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Do not understand the answer (M1 vectors).

Howdy :smile:.

I do not understand how to do question 7 (c) (ii). The parallel to (-i - 3j) question.

I have attached the question and answer.

Could someone explain it please? (I understand the other questions).

ThanksScreen Shot 2017-04-29 at 08.43.34.png
The reasoning is that anything moving parallel to the vector (-i -3j) will have i and j components in the same proportion, i.e. (i component) : ( j component) = (-1) : (-3). The reason for the plus-or-minus sign in the mark scheme is to cover the case of P moving parallel to (-i -3j) but in the opposite direction. (This still counts as parallel).
Reply 2
Original post by makin
Howdy :smile:.

I do not understand how to do question 7 (c) (ii). The parallel to (-i - 3j) question.

I have attached the question and answer.

Could someone explain it please? (I understand the other questions).

ThanksScreen Shot 2017-04-29 at 08.43.34.png



If you want to be able to answer these style of questions, you can just learn this formulae. It would be easier to just learn this as you will learn more about why it works in C4 vectors

[ x(t) / y(t) ] = [xi / yj ]
divide the x component by the y component (function of time)
this is equal to the x divided by y of the vector its parallel.
Original post by makin
Howdy :smile:.

I do not understand how to do question 7 (c) (ii). The parallel to (-i - 3j) question.

I have attached the question and answer.

Could someone explain it please? (I understand the other questions).

Thanks


To do this from first principles, note that if the velocity is parallel to i3j-\bold{i}-3\bold{j}, then we must have:

v=k(i3j)=ki3kj\bold{v} = k(-\bold{i}-3\bold{j}) = -k\bold{i}-3k\bold{j} for some real number kk

so that:

(12t)i+(3t3)j=ki3kj12t=k,3t3=3k(1-2t)\bold{i} +(3t-3)\bold{j} = -k\bold{i}-3k\bold{j} \Rightarrow 1-2t=-k, 3t-3=-3k

by equating components. You can then get the quoted result by dividing the equations.
So what this is saying is that you have a vector v which is parallel to the vector (-i-3j) shown in the linked diagram.

The first thing you need to do is find out what direction this vector is in. You have two options, either find the value of theta, or simply use the gradient. For this question i would recommend using the gradient for simplicity.

So the gradient from (0,0) to (-1,-3) is -3/-1 = 3

This tells us that because our vector v is parallel to this, it will also have a gradient of 3. Now look at this second diagram where vector v is also shown. All you need to do now is find the gradient of vector v and equate this to 3, solving algebraically for the unknown t.

so the gradient of v = (3t - 3)/(1 - 2t) = 3

3t -3 = 3(1 - 2t)
3t - 3 = 3 -6t
9t = 6
t=6/9 = 2/3 = 0.67

Hope that helped, have fun on your mathematical journey
Reply 5
Original post by TheEngineer12345
So what this is saying is that you have a vector v which is parallel to the vector (-i-3j) shown in the linked diagram.

The first thing you need to do is find out what direction this vector is in. You have two options, either find the value of theta, or simply use the gradient. For this question i would recommend using the gradient for simplicity.

So the gradient from (0,0) to (-1,-3) is -3/-1 = 3

This tells us that because our vector v is parallel to this, it will also have a gradient of 3. Now look at this second diagram where vector v is also shown. All you need to do now is find the gradient of vector v and equate this to 3, solving algebraically for the unknown t.

so the gradient of v = (3t - 3)/(1 - 2t) = 3

3t -3 = 3(1 - 2t)
3t - 3 = 3 -6t
9t = 6
t=6/9 = 2/3 = 0.67

Hope that helped, have fun on your mathematical journey


Thank you for taking the time to give me such a detailed explanation. I fully understand it now and can answer similar questions :biggrin:. Have a great day.

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