VCE Specialist 3/4 Question Thread!

[speedy]

I usually use dot product but if the cosine is something that doesn't easily give an angle (e.g. if it's cos(theta)=3/5) then I go for the separate angles thing.

But does separating the angles always work? If so, surely that's way easier.

[lolalol]

Hey guys, just wanted to check:

If z=rcis(theta)
Then Arg(zbar)=-theta
and Arg(1/z)=-theta

Is this all okay?

[keltingmeith]

Hey guys, just wanted to check:

If z=rcis(theta)
Then Arg(zbar)=-theta
and Arg(1/z)=-theta

Is this all okay?

Depends on if you're trying to rote learn that or if you know why it's true.

(it is correct, though)

[BLACKCATT]

When integrating --> logs, do we leave the modulus sign off for the final answer

[keltingmeith]

When integrating --> logs, do we leave the modulus sign off for the final answer

ONLY if you know the initial conditions.

And you must remember to take the right form based on those initial conditions.

[psyxwar]

hey for spesh exam 1, can i leave my answers in terms of g when not told explicitly what to do, or should I always multiply it out? like 2013 exam 1 I wrote 5g, answers had 5g=49

[infecthead]

hey for spesh exam 1, can i leave my answers in terms of g when not told explicitly what to do, or should I always multiply it out? like 2013 exam 1 I wrote 5g, answers had 5g=49

If the answer has something like 5g=49 then both forms are acceptable answers. I was also told by my teacher that leaving it as g is acceptable in exam 1.

[psyxwar]

If the answer has something like 5g=49 then both forms are acceptable answers. I was also told by my teacher that leaving it as g is acceptable in exam 1.

thanks, just double checking lol, paranoid af

[Mieow]

Question 3
Relative to an origin O, point A has cartesian coordinates (1, 2, 2) and point B has cartesian coordinates
(–1, 3, 4).
b. Show that the cosine of the angle between the vectors OA and AB is 4/9
c. Hence find the exact area of the triangle OAB.

How would I find the area of this triangle? I know the formula is 0.5bc sin(A), and I found sin(theta) but the answer I got didn't match the assessor's reports

[RKTR]

How would I find the area of this triangle? I know the formula is 0.5bc sin(A), and I found sin(theta) but the answer I got didn't match the assessor's reports

i did this just now. what did you get for sin(theta)?

[lzxnl]

When naming things,

do you go clockwise or anticlockwise?

Eg. Parallelogram ABCD

in clockwise direction, I have
A  B
C  D

Is that right?

Should it be?

A B
D C ?

Generally they mean
A     B
D     C

So that you read ABCD in a circle

How would I find the area of this triangle? I know the formula is 0.5bc sin(A), and I found sin(theta) but the answer I got didn't match the assessor's reports

So OA = (1,2,2) and AB = (-2,1,2). Lengths are both 3.
cos(angle) = dot product/9 = (-2+2+4)/9 = 4/9
sin(angle) = sqrt(1-cos^2 angle) = sqrt(1-16/81) = sqrt(65/81) = sqrt(65)/9
Etc

[auds]

Where does the v0^2 constant come from ?

[lzxnl]

Look carefully and you'll see that you have a definite integral there. Maybe that'll help.

Hint: if dv/dx = f'(x), v2 - v1 = f(x2) - f(x1)

[BLACKCATT]

How to solve this one?

[Robert123]

How to solve this one?

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.