Specialist's Specialist Thread

[Special At Specialist]

This is my specialist maths question thread for all of the problems that I am having trouble with. Please try to help me as much as possible. Thank you in advance

ABCD is a quadrilateral in which AB is parallel to and 3 times the length of CD. If CD = a and AD = b then BC in terms of a and b is equal to:
A) 3b
B) b - 4a
C) -3b
D) 2a - b
E) 2a + b

I chose option B but apparently I'm wrong.

[rife168]

E) 2a + b

A handy way to think about this is to trace the known vectors in order to get from B to C.
so (BA = 3a) + (AD = b) + (DC = -a)
therefore, 3a + b + (-a)
 = 3a + b - a
 = 2a + b

[Special At Specialist]

Thanks fletch-j! I have a new problem:

a, b and c are non-zero position vectors. The angle between vectors a and b is pi/4 and the angle between vectors a and c is pi/3. For the vectors to be linearly dependent, the angle between vectors b and c must be:
A) 5pi/12
B) 7pi/12
C) pi/2
D) pi/12 or 7pi/12
E) 5pi/12 or 7pi/12

I chose option A but I was wrong. Please help me solve this problem.
I think the answer was meant to be either D or E, but I'm not sure.

[yawho]

D

[Special At Specialist]

Thanks, but would you care to explain why?

[xZero]

for linearly dependency, a=mb+nc right? If you draw a line and call it a, you can draw vector b and c since you have the angle from vector a. To find the angle between b and c all you do is subtract the angle between a and b from the angle between a and c, which is pi/12. As for 7pi/12, remember they are vectors, they can be facing the same or opposite direction, if they are facing the same direction its pi/12 and if they are in opposite direction its 7pi/12.

[yawho]

For them to be linearly dependent all 3 position vectors must be on the same plane. There are only 2 possible arrangements of the vectors from O, either in order of a, b, c or c, a, b. In the first case the angle is pi/12, second case 7pi/12.

[Special At Specialist]

Thanks, that kind of makes sense.

Now I have another problem:

Would we be expected to solve integrals like this in a tech free exam? How would I go about such a problem? I can't use substitution because there is nothing on the numerator and I can't use partial fractions since I cannot factorise the denominator.

[pi]

No, that's too complicated for an exam, Is probably fine for an exam, maybe a 4 mark question

I'm struggling with it myself. edit: remembered how to do stuff like this thanks to the formula sheet =.=

[Special At Specialist]

Thanks pi! Using your method, I actually managed to prove a general case of:
where

http://img830.imageshack.us/img830/3774/mathsproblem16.png

Moderator action: removed real name, sorry for the inconvenience

[Special At Specialist]

Two vectors are given by a = -i - 2j + k and b = 2i + j + k. Find the acute angle between the vectors.

Usually I would just do the dot product, but this is what happens:
a . b = -2 - 2 + 1
a . b = -3
|a| = sqrt(1 + 4 + 1)
|a| = sqrt(6)
|b| = sqrt(4 + 1 + 1)
|b| = sqrt(6)
a . b = |a||b|cos(θ)
-3 = 6cos(θ)
cos(θ) = -1/2
θ = 2pi/3 or θ = 4pi/3

But an acute angle is where 0<θ<pi/2. Please help me!

[yawho]

Two vectors are given by a = -i - 2j + k and b = 2i + j + k. Find the acute angle between the vectors.

Usually I would just do the dot product, but this is what happens:
a . b = -2 - 2 + 1
a . b = -3
|a| = sqrt(1 + 4 + 1)
|a| = sqrt(6)
|b| = sqrt(4 + 1 + 1)
|b| = sqrt(6)
a . b = |a||b|cos(θ)
-3 = 6cos(θ)
cos(θ) = -1/2
θ = 2pi/3 or θ = 4pi/3

But an acute angle is where 0<θ<pi/2. Please help me!

The obtuse angle is 2pi/3, so the acute angle is pi - 2pi/3 = pi/3

[Special At Specialist]

How did you come to that conclusion? Did you move the vectors around?

[Lasercookie]

How did you come to that conclusion? Did you move the vectors around?

Just looking at the angles here. Maybe a diagram will help:


If you look at the x-axis, 2pi/3 is the angle from the positive x-axis. So if you want to figure out the angle from the negative x-axis (the red bit in the diagram) and you know that the x-axis is a straight line (180 degrees, or pi radians), so pi - 2pi/3 would make sense.

I guess you can just think of it circular functions type stuff. 2pi/3 is in the second quadrant and then find the equivalent angle in the first quadrant.

[Special At Specialist]

What is a co-domain? I always see functions expressed something like this:
f : [5, inf) -> R, f(x) = 2 + sqrt(x - 5)
So you can see from this function that the domain is [5, inf) and the range is [2, inf). But the range isn't written down; only the domain and co-domain are.
What is the co-domain? Why is it just R? Can you please give me an example of a function where the co-domain isn't R?