.r + p.q − p.r b = p.q + p.r + q.r − q.p = 2p.q = p.q + p.r + q.r − p.q = p.r + q.r = (p + q).r

C4 VECTORS Answers - Worksheet D page 2

Solomon Press

9 (5i − 3j + 2k).(3i + j − 6k) = 15 − 3 − 12 = 0 ∴ perpendicular 10 BA = (3i + 4j − 6k) − (i + 5j − 2k) = 2i − j − 4k BC = (8i + 3j + 2k) − (i + 5j − 2k) = 7i − 2j + 4k BA . BC = (2i − j − 4k).(7i − 2j + 4k) = 14 + 2 − 16 = 0 ∴ BA and BC are perpendicular ∴ ∠ABC = 90° 11 a (2i + 3j + k).(ci − 3j + k) = 0 b (−5i + 3j + 2k).(ci − j + 3ck) = 0 2c − 9 + 1 = 0 −5c − 3 + 6c = 0 c = 4 c = 3

c (ci − 2j + 8k).(ci + cj − 3k) = 0 d (3ci + 2j + ck).(5i − 4j + 2ck) = 0 c2 − 2c − 24 = 0 15c − 8 + 2c2 = 0 (c + 4)(c − 6) = 0 2c2 + 15c − 8 = 0 c = −4, 6 (2c − 1)(c + 8) = 0 c = −8, 1

2

12 a 122

−

= 1 4 4+ + = 3, 814

−

= 64 1 16+ + = 9, 122

−

.814

−

= 8 + 2 + 8 = 18

∴ cos θ = 183 9× = 2

3

b 412

−

= 16 1 4+ + = 21 , 2

36

− −

= 4 9 36+ + = 7, 412

−

.2

36

− −

= −8 + 3 + 12 = 7

∴ cos θ = 721 7×

= 121

or 121 21

c 121

−

= 1 4 1+ + = 6 , 17

2

−

= 1 49 4+ + = 3 6 , 121

−

.17

2

−

= 1 − 14 − 2 = −15

∴ cos θ = 156 3 6−×

= 56−

d 53

4

−

= 25 9 16+ + = 5 2 , 341

− −

= 9 16 1+ + = 26 , 53

4

−

.341

− −

= 15 + 12 − 4 = 23

∴ cos θ = 235 2 26×

= 2310 13

or 23130 13

13 a (3i − 4k) = 9 16+ = 5 b (2i − 6j + 3k) = 4 36 9+ + = 7 (7i − 4j + 4k) = 49 16 16+ + = 9 (i − 3j − k) = 1 9 1+ + = 11 (3i − 4k).(7i − 4j + 4k) = 21 + 0 − 16 = 5 (2i − 6j + 3k).(i − 3j − k) = 2 + 18 − 3 = 17 ∴ angle = cos−1 5

5 9× = cos−1 19 = 83.6° ∴ angle = cos−1 17

7 11× = 42.9°

c (6i − 2j − 9k) = 36 4 81+ + = 11 d (i + 5j − 3k) = 1 25 9+ + = 35 (3i + j + 4k) = 9 1 16+ + = 26 (−3i − 4j + 2k) = 9 16 4+ + = 29 (6i − 2j − 9k).(3i + j + 4k) = 18 − 2 − 36 = −20 (i + 5j − 3k).(−3i − 4j + 2k) = −3 − 20 − 6 = −29 ∴ angle = cos−1 20

11 26−×

= 110.9° ∴ angle = cos−1 2935 29

−×

= cos−1 ( 2935− ) = 155.5°


Solomon Press

14 a BA = (7 + 1)i + (2 − 6)j + (−2 + 3)k = 8i − 4j + k BC = (−3 + 1)i + (1 − 6)j + (2 + 3)k = −2i − 5j + 5k b BA = 64 16 1+ + = 9 BC = 4 25 25+ + = 3 6 BA . BC = (8i − 4j + k).(−2i − 5j + 5k) = −16 + 20 + 5 = 9 ∴ ∠ABC = cos−1 9

9 3 6× = cos−1 1

3 6 = 82.2°

c = 12 × 9 × 3 6 × sin 82.18° = 32.8

15 a AB = (4i + 3j − 2k) − (3i − 2j − k) = i + 5j − k AC = (2i − j) − (3i − 2j − k) = −i + j + k AB = 1 25 1+ + = 3 3 AC = 1 1 1+ + = 3 AB . AC = (i + 5j − k).(−i + j + k) = −1 + 5 − 1 = 3 cos (∠BAC) = 3

