worksheet 7.1 introduction to vectors name: · © john wiley & sons australia, ltd 2009 1...

© John Wiley & Sons Australia, Ltd 2009 1

WorkSHEET 7.1 Introduction to vectors Name: _________________________ 1

From the figure above, express the following vectors in terms of .

(a) the vector from A to D (b) the vector from C to F (Give 2 answers for each vector).

(a)

(b)

2

2 A hiker travels 3 km north, then 4 km west. How far is the hiker from the starting point? What is the bearing of the resultant displacement?

The hiker is 5 km from the starting point on a bearing N 53.1° W or (360 – 53.1)°T = 306.9°T.

2

~~~~~ and ,,, edcba~~~~~~ or 2 deccbaAD --++=

~~~~~~ or 2 bacedcCF --++=

°=

=

=+=

1.53

34tan

km 543 22

q

q

d

Maths Quest Maths C Year 11 for Queensland Chapter 7: Introduction to vectors WorkSHEET 7.1


3 find:

(a)

(b)

4

4 Find the distance between the ends of the vectors given by

.

I could also say the same question as … What is the distance between the points represented by the position vectors … Or I could say it more abstractly as … Calculate !𝑎𝑏$$$$⃑ !.

The distance is 13 units.

2

5 Let 𝑢' = 3𝚤̂ − 4𝚥̂ + 𝑘1 and 𝑣' = 2𝚤̂ + 2𝚥̂ − 3𝑘1 Find .

2

6 Find the value of m if is

perpendicular to .

,74 If~~~jiu -=

~ u

^~

u

( )

06.865

4916

74

74 (a)

22~

~~~

»=

+=

-+=

-=

u

jiu

÷÷ø

öççè

æ-=

-=

=

~~

~~

~

~^~

74651

65

74

(b)

ji

ji

u

uu

~~~~~~ 95 and 310 jibjia -=+=

13125

125

95310

22

~~

~~~~~~

=+=

+=

÷øö

çèæ --+=-

ji

jijiba

~~ .vu

5386

)3)(1()2)(4()2)(3(

322.43. ~~~~~~~~

-=--=

-+-+=

÷øö

çèæ -+÷øö

çèæ +-= kjikjivu

~~~~ 432 kjmimu +-=

~~~~ 32 kjiv ++=

31240124

0. if toangles rightat is 124

1262

32.432.

~~~~

~~~~~~~~

===+-

=+-=

+-=

÷øö

çèæ ++÷øö

çèæ +-=

mm

m

vuvum

mm

kjikjmimvu



7 Find the angle between the vectors .

The angle between is approximately

.

3

8 If calculate

(a)

(b)

(a)

= 2( ) +

=

(b)

= - 2( )

=

2

9 A delivery boy travels 2 km north and then 3 km west and then another 2 km north. (a) Use vectors to indicate these movements. (b) Give the magnitude of his displacement

from his starting position.

2

~~~~~~ 34 and 3 jivjiu -=+-=

( ) ( )

!6.161103cos

10310515cos

cos10515

cos2510312

cos3413

34.3

cos.

1

2222

~~~~

~~

=

÷÷ø

öççè

æ-=

-=

-=

=-

´´=--

´-+´+-=

÷øö

çèæ -÷

øö

çèæ +-

=

-q

q

q

q

q

q

jiji

uvvu

~~ and vu°6.161

~~~~~~~~ 2 and2 kjibkjia +-=++=

~~2 ba +

~~ 2 ab -

~~2 ba +

~~~ 2 kji ++ ~~~2 kji +-

~~~3 3 4 kji ++

~~ 2 ab -

~~~2 kji +- ~~~ 2 kji ++

~~ 5 kj --

~~~2 3 2 (a) jij +-

km5

43 (b)~~

=

+-= ji



10 Calculate the dot product of the vectors shown below.

1

11 A plane flys 50 km north west from airport A and lands at airport B. Then takes off and flys a further 80 km at a bearing of 𝑆457𝑊 before landing at airport C. The plane now has to head back home to airport A. Determine the straight line distance the plane has to fly home and the bearing it has to take. ** this one has my solution, so this is what your exam setting out would look like! *** the text book should have made the numbers so that the bearing came from an exact value triangle, hence no calculator … but they didn’t :-( You can see how my solution carries exact values where ever possible!

Draw a diagram … Because the question asks for the bearing “home”, make the resultant vector from C to A, and our Resultant Vector becomes -b - a. Now, convert to vector form using rule;

𝒂 = |𝑎| cos 𝜃 𝑖 + |𝑎| sin 𝜃 𝑗 so,

𝒂 = 50 cos 135 𝑖 + 50 sin 135 𝑗

∴ 𝑎 = −50√2

𝑖 +50√2

𝑗

and 𝒃 = 80 cos 225 𝑖 + 80 sin 225 𝑗

∴ 𝑏 = −80√2

𝑖 −80√2

𝑗

Now, (from diagram) 𝒗 = −𝒃 − 𝒂

𝑣 = −J−50√2

𝑖 +50√2

𝑗K − J−80√2

𝑖 −80√2

𝑗K

=130√2

𝑖 +30√2

𝑗

and use |𝑎| = L𝑥N + 𝑦N

|𝑣| = PJ130√2

KN

+ J30√2KN

= 10√89

further use 𝜃 = tanTU

𝑦𝑥

𝜃 = tanTU30√2130√2

≈ 137

Therefore the plane needs to fly approx. 94 kilometers at a bearing of 777𝑇 to get home.

42.3)110cos(52

cos.~~~~

-=´´=

=

!

qbaba

worksheet 7.1 introduction to vectors name: · © john wiley & sons australia, ltd 2009 1...

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