VCE Specialist 3/4 Question Thread!

[Kanye East]

Can anyone help express this in Cartesian form?
I keep getting -sqrt(3) + 2i/sqrt(2)

but the solutions says it is

Spoiler

-sqrt(3) + i

[e^1]

Can anyone help express this in Cartesian form?
I keep getting -sqrt(3) + 2i/sqrt(2)

but the solutions says it is

Spoiler

-sqrt(3) + i

[lzxnl]

If you're not sure, it's always a good idea to check the magnitude of your complex number. (abs(-sqrt 3 + 2i/sqrt 2))^2 = 3+4/2=5
But the original number has modulus 2, not sqrt 5. Something has to be wrong.

[TimmyC]

Hey guys I am really struggling with the steps of finding the vector resolute of a perpendicular to b. hehe trying to show unit vectors
        ^^
a-(a.b)b= (3i+2j-k)-9/25(4i+3k)

I would be extremely greatful if someone could talk me through it or something! Thanks in advance!

[BubbleWrapMan]

You seem to be on the right track, you just need to collect like terms.

[TimmyC]

You seem to be on the right track, you just need to collect like terms.

I am sorry to say but I can't get to the answer
It has to be more simple than I'm approaching it

[BubbleWrapMan]

Show me your approach?

[TimmyC]

Show me your approach?

This has to be 99% dreadfully wrong, but I am multiplying vector A by 25 and B by -9...

somehow get the correct answer but it makes no sense to me LOL

[BubbleWrapMan]

Okay let's start from the start. What's and what's ? And what's the final answer you're given?

[TimmyC]

Thank you!!
A= 3i+2j-k
B= 4i+3k

    ^ ^
(a.b)b= 9/25(4i+3k)

The answer given for A perpendicular to B is 1/25(39i+50j-52k)

[BubbleWrapMan]

[TimmyC]

Oh so that is actually how you do it... Thank you so so much!!

[Holmes]

From 2007 spec exam, I don't understand why the answer to this is B.
As the numbers are all z, does that mean the conjugate root theorem doesn't apply.? I was thinking of looking at 'equally spaced roots', but then I thought C was pretty equal as well as B. Or am I just not looking at it right?
Thanks so much for helping out.

[brightsky]

firstly, the fundamental theorem of algebra tells us that there are five solutions to the polynomial equation. now, all the coefficients in the given polynomial are real. so by the conjugate root theorem, if a + bi is a solution, then a -bi must be a solution as well. this rules out pretty much all the options except for B. and no, the roots do not necessarily have to be equally spaced. this only applies to equations of form z^n = k.

[Holmes]

firstly, the fundamental theorem of algebra tells us that there are five solutions to the polynomial equation. now, all the coefficients in the given polynomial are real. so by the conjugate root theorem, if a + bi is a solution, then a -bi must be a solution as well. this rules out pretty much all the options except for B. and no, the roots do not necessarily have to be equally spaced. this only applies to equations of form z^n = k.

Thanks. I get it. Also, I just realised that B and C have the same sort of symmetry, but in B, the dot is lying on the Re(z) axis, so it is a real solution by itself. In C, the lone dot is on the Im(z) axis, so it has to be mirrored by another solution according to the conjugate root theorem, which it isn't.

One last question though, the coefficients in the polynomial are all z's, aren't they?, meaning they're not real coefficients.
Thanks.