Year 12s: what will be the first way you research your future options? >>

AEA June 2011

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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

I can't remember the question as I spent ages worrying about Q6... I just remember my answer :P

Reply 141

Ultimate1

Loooool who did the 1 mark vector Q on finding A lies on the line and didn't do the rest? :tongue:

Reply 142

superkinetic

Original post by Agent Fiasco

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

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

Yeah you're to use dot product, to prove

\vec{AP} \cdot \vec{AB} = 0

, as cos90=0. But then you got to find B first which depends on the P' from (a). That is the only method I can think of.

Reply 143

implus

What were people's P' Vector? Mine was [text]\begin{pmatrix}
-2\\
13\\
5\end{pmatrix}

Reply 144

sb2star

Original post by Ultimate1

Loooool who did the 1 mark vector Q on finding A lies on the line and didn't do the rest? :tongue:

haha I did that! :biggrin:

that made my happy :P

Reply 145

Ultimate1

Original post by implus

What were people's P' Vector? Mine was [text]\begin{pmatrix}
-2\\
13\\
5\end{pmatrix}

Did you find t by getting the perpundicular distance from P to the line right? What did you get T as??

Reply 146

superkinetic

Original post by implus

What were people's P' Vector? Mine was [text]\begin{pmatrix}
-2\\
13\\
5\end{pmatrix}

I didn't get this. If you manage to prove (c) correctly with this value of P', then your P' is correct.

Reply 147

Agent Fiasco

Original post by superkinetic

Yeah you're to use dot product, to prove

\vec{AP} \cdot \vec{AB} = 0

, as cos90=0. But then you got to find B first which depends on the P' from (a). That is the only method I can think of.

I managed to work out B as B was on line l right? so i made the point b as eqaution with T and then solved it for

\vec{AP} \cdot \vec{AB} = 0

and worked out B would that be correct?

Reply 148

implus

Original post by Ultimate1

Did you find t by getting the perpundicular distance from P to the line right? What did you get T as??

I think it was

\frac{7}{2}

X

vector for which lied on the line L and such at PX was perpendicular to the line L was

\begin{pmatrix}[br]\frac{-9}{2}\\ [br]\frac{15}{2}\\ [br]6\end{pmatrix}

Reply 149

Ultimate1

Original post by implus

I think it was

\frac{7}{2}

X

vector for which lied on the line L and such at PX was perpendicular to the line L was

\begin{pmatrix}[br]\frac{-9}{2}\\ [br]\frac{15}{2}\\ [br]6\end{pmatrix}

WTF I got 158/69. Did you do like p.x =0 right? Arghhh I was on the right track but was getting weird number for t

Reply 150

superkinetic

Original post by Agent Fiasco

I managed to work out B as B was on line l right? so i made the point b as eqaution with T and then solved it for

\vec{AP} \cdot \vec{AB} = 0

and worked out B would that be correct?

But you are to find position vector B before concluding that

\vec{AP} \cdot \vec{AB} = 0

, not using

\vec{AP} \cdot \vec{AB} = 0

to work out position vector B. I dunno if the examiners will allow you to work out position vector B by using facts from subsequent parts of the qn.

Original post by Ultimate1

WTF I got 158/69. Did you do like p.x =0 right? Arghhh I was on the right track but was getting weird number for t

I think it's pure carelessness. I got 50t=175 which leads to t=7/2. But I think I am careless enough not to calculate P' properly.

Reply 151

Agent Fiasco

but didnt they tell us that APB was 90 degrees or did they ask to prove i fink i read the question wrong :frown:

Reply 152

superkinetic

Original post by Agent Fiasco

but didnt they tell us that APB was 90 degrees or did they ask to prove i fink i read the question wrong :frown:

If I didn't recall wrongly, the qn ask you to find position vector B before proving the angle is 90 degrees.

Reply 153

cazzy-joe

Original post by Ultimate1

Loooool who did the 1 mark vector Q on finding A lies on the line and didn't do the rest? :tongue:

Yayyy yayy me too :biggrin:

I'm so happy someone is accompanying me :biggrin:

Reply 154

Ultimate1

Weren't the equation to find t something like -7(-13-5t) + 2(3-3t) + 7(something) = 0

Reply 155

Raimu

So what are the chances of any of you getting the grade that you want. I think it's probably 0 for me, which sucks.

Reply 156

Agent Fiasco

Original post by Raimu

So what are the chances of any of you getting the grade that you want. I think it's probably 0 for me, which sucks.

I think I didn't do well enough either to get the merit I need. The paper was very hard, when I did the past papers I usually got around 53 marks but this one I will be lucky to get 40 :mad:

Reply 157

Quadratic

Just scanned through the posts but didn't see it: did anyone do Q1? I was convinced that was a typo. Oh and I'll be lucky if I got a merit with what I answered >.<

Also are you guys still waiting on the paper? I can scan one in at 6:00 tonight if no ones done it before then.

Apologies if either of the above have already been posted; just scanned through the recent posts quickly on my way out the door.

Reply 158

sb2star

Original post by Quadratic

This is what implus posted...

Original post by implus

This is how I did question 1:

$let \ \theta +35^{\circ} = A \\ and \ let \ \theta -53^{\circ} = B \\tanA=CotB \\ \frac{SinA}{CosA}=\frac{CosB}{SinB} \\ SinASinB=CosACosB \\ CosACosB-SinASinB=0 \\ Cos(A+B)=0 \\ Cos(2\theta -18^{\circ})=0 \\ 2\theta -18^{\circ}= 90^{\circ}, 270^{\circ} \\ \theta = 54^{\circ},144^{\circ}$