VCE Specialist 3/4 Question Thread!

[BLACKCATT]

So the answer is in the second quadrant, therefore the edge of the principal arguments will occur when
Z^3>cis(Pi/2)
Z^3<cis (Pi)
Solving for z will give the region (Pi/6, pi/3) (5pi/6,pi) and (-5pi/6,-2pi/3)
So the answer would be the Union of these, since none of them include all three regions, all of them are technically wrong.

Thank you!!

[Valyria]

Hey,

Could someone please explain how to apply the scalar resolute here or an alternative way to do part (ii)?

Thanks in advance

[Zealous]

Hey,

Could someone please explain how to apply the scalar resolute here or an alternative way to do part (ii)?

Thanks in advance

This is the extremely lazy method of doing it:
(Let blah=unit vector perpendicular to OA and OC)


But instead of using the resolute you can just use simultaneous equations given the dot product of OA and blah is 0, the dot product of OC and blah is 0, and the magnitude of blah is 1. Because it is a "show that", you'd probably wanna do it by hand.

Spoiler

[Valyria]

This is the extremely lazy method of doing it:
(Let blah=unit vector perpendicular to OA and OC)
(Image removed from quote.)

But instead of using the resolute you can just use simultaneous equations given the dot product of OA and blah is 0, the dot product of OC and blah is 0, and the magnitude of blah is 1. Because it is a "show that", you'd probably wanna do it by hand.

Spoiler

I'm fine with this part, it was part (ii) I needed help with; 'find the height'. Thanks anyways Zealous

[Robert123]

I'm fine with this part, it was part (ii) I needed help with; 'find the height'. Thanks anyways Zealous

First off, this pyramid is on an angle so the base isn't lying flat on the x and z axis. If it was, the height will be given by the y component of the top of the pyramid. So what you would need to do is find a vector perpendicular to the base (oh look, that is what the answer to part I is, what a coincidence). From there, you need to find out how much of the OD vector goes in that direction. To do this, you take the scalar resolute of OD in the direction of your results from part i.
Does that clarify what you have to do and why? Will you be find in doing the maths?

[Mieow]

This is the extremely lazy method of doing it:
(Let blah=unit vector perpendicular to OA and OC)
(Image removed from quote.)

But instead of using the resolute you can just use simultaneous equations given the dot product of OA and blah is 0, the dot product of OC and blah is 0, and the magnitude of blah is 1. Because it is a "show that", you'd probably wanna do it by hand.

Spoiler

What is that 'norm' function for that last equations with blah?

[theshunpo]

What is that 'norm' function for that last equations with blah?

norm(blah) is the magnitude of the vector. Since that's the unit vector, the magnitude is equal to 1.

[M_BONG]

Is there a quick way to expand, to find cartesian equation on CAS or by hand?

| z - u | = |z + v|


Without having to do all by hand? I know it's the perp. bisector, but just can't seem to find a shortcut with expanding these... Anyone?

[Mieow]

Find the value(s) of n such that Re(=0), where z=

Can someone help me with this please?

[Robert123]

Can someone help me with this please?

Hint, convert to polar form. Re(z) occurs when arg(z)=pi/2 or arg (z)=-pi/2, from that, use general solution strategies to find the general solutions for n.
Does that clarify how to approach these type of questions?

Is there a quick way to expand, to find cartesian equation on CAS or by hand?

| z - u | = |z + v|


Without having to do all by hand? I know it's the perp. bisector, but just can't seem to find a shortcut with expanding these... Anyone?

Well for the exam, you can have z=x+yi predefined in your CAS then use solve for y. Personally though, I can do it by hand relatively quickly though as those type of questions become very repeatitive and formulaic.
If that annoy you too much, you could easily have some predetermined rules in your rule book by having u=a+bi and v=c+di, then find the Cartesian equation with those variables. Then when you get in the exam, all you would have to do would sub in the values and there is your answer.

[Zealous]

Is there a quick way to expand, to find cartesian equation on CAS or by hand?
| z - u | = |z + v|
Without having to do all by hand? I know it's the perp. bisector, but just can't seem to find a shortcut with expanding these... Anyone?

If that annoy you too much, you could easily have some predetermined rules in your rule book by having u=a+bi and v=c+di, then find the Cartesian equation with those variables. Then when you get in the exam, all you would have to do would sub in the values and there is your answer.

Hahaha I was bored in class one day so I did it with a few friends and chucked it in my bound reference. So if we have the complex equation , the corresponding cartesian equation will be:



It may or may not show up on the exam but looks cool anyway.

Edit - Here's a graph form: https://www.desmos.com/calculator/yd8n2juo4g, pretty fun to play around with.

[jessss0407]

Hi!

Could someone help me with the following:

Prove that
sec(x) + cosec(x)cot(x) = sec(x)cosec²(x)

Thanks!

[Phy124]

Hi!

Could someone help me with the following:

Prove that
sec(x) + cosec(x)cot(x) = sec(x)cosec²(x)

Thanks!

[Bestie]

hello
can someone please help me with this q?
Why is the F ommited the question did say initial push of force F? is it cause its initial???

[psyxwar]

hello
can someone please help me with this q?
Why is the F ommited the question did say initial push of force F? is it cause its initial???

Force is no longer acting on it, so yeah, it's cuz it was initial.