VCE Specialist 3/4 Question Thread!

[ashoni]

(Try attempting the problem with the help before looking in the spoilers)

11. Combine each component.

Then for the vector to be parallel to the axis, the and components must be zero. So you can equate these two components to zero, get two equations with two unknowns, and solve for and .

Spoiler

17. Draw out the triangle to help you see the situation, then put the arrows for the vectors in the right direction, i.e. for vector AB, the arrow from A to B. Since is parallel to , .
Then


Now for the angle B to be a right angle, the dot product of the two vectors that meet at B must be zero, that is the dot product of and is zero.

Spoiler

But

Hope that helps

EDIT: Fixed the mistake as pointed out by 507

ahhh makes so much sense! thank you for taking your time to work it out for me b^3!

[zvezda]

Because you have a relation, and are given a value of , there are only two values of (in this case) that will correspond to that . So you have to take the positive one.

What you're doing is you're finding a for any value of , but evaluating the derivative with . You won't have all those values of the derivative on that curve, just those that have the value that corresponds to .

Ok fair enough, thanks for that b3

[ashoni]

need help with question... see attachment

[brightsky]

5. a) BL = 2/3 BC = 2/3 (c-b), so OL = OB + BL = b + 2/3 (c-b) = 1/3 b + 2/3 c
b) a = OA = -OL = -1/3 b - 2/3 c
so LHS = 3a + b + 2c = 3(-1/3 b - 2/3 c) + b + 2c = -b - 2c + b + 2c = 0 = RHS, as req
c) i) BO = - OB = -b
BM = BC + CM = c-b + 3/5 CA = c-b + 3/5 (a-c) = c-b + 3/5a - 3/5 c = 2/5 c - b + 3/5 (-1/3b-2/3c) = 2/5c - b - 1/5 b -2/5c = -6/5 b
so BM = 6/5 BO so B, O, M are collinear
ii) BO: OM = 1:1/5 = 5:1
d) AB = b-a = b-(-1/3b - 2/3c) = 4/3b + 2/3c
AN = k(4/3b + 2/3c) = 4/3 k b + 2/3 k c
ON = OA + AN = a + 4/3k b + 2/3 k c = (-1/3b - 2/3c) + 4/3k b + 2/3 k c = (4/3 k -1/3) b + (2/3 k - 2/3) c
we want ON to be parallel to OC. so 4/3 k - 1/3 = 0, k = 1/4
so AN = 1/3 b + 1/6 c = 1/4 AB
so AN:NB = 1:3

hopefully no errors

[tote.moore]

I'am wondering how to derive inverse trig functions such as cos^-1((3x)/2)).. The textbook give the basic formula for working with easy ones such as cos^-1(x/2), also any tips on graphing the derivatives of inverse trig functions.

[ashoni]

brightsky, could you explain how you got to

a = OA = -OL = -1/3 b - 2/3 c  from 5b.

I found the wording in the question quite confusing

[Hancock]

5. a) BL = 2/3 BC = 2/3 (c-b), so OL = OB + BL = b + 2/3 (c-b) = 1/3 b + 2/3 c
b) a = OA = -OL = -1/3 b - 2/3 c
so LHS = 3a + b + 2c = 3(-1/3 b - 2/3 c) + b + 2c = -b - 2c + b + 2c = 0 = RHS, as req
c) i) BO = - OB = -b
BM = BC + CM = c-b + 3/5 CA = c-b + 3/5 (a-c) = c-b + 3/5a - 3/5 c = 2/5 c - b + 3/5 (-1/3b-2/3c) = 2/5c - b - 1/5 b -2/5c = -6/5 b
so BM = 6/5 BO so B, O, M are collinear
ii) BO: OM = 1:1/5 = 5:1
d) AB = b-a = b-(-1/3b - 2/3c) = 4/3b + 2/3c
AN = k(4/3b + 2/3c) = 4/3 k b + 2/3 k c
ON = OA + AN = a + 4/3k b + 2/3 k c = (-1/3b - 2/3c) + 4/3k b + 2/3 k c = (4/3 k -1/3) b + (2/3 k - 2/3) c
we want ON to be parallel to OC. so 4/3 k - 1/3 = 0, k = 1/4
so AN = 1/3 b + 1/6 c = 1/4 AB
so AN:NB = 1:3

hopefully no errors

Learn LaTex Bro. (good job though)

[polar]

I'am wondering how to derive inverse trig functions such as cos^-1((3x)/2)).. The textbook give the basic formula for working with easy ones such as cos^-1(x/2), also any tips on graphing the derivatives of inverse trig functions.

use the chain rule

[tote.moore]

Thanks polar! i can now do it

[Homer]

it is know that and are both roots of the equation



where b, c, d, e are all integers. Find b, c, d, and e.

I dunno how to get the fourth root :/

[deleted]

Can't you just use 1-√2 and 1-√2i as the other roots?

And when you expand (z-1+√2)(z-1-√2)(z-1-√2i)(z-1+√2i) you get

z^4-4z^3+6z^2-4z-3

Therefore
b=-4 c=6 d=-4 e=-3

[Homer]

I did use 1-√2i but i dont think 1-√2 is right, since you cant really get the conjugate of 1+√2, its just a number (1+√2 +0i)

but i might be wrong thanks anyways

[nubs]

Kevin is right, you can use 1-√2

The conjugate of an irrational root is also a solution

The irrational conjugate roots theorem says:

Let p(x) be any polynomial with rational coefficients. If
a + b*sqrt(c) is a root of p(x), where sqrt(c) is irrational and
a and b are rational, then another root is a - b*sqrt(c).

There are a number of complications on top of that though which I won't get into, which is why I'd be pretty surprised if it were to appear on an exam. I do remember it being on an A+ study guide or something though

[Homer]

oh wow! never knew dat! thanks guys

[Homer]

 how do you guys approach questions related to vectors in geometry. Im finding it really difficult