WorkSHEET 8.2 Vector applications Name:

© John Wiley & Sons Australia, Ltd 2009 1

WorkSHEET 8.2 Vector applications Name: _________________________ 1 Bob is located at position 3𝑖 − 2𝑗 and Annie is

located at position −𝑖 + 𝑗. How far apart are they?

Set; 𝒂 = 3𝑖 − 2𝑗

and 𝒃 = −𝑖 + 𝑗

Use; |𝑎| = ,𝑥. + 𝑦.

Find the vector that joins Annie to Bob … this

Resultant Vector is represented as a Vector sum;

𝑅 = 𝒃 − 𝒂

So, 𝐴𝐵3333 = (−𝑖 + 𝑗) − (3𝑖 − 2𝑗)

= −𝑖 + 𝑗 − 3𝑖 + 2𝑗

∴ 𝐴𝐵3333 = −4𝑖 + 3𝑗

Now, by sub;

|𝐴𝐵| = ,−4. + 3.

= √25

∴ |𝐴𝐵| = 5

1

Maths Quest Maths C Year 11 for Queensland Chapter 8: Vector applications WorkSHEET 8.2


2 Abby pushes Gertrude with force of 2𝑖 + 3𝑗. At the same time, Billie also pushes Gertrude with a force of 5𝑖 − 2𝑗. If Gertrude’s friend is sick of other students pushing Gertrude around, what force with the friend need to exert so that Gertrude didn’t move?

Set; 𝑭𝒂 = √3𝑖 + 𝑗

and 𝑭𝒃 = √3𝑖 − 3𝑗

Because of the state of equilibrium, we can say;

𝐹= + 𝐹> + 𝑅 = 0 Where, R is the force needed to keep Gertrude

stationary By sub,

√3𝑖 + 𝑗 + √3𝑖 − 3𝑗 + 𝑅 = 0

𝑅 = −2√3𝑖 + 2𝑗 Use;

|𝑎| = ,𝑥. + 𝑦.

|𝑅| = @−2√3.+ 2. = 4

𝜃 = tanEF

𝑦𝑥

Here,

𝜃 = tanEF2

−2√3

= tanEF−1√3

= 150

The friend pushes with a force of 4N at an

angle of 150 degrees to the x axis. ** could you tell this was a question I made?

Small numbers, came out exactly! No need for calculator. Multiple things happening in the one question.

It should have ended with a conversion to a bearing too L

I wish I knew why these words indent like this when it goes to a new line? Its’ annoying!

2



3 A body of mass 25 kg rests on a horizontal table. The coefficient of friction between the table and the object is 0.4. Determine the magnitude of the least horizontal force that must be applied to the object in order for it to move.

Draw a Diagram! Use,

𝑭 = 𝑚𝒂 and 𝑭J = 𝝁𝑵 Weight of the object becomes;

𝒘 = 25𝑔 Because of the state of Equilibrium we can say;

𝑵 = 25𝑔 Now, by sub;

𝐹J = 0.4 × 25𝑔

∴ 𝐹J = 10𝑔𝑁𝑒𝑤𝑡𝑜𝑛𝑠

3

4 A 20 kg object is on a smooth surface inclined at 30° to the horizontal. It is stopped from sliding by a rope pulling parallel to the surface. Determine the Tension on the rope.

Draw a diagram! Use,

𝑭 = 𝑚𝒂 Clearly,

𝒘 = 20𝑔 Because of the state of Equilibrium we can say the tension on the rope is equal to the component of the weight vector, acting parallel to the slope. As such;

𝑇 = 20𝑔 sin 30

= 10𝑔 So, the Tension on the rope is 10g Newtons

6



5 A student brought their new cute puppy (weighing 4 kg) in for show and tell. The maths teacher put the puppy on a table and started to lift one side. It wasn’t until the table reached an angle of 60 degrees that the puppy was on the ‘verge’ of sliding off. Determine the coefficient of friction between the puppy and the table.

Draw a Diagram! Use,

𝑭 = 𝑚𝒂 Weight of the object becomes … 𝒘 = 4𝑔 Break up this weight vector into components parallel and perpendicular to the face of the table;

𝑵 = 4𝑔 cos 60 = 2𝑔 and

𝑭∥ = 4𝑔 sin 60 = 2𝑔√3 Because of the state of equilibrium, the force of Friction must equal the force parallel to the table; hence

𝑭J = 𝑭∥ Using,

𝑭J = 𝝁𝑵 By sub;

𝝁 × 2𝑔 = 2𝑔√3

∴ 𝝁 = √3 ** Did you spot this as one of my questions? Do you see those justification statements in the working? Nice hey!

2



6 The velocities of two planes are .

Calculate: (a) the velocity of plane 1 relative to plane 2 (b) the velocity of plane 2 relative to plane 1.

I’ll leave the text book solutions here, but I think they are “Ugly” J ?