3 3 3× = 1

3

b sin2 (∠BAC) = 1 − ( 13 )2 = 8

9

sin (∠BAC) = 89 = 2

3 2

area = 12 × 3 3 × 3 × 2

3 2 = 3 2

16 a 44

2

−

= 16 16 4+ + = 6, 806

−

= 64 36+ = 10, 44

2

−

.806

−

= 32 + 0 − 12 = 20

∴ acute angle = cos−1 206 10× = cos−1 1

3 = 70.5°

b 61

18

− −

= 36 1 324+ + = 19, 4123

−

= 16 144 9+ + = 13, 61

18

− −

.4123

−

= 24 + 12 − 54 = −18

∴ acute angle = cos−1 1819 13

−× = cos−1 18

247 = 85.8°

c 11

3

−

= 1 1 9+ + = 11 , 25

3

−

= 4 25 9+ + = 38 , 11

3

−

.25

3

−

= 2 + 5 + 9 = 16

∴ acute angle = cos−1 1611 38×

= cos−1 1611 38

= 38.5°

d 46

7

− −

= 16 36 49+ + = 101 , 518

− −

= 25 1 64+ + = 90 , 46

7

− −

.518

− −

= −20 + 6 − 56 = −70

∴ acute angle = cos−1 70101 90

−×

= cos−1 70101 90

= 42.8°


Solomon Press

17 a AB = (6i + 5j + k) − (5i + 8j − k) = i − 3j + 2k ∴ r = 5i + 8k − k + λ(i − 3j + 2k) b 5 + λ = 4 − 5µ (1) 8 − 3λ = −3 + µ (2) −1 + 2λ = 5 − 2µ (3) 3×(1) + (2) ⇒ 23 = 9 − 14µ µ = −1, λ = 4 check (3) −1 + 2(4) = 5 − 2(−1) true ∴ intersect pos. vector of int. = 9i − 4j + 7k c (i − 3j + 2k) = 1 9 4+ + = 14 (−5i + j − 2k) = 25 1 4+ + = 30 (i − 3j + 2k).(−5i + j − 2k) = −5 − 3 − 4 = −12 acute angle = cos−1 12

14 30−×

= cos−1 6105

= 54.2° (1dp)

18 λ = 2

3x − =

2y = 5

6z +−

µ = 44

x −−

= 17

y + = 34

z −−

x = 2 + 3λ, y = 2λ, z = −5 − 6λ x = 4 − 4µ, y = −1 + 7µ, z = 3 − 4µ r = 2i − 5k + λ(3i + 2j − 6k) r = 4i − j + 3k + µ(−4i + 7j − 4k) (3i + 2j − 6k) = 9 4 36+ + = 7 (−4i + 7j − 4k) = 16 49 16+ + = 9 (3i + 2j − 6k).(−4i + 7j − 4k) = −12 + 14 + 24 = 26 ∴ acute angle = cos−1 26

7 9× = cos−1 2663 = 65.6°

19 a 7 + 2λ = −4 + 5µ (1) −λ = 7 − 4µ (2) −2 + 2λ = −6 − 2µ (3) (1) − (3) ⇒ 9 = 2 + 7µ µ = 1 ∴ A (1, 3, −8) b (2i − j + 2k) = 4 1 4+ + = 3 (5i − 4j − 2k) = 25 16 4+ + = 3 5 (2i − j + 2k).(5i − 4j − 2k) = 10 + 4 − 4 = 10 acute angle = cos−1 10

3 3 5× = cos−1 ( 2

9 5 ) = 60.2° (1dp)

c 7 + 2λ = 5 ⇒ λ = −1 sub. λ = −1 in eqn for l r = 7i − 2k − (2i − j + 2k) = 5i + j − 4k ∴ B lies on l d AB = (5i + j − 4k) − (i + 3j − 8k) = 4i − 2j + 4k AB = 16 4 16+ + = 6 ∴ dist. of B from m = 6 sin 60.20° = 5.21 (3sf)


Solomon Press

20 a AB = (11i + 5j + k) − (9i + 6j) = 2i − j + k OC = 9i + 6j + λ(2i − j + k) = OA + λ AB ∴ C lies on l b [(9 + 2λ)i + (6 − λ)j + λk].[2i − j + k] = 0 2(9 + 2λ) − (6 − λ) + λ = 0 λ = −2 c sub. λ = −2 in OC ∴ 5i + 8j − 2k 21 a [(−4 + λ)i + (2 + 3λ)j + (7 − 4λ)k].[i + 3j − 4k] = 0 (−4 + λ) + 3(2 + 3λ) − 4(7 − 4λ) = 0 λ = 1 ∴ (−3, 5, 3) b [(7 + 6λ)i + (11 − 9λ)j + (−9 + 3λ)k].[6i − 9j + 3k] = 0 6(7 + 6λ) − 9(11 − 9λ) + 3(−9 + 3λ) = 0 λ = 2

