# math vectors help

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I was doing pure math for O levels and came across this vector question while revising. it looks very simple but I realised they have given relationship in terms of the magnitudes and not vectors so my question is could anyone explain to me how part A and B can be done atleast. I need the answer ASAP

https://imgur.com/a/Z4iLQcC

that's the question

please anyone help will be appreciated

https://imgur.com/a/Z4iLQcC

that's the question

please anyone help will be appreciated

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#2

(Original post by

I was doing pure math for O levels and came across this vector question while revising. it looks very simple but I realised they have given relationship in terms of the magnitudes and not vectors so my question is could anyone explain to me how part A and B can be done atleast. I need the answer ASAP

https://imgur.com/a/Z4iLQcC

that's the question

please anyone help will be appreciated

**brainmaster**)I was doing pure math for O levels and came across this vector question while revising. it looks very simple but I realised they have given relationship in terms of the magnitudes and not vectors so my question is could anyone explain to me how part A and B can be done atleast. I need the answer ASAP

https://imgur.com/a/Z4iLQcC

that's the question

please anyone help will be appreciated

For part A since AB is double OA then to get for A to B in terms of a and e you have to do move up (from point A) by two a (so +2a) and then down two e (so +2e) to get to point B. So overall your answer is simply AB=2(a+e)

Hope this helped!

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(Original post by

Happy to help!

For part A since AB is double OA then to get for A to B in terms of a and e you have to do move up (from point A) by one a (so a) and then down one e (so e) to get to point B. So overall your answer is simply AB= a e

As for part B, its simply BE= -2a, why? Because to get from B to A you have to do the opposite of part A -a-e, when you are at point A you can get to e by doing -a e. Overall you get -a-a e-e or -2a. So B= -2a

Hope this helped!

**sheikh_youssef**)Happy to help!

For part A since AB is double OA then to get for A to B in terms of a and e you have to do move up (from point A) by one a (so a) and then down one e (so e) to get to point B. So overall your answer is simply AB= a e

As for part B, its simply BE= -2a, why? Because to get from B to A you have to do the opposite of part A -a-e, when you are at point A you can get to e by doing -a e. Overall you get -a-a e-e or -2a. So B= -2a

Hope this helped!

this is the marking scheme;

AB = 2OA --> OC=3OA

vector OC = 3 (a e) --> vector AB= 2(a e) that's for part A..

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#4

(Original post by

Happy to help!

For part A since AB is double OA then to get for A to B in terms of a and e you have to do move up (from point A) by one a (so +a) and then down one e (so +e) to get to point B. So overall your answer is simply AB= a+e

**sheikh_youssef**)Happy to help!

For part A since AB is double OA then to get for A to B in terms of a and e you have to do move up (from point A) by one a (so +a) and then down one e (so +e) to get to point B. So overall your answer is simply AB= a+e

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#5

(Original post by

I did what you did but the marking scheme says something different;

this is the marking scheme;

AB = 2OA --> OC=3OA

vector OC = 3 (a e) --> vector AB= 2(a e) that's for part A..

**brainmaster**)I did what you did but the marking scheme says something different;

this is the marking scheme;

AB = 2OA --> OC=3OA

vector OC = 3 (a e) --> vector AB= 2(a e) that's for part A..

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(Original post by

a+e is a vector in the direction of AB, but it's length is the same as that of OA or OE. If you think of an eqilateral triangle....

**ghostwalker**)a+e is a vector in the direction of AB, but it's length is the same as that of OA or OE. If you think of an eqilateral triangle....

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(Original post by

Wouldn't OC be the same as 2a+2e?

**sheikh_youssef**)Wouldn't OC be the same as 2a+2e?

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#8

**brainmaster**)

I was doing pure math for O levels and came across this vector question while revising. it looks very simple but I realised they have given relationship in terms of the magnitudes and not vectors so my question is could anyone explain to me how part A and B can be done atleast. I need the answer ASAP

https://imgur.com/a/Z4iLQcC

that's the question

please anyone help will be appreciated

**a**and the vector

**e**have opposite vertical components, so if we add them, the vertical components will cancel out and we will get just a horizontal vector. Now, what is the length of this horizontal vector? Well, angle EOA is 120 degrees, so COA is 60 degrees by symmetry. Drop a perpendicular from A that meets OC at D, and by trigonometry in triangle DOA, we have OD = OA cos 60 = OA/2. Thus the horizontal component of vector

**a**has length OA/2, and by symmetry so does the horizontal component of vector

**e**, so when we add them,

**a**+

**e**has a horizontal component with.length OA/2 + OA/2 = OA. Thus, as we already showed that

**a**+

**e**has no vertical component, we can say that it is a horizontal vector with length OA, and

**AB**is a horizontal vector with length AB = 2*OA (from the question), so we deduce that

**AB**= 2*(

**a**+

**e**).

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#9

(Original post by

could you please explain fully ?

