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# AEA June 2011 watch

1. (Original post by Ultimate1)
How did you do it?

For the reflection you had to find the perpundicular distance to the line first and then subtract this distance from the line to the get the reflection, bit it wasn't working out for me
I knew you had to find the perpendicular distance but I didn't know how you were supposed to??? it's not in my C3, C4 but actually in FP3... so I didn't think I need to use FP3... and I don't like FP3 vectors so didn't wanna use it anyway :P
2. I evntually found T to be something like 7/2, but even then then the position vector of the perpendicular point wasn't a nice number at all and didn't work out that well.

How many marks do you think you guys got overall? I think I may have scrapped 20
3. (Original post by Ultimate1)
Thanks. Ah man don't give up hope you may have got the D/2. What's your insurance?
my insurance is Southampton AAB I'll probably be going there :P but fingers crossed for D/2!!!
4. (Original post by Raimu)
I evntually found T to be something like 7/2, but even then then the position vector of the perpendicular point wasn't a nice number at all and didn't work out that well.

How many marks do you think you guys got overall? I think I may have scrapped 20
I had no problems with Q2,3,4 and 5... so hopefully I didn't make any mistakes in those...
5. (Original post by sb2star)
I knew you had to find the perpendicular distance but I didn't know how you were supposed to??? it's not in my C3, C4 but actually in FP3... so I didn't think I need to use FP3... and I don't like FP3 vectors so didn't wanna use it anyway :P
I worked by finding the vector starting from P that is perpendicular to the directional vector of L. After finding the vector, I added 2 of this vector to P to find P', as the distance from P to L is the same as distance from P' to L. But my t is a fraction and I guess I messed up somewhere. The worse part is the subsequent parts of Qn 6 depends on the (a), so I messed up the entire qn.

Anyway I don't think this is further math right? You just have to use the dot product.
6. This is how I did question 1:

$let \ \theta +35^{\circ} = A \\ and \ let \ \theta -53^{\circ} = B \\tanA=CotB \\ \frac{SinA}{CosA}=\frac{CosB}{SinB} \\ SinASinB=CosACosB \\ CosACosB-SinASinB=0 \\ Cos(A+B)=0 \\ Cos(2\theta -18^{\circ})=0 \\ 2\theta -18^{\circ}= 90^{\circ}, 270^{\circ} \\ \theta = 54^{\circ},144^{\circ}$
7. (Original post by implus)
This is how I did question 1:

$let \ \theta +35^{\circ} = A \\ and \ let \ \theta -53^{\circ} = B \\tanA=CotB \\ \frac{SinA}{CosA}=\frac{CosB}{SinB} \\ SinASinB=CosACosB \\ CosACosB-SinASinB=0 \\ Cos(A+B)=0 \\ Cos(2\theta -18^{\circ})=0 \\ 2\theta -18^{\circ}= 90^{\circ}, 270^{\circ} \\ \theta = 54^{\circ},144^{\circ}$
I did not think of that at all Thanks for the solution!
8. that was very hard papers couldnt do too many question. btw did anyone manage to do number o wat was the answer i managed to get 2+ln2 which i assume is wrong since the two constants were a and b

i completely messed up question 3 and the vector one. how was number 3 done???
9. (Original post by superkinetic)
I worked by finding the vector starting from P that is perpendicular to the directional vector of L. After finding the vector, I added 2 of this vector to P to find P', as the distance from P to L is the same as distance from P' to L. But my t is a fraction and I guess I messed up somewhere. The worse part is the subsequent parts of Qn 6 depends on the (a), so I messed up the entire qn.

Anyway I don't think this is further math right? You just have to use the dot product.
Same my t too was a fraction like 158/69
10. (Original post by superkinetic)
I worked by finding the vector starting from P that is perpendicular to the directional vector of L. After finding the vector, I added 2 of this vector to P to find P', as the distance from P to L is the same as distance from P' to L. But my t is a fraction and I guess I messed up somewhere. The worse part is the subsequent parts of Qn 6 depends on the (a), so I messed up the entire qn.