3 ∴ (11, 5, −7)

Solomon Press

VECTORS C4 Answers - Worksheet E

1 a 1 + 2λ = 7 ∴ λ = 3 2 a AB =

506

−

− 164

= 46

10

− −

p − 3λ = −1 ∴ p = 8 ∴ r = 164

+ s46

10

− −

−4 + qλ = −1 ∴ q = 1 b 1 + 4s = 5 + t (1) b 2i + j − 3k = 4 1 9+ + = 14 6 − 6s = −5 − 4t (2) −4i + 5j − 2k = 16 25 4+ + = 45 4×(1) + (2) ⇒ 10 + 10s = 15 (2i + j − 3k).(−4i + 5j − 2k) s = 1

2

= −8 + 5 + 6 = 3 ∴ pos. vector of C = 331

−

θ = cos−1 314 45

= 83.1° (1dp) c pos. vector of mid-point of AB

= OA + 12 AB

= 164

+ 12

46

10

− −

= 331

−

∴ C is mid-point of AB

3 a PQ = (3i + j) − (5i − 2j + 2k) 4 a 5 + 2λ = 7 − µ (1) = −2i + 3j − 2k −λ = −3 + µ (2) ∴ r = 5i − 2j + 2k + λ(−2i + 3j − 2k) 1 + 2λ = 7 − 2µ (3) b 5 − 2λ = 4 + 5µ (1) (1) + (2) ⇒ 5 + λ = 4 −2 + 3λ = 6 − µ (2) λ = −1, µ = 4 2 − 2λ = −1 + 3µ (3) check (3) 1 + 2(−1) = 7 − 2(4) (1) − (3) ⇒ 3 = 5 + 2µ true ∴ intersect µ = −1, λ = 3 pos. vector of int. = 3i + j − k check (2) −2 + 3(3) = 6 − (−1) b diagonals bisect each other true ∴ intersect let M be point of intersection pos. vector of int. = −i + 7j − 4k ∴ AM = (3i + j − k) − (9i − 2j + 5k) c −2i + 3j − 2k = 4 9 4+ + = 17 = −6i + 3j − 6k 5i − j + 3k = 25 1 9+ + = 35 OC = OA + 2 AM (−2i + 3j − 2k).(5i − j + 3k) = (9i − 2j + 5k) + 2(−6i + 3j − 6k) = −10 − 3 − 6 = −19 = −3i + 4j − 7k θ = cos−1 19

17 35− = 38.8° c area of triangle ABC = 1

2 × 54 = 27

AC = 2(−6i + 3j − 6k) = 6(−2i + j − 2k) AC = 6 4 1 4+ + = 18 let distance of B from l1 = d ∴ 1

2 × 18 × d = 27 d = 3

C4 VECTORS Answers - Worksheet E page 2

Solomon Press

5 a AB = (2i − j + 2k) − (4i + 2j − 4k) 6 a AB = 418

−

− 51

10

− −

= 1

22

−

= −2i − 3j + 6k ∴ r = 51

10

− −

+ λ1

22

−

∴ r = 4i + 2j − 4k + λ(−2i − 3j + 6k) b 5 − λ = 0 ⇒ λ = 5 b r = 4i − 7j − k + µ(6j − 2k) sub. λ = 5 in l

c −7 + 6µ = 2 ⇒ µ = 32 r =

090

∴ C (0, 9, 0)

sub. µ = 32 in l2 c OD =

51 2

10 2

λλλ

− − + − +

r = 4i − 7j − k + 32 (6j − 2k) OD .

122

−

= 0

= 4i + 2j − 4k ∴ A lies on l2 −(5 − λ) + 2(−1 + 2λ) + 2(−10 + 2λ) = 0 d −2i − 3j + 6k = 4 9 36+ + = 7 9λ − 27 = 0

6j − 2k = 36 4+ = 40 λ = 3, OD = 254

−

(−2i − 3j + 6k).(6j − 2k) ∴ D (2, 5, −4) = 0 − 18 − 12 = −30 d OD = 4 25 16+ + = 45 = 3 5 θ = cos−1 30

7 40− = 47.3° (1dp) CD = 4 16 16+ + = 6

area = 12 × 6 × 3 5 = 9 5

C4 VECTORS Answers - Worksheet E page 3

Solomon Press

7 a −6 + 4s = 6 ⇒ s = 3 8 a AB = (4i + 6j + 2k) − (4i + 5j + 6k) sub. s = 3 in l1 = j − 4k

r = 162

− −

+ 3041

−

= 165

−

∴ r = 4i + 5j + 6k + λ(j − 4k)

