VCE Specialist 3/4 Question Thread!

[Greatness]

Is question 1 asking for a vector resolute or something similar to that? lol i cant remember XD

[Mr. Study]

Is question 1 asking for a vector resolute or something similar to that? lol i cant remember XD

:O, It's asking for a 'new' vector, vector C that is the vector component of vector a perpendicular to vector b.

:S, This is getting confusing. xD

[b^3]

Regarding the first question.

It does involve using vector resolutes.

So what you are being asked to do is find "the vector component of a perpendicular to b". I.e. find the perpendicular resolute. That is the dotted line in the diagram (ignore the |A|cos(theta), not needed).

Lets make where that right angle is point C.
I.e. we are looking for OC.

The vector resolute of a in the direction of b is

Unlatex'd. OC=[(a.b)/(b.b)]*b

That is representing the vector that is in the same direction as b, but has the lenght from the bottom right most point to where the right angle is, i.e. the projection.

Now we want the vector that is perpendicular, not parallel to b.

So that would be the dotted line, now the dotted line will be 

Unlatex'd. CA=OA-OC=a-[(a.b)/(b.b)]*b

Hope that helps, probably didn't explain it well.

[Greatness]

yeah i think that is the vector resolute. read the theory on it in the textbook or look at some examples.

[Mr. Study]

Regarding the first question.

It does involve using vector resolutes.
(Image removed from quote.)
So what you are being asked to do is find "the vector component of a perpendicular to b". I.e. find the perpendicular resolute. That is the dotted line in the diagram (ignore the |A|cos(theta), not needed).

Lets make where that right angle is point C.
I.e. we are looking for OC.

The vector resolute of a in the direction of b is

Unlatex'd. OC=[(a.b)/(b.b)]*b

That is representing the vector that is in the same direction as b, but has the lenght from the bottom right most point to where the right angle is, i.e. the projection.

Now we want the vector that is perpendicular, not parallel to b.

So that would be the dotted line, now the dotted line will be 

Unlatex'd. CA=OA-OC=a-[(a.b)/(b.b)]*b

Hope that helps, probably didn't explain it well.

I see, Ah! The other text book I use ALWAYS says 'Find the vector resolute of a perpendicular to b' but this one 'jazzed' it up.

Thanks swarley and b^3.

EDIT: Just this really quick question

What is the quickest way to 'split' up this: 7pi/12? I can easily to the really simple stuff like pi/12 or 2pi/3. Hopefully that isn't too much of a basic question but thank you.

[soccerboi]

Hi, could somebody help me with this question, describe the steps i should take or even show me the steps if possible

Solve this equation over C: z²=8+15i

Thanks!

[Mr. Study]

Move to one side and use the quadratic formula. ^

I cant do it right now as I'm at the state library and I have less than 1 minute remaining, sorry.

Sorry, thought there was a Z for 8. :S My bad.

[John President]

Hi, could somebody help me with this question, describe the steps i should take or even show me the steps if possible

Solve this equation over C: z²=8+15i

Thanks!

Given that 15^2 + 8^2 = 289 = 17^2, I suspect that they want you to convert 8+15i into polar form (modulus-argument form) before using De Moivre's theorem to find the roots. It looks like you'll need a calculator though - and don't forget that there will be two solutions!

[soccerboi]

its suppose to be calc free so i dono how to get the argument for it by hand

[brightsky]

z^2 = 8 + 15i
(x+yi)^2 = 8 + 15i
x^2 - y^2 + 2xyi = 8 + 15 i
x^2 - y^2 = 8...(1)
2xy = 15 ...(2)
from (2), y = 15/(2x)
sub into (1)
x^2 - (15/(2x))^2 = 8
x^2 - 225/4x^2 = 8
4x^4 - 32x^2 - 225 =0
(2x^2 - 25) (2x^2 + 9)=0
2x^2 = 25/2 or 2x^2 + 9 = 0 (no sol for this one since x E R)
x^2 = 25/4
x = +-5/2
so y = 15/(+-5) = +-3

[Bhootnike]

spec prac exam question attached

question is, how do you do it!?
i got B, but realised i actually didn't get any of the options listed! -        A/[3(x+c)]  +  B/(x+c)^2


the answer is D. what do you do to get rid of the 3?

[xZero]

 a/[3(x+c)]  +  B/(x+c)^2, let A=a/3 => A/(x+c) + B/(x+c)^2

[Planck's constant]

the answer is D. what do you do to get rid of the 3?

You dont.
Its absorbed in A and B

[#1procrastinator]

For the complex numbers u = a+bi, v=c+di and w=p+qi prove that:

a) u+v=v+u
b) (u+v)+w = u(v+w)
c) uv=vu
d) (uv)w=u(vw)

Here's my attempt at a) and c)

1) (a+bi)(c+di)
2) (ac-bd)+(ad+bc)i
3) ??

c)
1) (a+bi) + (c+di)
2) (a+c) + (b+d)i by definition of addition
3) ??

hehe

[Mr. Study]

Just this vector q's at the bottom.

I got =6m-2m.

=-
=-(6m-2m()
Not too sure about this next line:
=-6m)+(+2m)
=8-6m+(-4+2m)

Dot Product of .
=(6m-2m).((8-6m)+(-4+2m)
=(6m(8-6m)+(-2m(-3+2m)))
=(48m-36m²+(6m-4m²)
=48m-36m²+6m-4m²
=54m-40m²
=2m(27-20m)

I am 100% sure my working out is wrong. Could someone look over it and explain? I usually get mixed up with Dot products when there is a variable. :S

Thanks.

(This questions from the VCAA 2002 Exam 2)