STEP II 2006 Q6 (Vectors) Watch

velvetcap
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#1
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Hi, I just attempted Q6 from the 2004 STEP II and got the following vectors:

p=4a
q=b-3a
(these were both correct)

P=2a (this was correct)

Q=3a-b
R=-4a+2b+c
(according to the official answer booklet these are incorrect)

I would be very grateful if someone could point out why the last two vectors I obtained are wrong (why they don't satisfy all the conditions)

Thanks!
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metaltron
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(Original post by velvetcap)
Hi, I just attempted Q6 from the 2004 STEP II and got the following vectors:

p=4a
q=b-3a
(these were both correct)

P=2a (this was correct)

Q=3a-b
R=-4a+2b+c
(according to the official answer booklet these are incorrect)

I would be very grateful if someone could point out why the last two vectors I obtained are wrong (why they don't satisfy all the conditions)

Thanks!
Hi Could you post how you got your value of Q, or at least describe what you did? It would then be easier to see where you went wrong.

If you haven't seen it already there is also a STEP thread that you can post in if you want help, you might get a few more responses.

http://www.thestudentroom.co.uk/show...03970&page=152
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brianeverit
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(Original post by velvetcap)
Hi, I just attempted Q6 from the 2004 STEP II and got the following vectors:

p=4a
q=b-3a
(these were both correct)

P=2a (this was correct)

Q=3a-b
R=-4a+2b+c
(according to the official answer booklet these are incorrect)

I would be very grateful if someone could point out why the last two vectors I obtained are wrong (why they don't satisfy all the conditions)

Thanks!
P=\alpha a and Q=\beta (b-3a) for some values of \alpha, \beta
P+ Q + R =a + b + c so traking scalar product of both sides with a and b in turn leads to a paior of simultaneous equations in \alpha\ and\ \beta
which should give \alpha =2 \ and\ \beta=\frac{3}{2} leading to the required answers.
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velvetcap
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(Original post by brianeverit)
P=\alpha a and Q=\beta (b-3a) for some values of \alpha, \beta
P+ Q + R =a + b + c so traking scalar product of both sides with a and b in turn leads to a paior of simultaneous equations in \alpha\ and\ \beta
which should give \alpha =2 \ and\ \beta=\frac{3}{2} leading to the required answers.
Thanks for your help!

Do you know why my answer is wrong by any chance? As I said it satisfies all the conditions so I don't see why it would be wrong. Maybe there are multiple accepted answers? (I don't know if STEP do this or not)
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metaltron
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(Original post by velvetcap)
Thanks for your help!

Do you know why my answer is wrong by any chance? As I said it satisfies all the conditions so I don't see why it would be wrong. Maybe there are multiple accepted answers? (I don't know if STEP do this or not)
Ok, I think you've made the mistake of thinking that b.b = 5 but it actually = 5^2 = 25. Therefore R is not perpendicular to the plane
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velvetcap
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(Original post by metaltron)
Ok, I think you've made the mistake of thinking that b.b = 5 but it actually = 5^2 = 25. Therefore R is not perpendicular to the plane
I tried taking the dot product of the R vector I obtained with 2 vectors parallel to the plane and g0t 0 in both instances
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metaltron
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(Original post by velvetcap)
I tried taking the dot product of the R vector I obtained with 2 vectors parallel to the plane and g0t 0 in both instances
Ok so choose b.

R.b = -4a.b + 2b.b + b.c = -12 + 50 + 2 = 40

Your solution is essentially right, but you've made a careless slip and said b.b = 5 not 5^2
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velvetcap
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(Original post by metaltron)
Ok so choose b.

R.b = -4a.b + 2b.b + b.c = -12 + 50 + 2 = 40

Your solution is essentially right, but you've made a careless slip and said b.b = 5 not 5^2
Ah I finally get it now, thanks for your help
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metaltron
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(Original post by velvetcap)
Ah I finally get it now, thanks for your help
Good luck tomorrow
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velvetcap
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(Original post by metaltron)
Good luck tomorrow
Haha thanks I'll be needing it
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