VCE Specialist 3/4 Question Thread!

[TrueTears]

just equate the expression from part a) to 2 and solve for theta

[HERculina]

just equate the expression from part a) to 2 and solve for theta

yar i did but i got stuck in the last part....
how do you go from sintheta = 1/2 -----> answer: theta = pi/6 + 2kpie, 5pi/6 + 2kpi, k = intergers

[TrueTears]

just equate the expression from part a) to 2 and solve for theta

yar i did but i got stuck in the last part....
how do you go from sintheta = 1/2 -----> answer: theta = pi/6 + 2kpie, 5pi/6 + 2kpi, k = intergers

oh it's just general solutions, Re: Mathematical Methods Guides and Tips i used the exact same equation as an example lol

[HERculina]

just equate the expression from part a) to 2 and solve for theta

yar i did but i got stuck in the last part....
how do you go from sintheta = 1/2 -----> answer: theta = pi/6 + 2kpie, 5pi/6 + 2kpi, k = intergers

oh it's just general solutions, Re: Mathematical Methods Guides and Tips i used the exact same equation as an example lol

OHH kewl thanks. why haven't i ever been taught about this, is this part of the 1/2 methods book - maybe we skipped the exercise on it

[paulsterio]

Is there supposed to be a picture attached?

was that at me? LOL?! i'm lost!

[dc302]

Is there supposed to be a picture attached?

was that at me? LOL?! i'm lost!

Nah there was a post but it got deleted.

[Dominatorrr]

Express (2sqrt(3)+2i)/(1-3i) in polar form.

[abd123]

Express (2sqrt(3)+2i)/(1-3i) in polar form.

Hoped I helped .

[Dominatorrr]

hey thanks but the denominator, isn't the imaginary part -3 instead of sqrt(3) or did I miss some rule when going through the chapter? Otherwise the root3 would make the question a whole lot understandable and easier.

[abd123]

hey thanks but the denominator, isn't the imaginary part -3 instead of sqrt(3) or did I miss some rule when going through the chapter? Otherwise the root3 would make the question a whole lot understandable and easier.

Are you talking about whats inside the square root? if there was -3 in , than it would result as , because we know that as general rule that , so if we expand out .

Don't worry to much about it though, just skim through chapter on complex numbers you will understand a lot if you revisit some of the sub topics of complex numbers.

[Bhootnike]

hey thanks but the denominator, isn't the imaginary part -3 instead of sqrt(3) or did I miss some rule when going through the chapter? Otherwise the root3 would make the question a whole lot understandable and easier.

Are you talking about whats inside the square root? if there was -3 in , than it would result as , because we know that as general rule that , so if we expand out .

Don't worry to much about it though, just skim through chapter on complex numbers you will understand a lot if you revisit some of the sub topics of complex numbers.

I think what dominator was asking was,  in the denominator,  its : 1-3i?  You've used 1-root3 I

[Dominatorrr]

hey thanks but the denominator, isn't the imaginary part -3 instead of sqrt(3) or did I miss some rule when going through the chapter? Otherwise the root3 would make the question a whole lot understandable and easier.

Are you talking about whats inside the square root? if there was -3 in , than it would result as , because we know that as general rule that , so if we expand out .

Don't worry to much about it though, just skim through chapter on complex numbers you will understand a lot if you revisit some of the sub topics of complex numbers.

I think what dominator was asking was,  in the denominator,  its : 1-3i?  You've used 1-root3 I

Yeah that's what I meant

[Ravit]

need help!

Find the point P on the line x-6y=11 such that the vector OP is parallel to the vector 3i+j

[moekamo]

let P=(x,y)=(x,1/6(x-11)) from the equation of the line, then OP=xi+1/6(x-11)j

now OP=k(3i+j) since they are parallel, so xi+1/6(x-11)j=3ki+jk, then equate the i, j components, solve for x by eliminating k, then sub back into the line equation to get the y value.

you should get P=(-11, -11/3) i think...if i did it right...

[Ravit]

yes your right.
thank you very much