VCE Specialist 3/4 Question Thread!

[Eiffel]

ahh right, just didnt seem obvious didnt know hat i was thinking.

lol i did 2^2 for 2^1 ..

[kinslayer]

orrrrrrr, (u-v) = 0 orrrrrrr, (u+v) = 0

(0 = zero vector)

Don't forget this

Still perpendicular 

[lzxnl]

Well...the zero vector doesn't really have a direction...saying it's perpendicular doesn't mean anything geometrically at least. Of course, if orthogonality is defined by a zero inner product, the geometrical meaning doesn't matter.

[kinslayer]

Well...the zero vector doesn't really have a direction...saying it's perpendicular doesn't mean anything geometrically at least. Of course, if orthogonality is defined by a zero inner product, the geometrical meaning doesn't matter.

Yeah, depends what your definition of perpendicular is. Orthogonality certainly is defined in terms of the inner product everywhere I've seen it, and the convention is usually to take perpendicular and orthogonal to mean the same thing. eg: http://mathworld.wolfram.com/Perpendicular.html

It definitely doesn't make sense to talk about angles between any vector and the zero vector, though, since angles are defined in terms of the inner product in a particular way that isn't defined for vectors with zero magnitude.

[grannysmith]

If a question asks to find all roots of a complex number that is in cartesian form, do we give the answer in polar form or cartesian form (using de Moivre's)? Or it doesn't matter unless the question specifically states one or the other?

[Maths Forever]

The question will usually specify. But in most of the past VCAA exams, you are requested to leave your answer in Cartesian form.

[stockstamp]

I need help with the following please:

Simplify the following, giving your answer in the modulus–Argument form

(1 + √3i)³
_______
i(1 − i)⁵


thanks

[Maths Forever]

I need help with the following please:

Simplify the following, giving your answer in the modulus–Argument form

(1 + √3i)³
_______
i(1 − i)⁵


thanks

Firstly, convert the numerator and denominator into polar form:

The Numerator:
(1 + √3i)^3 = {√[1^2 + (√3)^2]^3} cis [3[arctan (√3 / 1)]]
= [1+3]^3/2 cis (3pi/3)
= 8 cis (pi) = 2^3 cis (pi)

The Denominator:
i (1-i)^5 = [cis(pi/2)] x {√[1^2 + (-1)^2]^5} cis [5[arctan (-1 / 1)]]
= cis (pi/2) x [(√2)^5]cis(-5pi/4)
= {2^5/2}cis(pi/2 -5pi/4) = {2^5/2}cis(-3pi/4) .....using polar form products

Divide numerator by denominator using polar form division:

[2^3 / 2^5/2] cis (pi - (-3pi/4) = 2^1/2 cis (7pi/4) = √2 cis (-pi/4) ......since -pi < Arg (z) ≤ pi

= 1-i (in Cartesian form) .... but I know the question specifies modulus-Argument form.

I hope this helps and good luck!

[SE_JM]

Hello,
I would really appreciate it if you could help me understand this 3D problem.
One of the things I seem to struggle is the 3D concepts because i'm not used to spatial thinking and am quite weak at it.
I have been battling with this example in my book. It has the solutions, but no matter how much I look at it, I cannot seem to understand it.
I'll attach the example. Could someone please help me?

Thank you!!

[pi]

What don't you understand?

For understanding the various planes, it's simply an exercise of connecting the dots. eg. for ACD', this doesn't correspond to a face of the rectangular prism, so you'll have to create this plane by drawing a line from A to the midpoint (M) of CD' (as they lie on a common face, this is easy). So you get a plane that completely passes through CD' and also the corner A in a diagonal fashion. For the plane DCD' that is simply the same plane as the top face of the rectangular prism. It's quite hard to explain with words, try making the box in real life to convince yourself

Getting better at this spatial stuff comes with time, but is a very important skill to have

(also yay for studying at 9am! )

[stockstamp]

Firstly, convert the numerator and denominator into polar form:

The Numerator:
(1 + √3i)^3 = {√[1^2 + (√3)^2]^3} cis [3[arctan (√3 / 1)]]
= [1+3]^3/2 cis (3pi/3)
= 8 cis (pi) = 2^3 cis (pi)

The Denominator:
i (1-i)^5 = [cis(pi/2)] x {√[1^2 + (-1)^2]^5} cis [5[arctan (-1 / 1)]]
= cis (pi/2) x [(√2)^5]cis(-5pi/4)
= {2^5/2}cis(pi/2 -5pi/4) = {2^5/2}cis(-3pi/4) .....using polar form products

Divide numerator by denominator using polar form division:

[2^3 / 2^5/2] cis (pi - (-3pi/4) = 2^1/2 cis (7pi/4) = √2 cis (-pi/4) ......since -pi < Arg (z) ≤ pi

= 1-i (in Cartesian form) .... but I know the question specifies modulus-Argument form.

I hope this helps and good luck!

Thanks a lot! writing √2⁵ as 2^5/2 was quite clever, and I certainly didnt think of it!

[Eiffel]

why is the answer E? i can see if a tangent is drawn at say just past 1,0 the gradient would be like 30 for example but i dont see why it fully excludes + - 2.Surely it will catch up at some time?

[Fishing]

Hey can someone please help me with a vector question?

Point A has a position vector of a= 2i +3j + k
Point C has a position vector of c = 3i -j -2k
Point D has a position vector of d= -3i + 4j +6k

a) Find cos <ACD and hence, find the closest distance from point A to the line CD.

I've already found (correct me if I'm wrong), though I'm not sure what do do next? Can anyone help? Thanks in advance.

[Conic]

For this hyperbola, we have (you can show this by using implicit differentiation, which is covered later in the course).

Lets see what happens if the gradient is 2. If the gradient is 2 at some point (x,y) on the hyperbola, then dy/dx=2, so



Substituting this into the equation of the hyperbola gives



Well, if we assume that the gradient is 2, then we get to a contradiction, so the gradient can't equal 2 at any point of the hyperbola. A similar thing happens if we have dy/dx=-2.

[Eiffel]

how can i go about this graphically as well? why does the graph of the hyperbola never catch up, so to speak, with the asymptotes?