**brainmaster**)could you please explain fully ?

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#13

(Original post by

hey I'm really sorry but after that explanation I did the others well but the last part I'm not getting.....plz help me do the whole question

**brainmaster**)hey I'm really sorry but after that explanation I did the others well but the last part I'm not getting.....plz help me do the whole question

What have you done so far? Any thoughts? Please don't say "nothing".

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(Original post by

I presume this is part c), since that's as far as your image goes, and that's what you mean by the "whole question"

What have you done so far? Any thoughts? Please don't say "nothing".

**ghostwalker**)I presume this is part c), since that's as far as your image goes, and that's what you mean by the "whole question"

What have you done so far? Any thoughts? Please don't say "nothing".

I said KPC = PQ and hED = EQ

like I wrote OQ = OA + 2/5AB + kPC

and then OQ = OE + hED

and then equated them and was using the rule of non parallel coefficients are equal however the ED is supposed to be 2 (a+e) since it's parallel to AB so I said that 2h (a+e) = EQ but the marking scheme says that ED = (a+e)....how?

here is the full question

https://imgur.com/a/Z4iLQcC

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#15

**brainmaster**)

hey I'm really sorry but after that explanation I did the others well but the last part I'm not getting.....plz help me do the whole question

**e**) by symmetry.

For part (d), since P, C, and Q are collinear, we know vector PQ = t * vector QC for some constant t. Thus we can get an expression for vector OQ as vector OP + vector PQ (where vector OP = vector OA + 2/5 * vector AB), and we can also get another expression for vector OQ as vector OE + vector EQ, but as Q lies on ED, we can write vector EQ as k * vector ED for some constant k. Now you can equate the two expressions for vector OQ, and then since vectors

**a**and

**e**are linearly independent (i.e. non-parallel), you can equate their coefficients to solve for k and t.

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(Original post by

For part (c), you have vector PC = vector PB + vector BC = 3/5*vector AB + vector BC; you know vector AB from part (a), and vector BC is the same as vector OE (i.e.

For part (d), since P, C, and Q are collinear, we know vector PQ = t * vector QC for some constant t. Thus we can get an expression for vector OQ as vector OP + vector PQ (where vector OP = vector OA + 2/5 * vector AB), and we can also get another expression for vector OQ as vector OE + vector EQ, but as Q lies on ED, we can write vector EQ as k * vector ED for some constant k. Now you can equate the two expressions for vector OQ, and then since vectors

**Prasiortle**)For part (c), you have vector PC = vector PB + vector BC = 3/5*vector AB + vector BC; you know vector AB from part (a), and vector BC is the same as vector OE (i.e.

**e**) by symmetry.For part (d), since P, C, and Q are collinear, we know vector PQ = t * vector QC for some constant t. Thus we can get an expression for vector OQ as vector OP + vector PQ (where vector OP = vector OA + 2/5 * vector AB), and we can also get another expression for vector OQ as vector OE + vector EQ, but as Q lies on ED, we can write vector EQ as k * vector ED for some constant k. Now you can equate the two expressions for vector OQ, and then since vectors

**a**and**e**are linearly independent (i.e. non-parallel), you can equate their coefficients to solve for k and t.
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#17

(Original post by

Yes did that however is vector ED = 2 (a+e) or a+e?

**brainmaster**)Yes did that however is vector ED = 2 (a+e) or a+e?

**a**+

**e**) from part (a).

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Vector OQ = vector OE +vector EQ

Vector OQ= e+p (a+e).....that's where I'm confused

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#19

My method was to resolve everything in the i and j directions(horizontal and vertical) e.g

**a**=|a|cos(60)**i**+|a|sin(60)**j**.
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(Original post by

My method was to resolve everything in the i and j directions(horizontal and vertical) e.g

**Dalek1099**)My method was to resolve everything in the i and j directions(horizontal and vertical) e.g

**a**=|a|cos(60)**i**+|a|sin(60)**j**.
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