# math vectors help

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#1
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
0
3 years ago
#2
(Original post by 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
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 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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#3
(Original post by 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!
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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3 years ago
#4
(Original post by 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
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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3 years ago
#5
(Original post by 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..
Wouldn't OC be the same as 2a+2e?
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#6
(Original post by 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....
could you please explain fully ?
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#7
(Original post by sheikh_youssef)
Wouldn't OC be the same as 2a+2e?
idk I really don't understand this question
0
3 years ago
#8
(Original post by 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
By symmetry (about the line OC), the vector 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).
0
3 years ago
#9
(Original post by brainmaster)
could you please explain fully ?
See attached. Look at the eqilateral triangle OAZ.

1
3 years ago
#10
(Original post by ghostwalker)
See attached. Look at the eqilateral triangle OAZ.

Ahhh I see... I'll edit my previous one so I don't mislead anyone
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#11
thank you all of you...I now understand it 1
#12
(Original post by ghostwalker)
See attached. Look at the eqilateral triangle OAZ.

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
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3 years ago
#13
(Original post by 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
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".
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#14
(Original post by 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've done part a b and c but the last part I've tried and I've done what the have done too
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
0
3 years ago
#15
(Original post by 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
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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#16
(Original post by 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.
Yes did that however is vector ED = 2 (a+e) or a+e?
0
3 years ago
#17
(Original post by brainmaster)
Yes did that however is vector ED = 2 (a+e) or a+e?
Vector ED = Vector AB (by symmetry) = 2(a + e) from part (a).
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#18
(Original post by Prasiortle)
Vector ED = Vector AB (by symmetry) = 2(a + e) from part (a).
the marking scheme says;
Vector OQ = vector OE +vector EQ
Vector OQ= e+p (a+e).....that's where I'm confused
0
3 years ago
#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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#20
(Original post by 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 .
could you explain this. ...I remember seeing this in C4 but not sure
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