Here are My solutions: (a) Think … we need:

𝒗`F/`. = 𝒗`F/b + 𝒗b/`. Velocity of Plane 1 over the ground

𝒗`F/b = ⟨200, −40⟩ Velocity of the ground rel to Plane 2

𝒗b/`. = ⟨−230, −60⟩ Now;

𝒗`F/`. = ⟨200, −40⟩ + ⟨−230, −60⟩ and

𝒗`F/`. = ⟨−30, −100⟩ So, the same as the text book solution, but using the standard relative velocity formula! (b)

𝒗`./`F = 𝒗`./b + 𝒗b/`F

𝒗`./b = ⟨230, 60⟩

𝒗b/`F = ⟨−200, 40⟩ Now;

𝒗`./`F = ⟨230, 60⟩ + ⟨−200, 40⟩ and

𝒗`./`F = ⟨30, 100⟩

2 60 230 and 40 200

~~2~~~1~ jivjiv +=-=

~~

2 rel 11 rel 2

~~

~~~~

212 rel 1

10030

(b)

10030

)60230()40200(

(a)

ji

vv

ji

jiji

vvv

~~

~~~

+=

-=

--=

+--=

-=



7 A ferry is heading north-east at 20 km/h. At the same time a nearby launch is travelling south at 15 km/h. Calculate the velocity of the ferry as seen by someone on the launch.

Using the cosine rule:

Using the sine rule:

48.3° - 45° = 3.3°

*** Did you get that question correct? Do you agree with this solution? If you did or if you didn’t, then read the next box!

4

I know you hate when I talk about the Rubric, but lets check the “Reasonableness of that solution”. The Ferry is heading North East at 20 km/h, so its “northerly” component of speed is 20 cos 45 = .f

√.=

14.14 km/h. If the launch is travelling South at 15 km/h and the Ferry has a northerly speed of 14.14 km/h …how on earth is the relative velocity between the 2 just 14.2 km/h. Its not reasonable at all! I reckon their messy use of the cosine rule has muddled them up. The Cosine Rule is for Maths B. Don’t even think about using the Sine or Cosine rule in your Maths C Exams! I’ll do the question again, using the standard relative velocity formula and see how I go … J

LFL rel F ~~~ vvv -=

km/h 2.14

45cos152021520 22

L rel F

=

´´-+= !

~v

!

!

!

!

3.48

2.1445sin15sin

2.1445sin15sin

2.1445sin

15sin

1

=

÷÷ø

öççè

æ ´=

´=

=

-q

q

q

E of S 3.3at km/h 2.14 ThereforeL rel F

!=~v



8 A ferry is heading north-east at 20 km/h. At the same time a nearby launch is travelling south at 15 km/h. Calculate the velocity of the ferry as seen by someone on the launch.

𝒗J/g = 𝒗J/h + 𝒗h/g

𝒗J/h = ⟨20 cos 45, 20 sin 45⟩

𝒗h/g = ⟨0, 15⟩

𝒗J/g = ⟨20 cos 45, 20 sin 45⟩ + ⟨0, 15⟩

𝒗J/g = ⟨14.14, 29.14⟩

|𝒗| = √14.14. + 29.14. = 32.39 km/h

𝜃 = tanEF32.3914.14 = 66.42j

convert to a bearing, that’s 23.58j𝑇.

Solution done in half the work, and it is actually correct … J Again check it is reasonable … if you can visualise yourself on the launch heading south at 15km/h, surely the ferry heading north east is heading away from you pretty quickly, and at a more acute angle that its own heading of 045j𝑇.

9 A boat can travel at 16 km/h in still water. The driver of the boat wants to cross a river at right angles to the bank. If the river is flowing at 4 km/h, at what angle should the driver head in order to travel in the desired direction?

The driver should head upstream at an angle of about 76° to the shore. *** I’m out of time … surely this is not what you would give me in the exam … ? Make sure you know how to set this one out Properly … if you are not sure, come see me!

2

!14164sin

»

=

q

q



10 Given D is the midpoint of AC and E is the midpoint of CB, Prove that DE and AB are parallel.

Clearly, 𝐴𝐵3333 = 𝐴𝐷3333 + 𝐷𝐶3333 + 𝐶𝐸3333 + 𝐸𝐵3333𝑒𝑞. 1

As D bisects AC →∴ 𝐴𝐷3333 = 𝐷𝐶3333 As E bisects CB →∴ 𝐶𝐸3333 = 𝐸𝐵3333 Sub into eq.1

𝐴𝐵3333 = 𝐷𝐶3333 + 𝐷𝐶3333 + 𝐶𝐸3333 + 𝐶𝐸3333

𝐴𝐵3333 = 2𝐷𝐶3333 + 2𝐶𝐸3333

𝐴𝐵3333 = 2(𝐷𝐶3333 +𝐶𝐸3333)𝑒𝑞. 2 Now further, clearly;

𝐷𝐸3333 = 𝐷𝐶3333 +𝐶𝐸3333𝑒𝑞. 3 By sub into eq.2,

𝐴𝐵3333 = 2𝐷𝐸3333 Because BC and DE are scalars;

𝐴𝐵3333 ∥ 𝐷𝐸3333

𝑄𝐸𝐷

11 Point D lies on the line segment AB and divides it in the ratio 2 : 1. Find an expression for the position vector of the point D, in terms of the position vectors of the points A and B.