Anyway I don't think this is further math right? You just have to use the dot product.
how did you find the vector starting from P that is perpendicular to the direction of L? so how did you find the direction vector for that perpendicular line? well I couldn't do part (a) at all so I called P' (x y z) and then did all the methods... not sure they'll give me method marks...

in FM you use cross products to find perpendicular distance etc... so I could probably have been able to do it with it, but after the weekend I forgot my FP3 vectors... :P
11. (Original post by Agent Fiasco)
that was very hard papers couldnt do too many question. btw did anyone manage to do number o wat was the answer i managed to get 2+ln2 which i assume is wrong since the two constants were a and b

i completely messed up question 3 and the vector one. how was number 3 done???
i got 1 + ln 4 hahaha. dear god, i think i got 0 in that
12. (Original post by Agent Fiasco)
that was very hard papers couldnt do too many question. btw did anyone manage to do number o wat was the answer i managed to get 2+ln2 which i assume is wrong since the two constants were a and b

i completely messed up question 3 and the vector one. how was number 3 done???
I got 2 + ln 4 ...

13. (Original post by Raimu)
I evntually found T to be something like 7/2, but even then then the position vector of the perpendicular point wasn't a nice number at all and didn't work out that well.
I got 7/2 for t, but my answer to (a) didn't make (c) work right.

(Original post by implus)
This is how I did question 1:

$let \ \theta +35^{\circ} = A \\ and \ let \ \theta -53^{\circ} = B \\tanA=CotB \\ \frac{SinA}{CosA}=\frac{CosB}{SinB} \\ SinASinB=CosACosB \\ CosACosB-SinASinB=0 \\ Cos(A+B)=0 \\ Cos(2\theta -18^{\circ})=0 \\ 2\theta -18^{\circ}= 90^{\circ}, 270^{\circ} \\ \theta = 54^{\circ},144^{\circ}$
That is pretty brilliant. I didn't realise the sum of cosine comes into play.

(Original post by sb2star)
how did you find the vector starting from P that is perpendicular to the direction of L? so how did you find the direction vector for that perpendicular line? well I couldn't do part (a) at all so I called P' (x y z) and then did all the methods... not sure they'll give me method marks...

in FM you use cross products to find perpendicular distance etc... so I could probably have been able to do it with it, but after the weekend I forgot my FP3 vectors... :P
Let X be a point on L where is perpendicular to L. Using dot product, . No need to use cross product, since you're not finding a vector perpendicular to 2 vectors.
14. (Original post by sb2star)
I got 2 + ln 4 ...

i think u must be right cause
how did u do it??

i expanded it than intergrated them each than put the limits i must have messed up some values
15. (Original post by superkinetic)
I got 7/2 for t, but my answer to (a) didn't make (c) work right.

That is pretty brilliant. I didn't realise the sum of cosine comes into play.

Let X be a point on L where is perpendicular to L. Using dot product, . No need to use cross product, since you're not finding a vector perpendicular to 2 vectors.
Hmmmmm I was getting t =158/69....
16. (Original post by sb2star)
I got 2 + ln 4 ...

Awesome, I've got 2 + ln 4 too!!
17. (Original post by sb2star)
I got 2 + ln 4 ...

I got this as well.
18. (Original post by Agent Fiasco)
i think u must be right cause
how did u do it??

i expanded it than intergrated them each than put the limits i must have messed up some values
Did you manage to do the second parts of the vector questions? It's about those kite, and have you managed to find B?
19. (Original post by sb2star)
I got 2 + ln 4 ...

(Original post by Ultimate1)
Hmmmmm I was getting t =158/69....
I might be wrong, as mentioned earlier, my value doesn't help me in proving part c correctly. I always thought vectors qn are cumbersome, as they're bound to trip ppl.
20. (Original post by superkinetic)

Let X be a point on L where is perpendicular to L. Using dot product, . No need to use cross product, since you're not finding a vector perpendicular to 2 vectors.

thats wat i did fbut for my T parameter i had a huge value somefin like 141/50 so i gave up on it

btw u know the next few question on vectors was one of them prove that angle APB is 90????

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