∴ P (1, 6, −5) lies on l1 b 4 = 1 + µ (1) b 1 = 4 + 3t ⇒ t = −1 5 + λ = 5 + µ (2) sub. t = −1 in l2 6 − 4λ = −3 − µ (3)

r = 441

− −

− 32

2

−

(1) ⇒ µ = 3

OQ = 123

− −

sub. (2) ⇒ λ = 3

c PQ = 0 64 4+ + = 68 = 2 17 check (3) 6 − 4(3) = −3 − (3)

32

2

−

= 9 4 4+ + = 17 true ∴ intersect

∴ OR = OQ ± 232

2

−

pos. vector of int. = 4i + 8j − 6k

= 5

27

− −

or 76

1

−

c (j − 4k) = 1 16+ = 17

(i + j − k) = 1 1 1+ + = 3 (j − 4k).(i + j − k) = 0 + 1 + 4 = 5 θ = cos−1 5

3 17 = 45.6° (1dp)

d let closest point be C OC = (1 + µ)i + (5 + µ)j + (−3 − µ)k AC = OC − OA = (−3 + µ)i + µj + (−9 − µ)k AC must be perpendicular to l2 ∴ AC .(i + j − k) = 0 (−3 + µ) + µ − (−9 − µ) = 0 µ = −2 ∴ OC = −i + 3j − k

Solomon Press

VECTORS C4 Answers - Worksheet F

1 a AB =

034

−

− 215

− −

= 2

41

−

2 a BA = (−4i + 2j − k) − (2i + 5j − 7k)

r = 215

− −

+ λ2

41

−

= −6i − 3j + 6k

b 2 − 2λ = 6 + aµ (1) BC = (6i + 4j + k) − (2i + 5j − 7k) −1 + 4λ = −5 − 3µ (2) = 4i − j + 8k −5 + λ = 1 + µ (3) BA = 36 9 36+ + = 9 (2) + 3×(3) ⇒ −16 + 7λ = −2 BC = 16 1 64+ + = 9 λ = 2, µ = −4 BA . BC = −24 + 3 + 48 = 27 sub. (1) 2 − 2(2) = 6 + a(−4) cos (∠ABC) = 27

9 9× = 13

−2 = 6 − 4a b AC = (6i + 4j + k) − (−4i + 2j − k) a = 2 = 10i + 2j + 2k point of intersection: (−2, 7, −3) OM = OA + 1

2 AC = (−4i + 2j − k) + 1

2 (10i + 2j + 2k) = i + 3j c BM = (i + 3j) − (2i + 5j − 7k) = −i − 2j + 7k BM . AC = (−i − 2j + 7k).(10i + 2j + 2k) = −10 − 4 + 14 = 0 ∴ BM perpendicular to AC d ∠ABC = cos−1 1

3 = 70.529 isosceles triangle ∴ ∠ACB = 1

2 (180 − 70.529) = 54.7° (1dp)

C4 VECTORS Answers - Worksheet F page 2

Solomon Press

3 a AB = 1173

−

− 953

−

= 220

4 a (7i − 5j − k) = 49 25 1+ + = 5 3

∴ r = 953

−

+ λ220

(4i − 5j + 3k) = 16 25 9+ + = 5 2

b OC = 9 25 2

3

λλ

+ + −

(7i − 5j − k).(4i − 5j + 3k) = 28 + 25 − 3 = 50

OC .220

= 0 cos (∠AOB) = 505 3 5 2×

= 26

= 13 6

2(9 + 2λ) + 2(5 + 2λ) + 0 = 0 b AB = (4i − 5j + 3k) − (7i − 5j − k) 8λ + 28 = 0 = −3i + 4k

λ = 72− , OC =

223

− −

AB . OB = (−3i + 4k).(4i − 5j + 3k)

c OC = 4 4 9+ + = 17 = −12 + 0 + 12 = 0

AC = 72

220

= 7 1 1+ = 7 2 ∴ AB perpendicular to OB

area = 12 × 17 × 7 2 = 20.4 c OC = 3

2 (4i − 5j + 3k) d AC = 7

2 AB = 6i − 152 j + 9

2 k

∴ area OAB : area OAC = 2 : 7 AC = (6i − 152 j + 9

2 k) − (7i − 5j − k) = −i − 5

2 j + 112 k

AC . OA = (−i − 52 j + 11

2 k).(7i − 5j − k) = −7 + 25

2 − 112 = 0

∴ AC perpendicular to OA d ∠CAO = 90° ∴ ∠ACO = 90° − ∠AOC = 90° − ∠AOB = 90° − cos−1 ( 1

3 6 ) = 54.7°

vectors worksheet a

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