3

~~

~~~

~~

~~

32

31

)(32

AB32

AB

ba

aba

ad

ab

+=

-+=

+=

-=

®

®



12 On the triangle ABC, point D lies on AB and divides it in the ratio 2 : 1. Point E lies on the side AC and divides it also in the ratio 2 : 1. Show that ED is parallel to BC.

From the previous question:

4

BC. toparallel is ED

BC32

)(32

31

32

31

32

DE

31

32

31

32

~~

~~~~

~~

~~~

~~~

\

=

-=

÷øö

çèæ +-÷

øö

çèæ +=

-=\

+=

+=

®

®

bc

abac

de

ace

abd



13 Use vector methods to prove that the diagonals of a parallelogram bisect each other.

Set: 𝐴𝐶3333 and 𝐵𝐷3333 are straight diagonals of parallelogram ABCD,

Hence,

𝐴𝐸3333 ∥ 𝐸𝐶3333 Therefore they are scalars of each other,

𝐴𝐸3333 = 𝑘𝐸𝐶3333 Further,

𝐵𝐸3333 ∥ 𝐸𝐷3333 Therefore they are scalare of each other,

𝐵𝐸3333 = 𝑚𝐸𝐷3333 Clearly,

𝐴𝐶3333 = 𝐴𝐸3333 + 𝐸𝐶3333 Now,

𝐴𝐷3333 = 𝐴𝐸3333 + 𝐸𝐷3333

𝐵𝐶3333 = 𝐵𝐸3333 + 𝐸𝐶3333 As,

𝐴𝐷3333 = 𝐵𝐶3333 We can say through substitution,

𝐴𝐸3333 + 𝐸𝐷3333 = 𝐵𝐸3333 + 𝐸𝐶3333 Rearrange,

𝐴𝐸3333 − 𝐸𝐶3333 + 𝐸𝐷3333 − 𝐵𝐸3333 = 0 Via substitution (scalars)

𝑘𝐸𝐶3333 − 𝐸𝐶3333 + 𝐸𝐷3333 − 𝑚𝐸𝐷3333 = 0

Factorise

𝐸𝐶3333(𝑘 − 1) + 𝐸𝐷3333(1 − 𝑚) = 0

Here, either BOTH 𝐸𝐶3333𝑎𝑛𝑑𝐸𝐷3333 are Null vectors (which is a trivial solution), or BOTH

(𝑘 − 1)𝑎𝑛𝑑(1 − 𝑚) = 0

To make this happen, 𝑘 = 1𝑎𝑛𝑑𝑚 = 1 Therefore,

𝐴𝐸3333 = 1𝐸𝐶3333 and

𝐵𝐸3333 = 1𝐸𝐷3333 Hence, point E is the midpoint of both diagonals.

𝑄𝐸𝐷



14 Vector Proofs can be hard an there can be different methods to prove: Use vector methods to prove that the diagonals of a parallelogram bisect each other.

Setting 𝑋 as the intersection of the diagnoals;

We are not specifying that 𝑋 is the midpoint, it is simply a certain ‘scalar” (use 𝑝&𝑞 as the scalars) along vector 𝑑w and 𝑐y. Hence;

𝑋 = 𝑎y + 𝑝𝑑w and

𝑋 = 𝑏w + 𝑎y − 𝑞𝑐y Clearly,

𝑎y + 𝑝𝑑w = 𝑏w + 𝑎y − 𝑞𝑐y and

𝑝𝑑w = 𝑏w − 𝑞𝑐y

Now as,

𝑑w = 𝑎y −𝑏w and

𝑐y = 𝑎y +𝑏w Through substitution,

𝑝{𝑎y −𝑏w | = 𝑏w − 𝑞{𝑎y +𝑏w |

𝑝𝑎y − 𝑝𝑏w = 𝑏w − 𝑞𝑎y − 𝑞𝑏w

𝑏w = 𝑝𝑎y − 𝑝𝑏w + 𝑞𝑎y + 𝑞𝑏w

𝑏w = (𝑝 + 𝑞)𝑎y + (𝑞 − 𝑝)𝑏w

Clearly vector 𝑎y and vector 𝑏w are in different directions, hence the coefficient of 𝑎y is 0; and as𝑏w = 𝑏w , the coefficient of 𝑏w is 1

𝑝 + 𝑞 = 0 → 𝑒𝑞. 1

𝑝 − 𝑞 = 1 → 𝑒𝑞. 2 Hence, solving simultaneously, we have,

𝑝 = F. and 𝑞 = − F

.

Hence 𝑋 is the midpoint of both diagonals QED !

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WorkSHEET 8.2 Vector applications Name:

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