[SecondHand]

Not sure what you have done on question two but when you get to x^-1a^-1 then you just use the given result. Question 3 means "prove all elements are self inverse, still not sure how to tackle it.

[SecondHand]

(Original post by ElMoro)
For the June 11 paper, question 5 part ii: why must k = -1?

EDIT: Nevermind, I was being an idiot

For anyone who is not clear, this is so that it is in the form.



If then on the RHS it is a function of u as well.

[Anon 17]

Yeah, I think question 3 follows on from one as one is proving a specific element is self inverse, and three all of them.

I'm just going to hope that these kind of questions are either nice or not prominent in the exam tommorow...

[ElMoro]

For this question part (ii)



How is that first step justified; surely they're assuming the result there?

[Killjoy-]

(Original post by ElMoro)
For this question part (ii)



How is that first step justified; surely they're assuming the result there?

For 3. ii) ?
They have removed the left x and the right y, to end up with in the middle, or .

[SecondHand]

write (xy)^n as xyxyxyxyxyxyxyxyxy and then take and x from the front and a y from the back and you are left with x(yxyxyxyxyxyxyxyxyx)y or x(yx)^n-1y

[ElMoro]

(Original post by Killjoy-)
For 3. ii) ?
They have removed the left x and the right y, to end up with in the middle, or .

Ohhhhh, I get it! Thank you!

Don't think i could think of that in an exam tbh

EDIT: Unfortunately, I've repped you too recently

[Killjoy-]

(Original post by ElMoro)
Ohhhhh, I get it! Thank you!

Don't think i could think of that in an exam tbh

EDIT: Unfortunately, I've repped you too recently

That's alright.
Did you see my response to the question you asked about the roots on the tan equation?

[Ree69]

Page 318 q12. Anyone care to help..?

[SecondHand]

12a)



Length = 22

12b)

Length of AB = 22
Length of OA =
length of OB =

Length of OB is a multiple of the direction vector

[Ree69]

(Original post by SecondHand)
12a)



Length = 22

12b)

Length of AB = 22
Length of OA =
length of OB =

Length of OB is a multiple of the direction vector

Cheers - that's a very swish way of working out the position vector of B!

[Purehakkai]

Is anyone else feeling majorly underprepared?

[SecondHand]

You have to be confident going in or there is no way you will be confident going out.

[Ree69]

...who on earth is confident going out of an exam?

[ElMoro]

(Original post by Killjoy-)
That's alright.
Did you see my response to the question you asked about the roots on the tan equation?

Yep I did. Thanks for that as well (sorry I thought I replied to it before )

[Purehakkai]

(Original post by Ree69)
...who on earth is confident going out of an exam?

I am for economics never maths tho

[Satsui no hado]

I'm just hoping the groups questions are relatively friendly :P

[Ree69]

Jan 12 Q6iii. I've got the answer but I'm not quite sure why it makes sense. Could someone explain this to me?

[Killjoy-]

(Original post by Ree69)
Jan 12 Q6iii. I've got the answer but I'm not quite sure why it makes sense. Could someone explain this to me?

You know a plane can be expressed parametrically with two direction vectors that lie on it.
(These vectors cannot be parallel - cannot be the same.)

In part i) you were asked to indirectly find the normal to the plane.
Now recall that the vector product of two direction vectors will give a direction vector that is perpendicular to both of them.

So the vector product of the normal to the plane and the direction vector after lambda should give c.

For a vector equation of a plane the two direction vectors in the plane may or may not be perpendicular, but the normal is obviously perpendicular to both of them in all forms of the vector equation, since these vectors lie in the plane.

I think there was another method to find c in the MS but I can't remember - I used the one above when doing the paper since it seemed quicker.

[Ree69]

(Original post by Killjoy-)
You know a plane can be expressed parametrically with two direction vectors that lie on it.
(These vectors cannot be parallel - cannot be the same.)

In part i) you were asked to indirectly find the normal to the plane.
Now recall that the vector product of two direction vectors will give a direction vector that is perpendicular to both of them.

So the vector product of the normal to the plane and the direction vector after lambda should give c.

For a vector equation of a plane the two direction vectors in the plane may or may not be perpendicular, but the normal is obviously perpendicular to both of them in all forms of the vector equation, since these vectors lie in the plane.

I think there was another method to find c in the MS but I can't remember - I used the one above when doing the paper since it seemed quicker.

Ahh... cheers