AQA FP4 25th May 2016 Discussion Thread

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5D Jan 2011...

I don't fully get why the line where 2 planes intersect is perpendicular to both planes

HELP

Well you know a point on the plane (4, 1, 9)
So if you consider the eq. of the plane in the form r.n=d
You are missing n and d, however in part a) you found an equation of the line of intersection between pi1 and pi2.
The direction vector of this line HAS to be perpendicular to both if you think about it.
Hence you just sub this vector (1,-12,2) and your point in and you will find the constant --> eq of the plane found

Reply 81

01binary

Original post by B_9710

The fact that the eigenvalue is repeated is not important, it's just a coincidence in this case. It doesn't have to be repeated. Don't think of it as splitting a vector, think of it as just finding 2 'random' vectors that satisfy the equations (the 3 repeated ones).

In that case then, how do you identify that this is in fact an invariant plane?
Is it because you have 3 of the same equations and you can find 2 vectors that satisfy in this case (x+y-z=0)?

Reply 82

Argylesocksrox

Original post by B_9710

You gonna post those answers? :biggrin:

Reply 83

B_9710

Original post by Argylesocksrox

You gonna post those answers? :biggrin:

Oh yes. You reminded me. I'll do it in a min.

Reply 84

B_9710

Original post by pic

5D Jan 2011...

I don't fully get why the line where 2 planes intersect is perpendicular to both planes

HELP

The line of intersection is parallel to both planes. This means that the line is perpendicular to the normal of both planes (which is why you do the vector product of the 2 normals as it gives you a vector that is perpendicular to both normals.) A vector that is perpendicular to the normal of a plane will be parallel to the plane itself.

Notice how the line of intersection lies in

\Pi_1

and

\Pi_2

so is parallel to both?

Reply 85

pic

Original post by 01binary

The direction vector of this line HAS to be perpendicular to both if you think about it.

I just can't visualise this... is the direction of the line of intersection between 2 planes ALWAYS going to be perpendicular to both planes?

Reply 86

01binary

Original post by pic

I just can't visualise this... is the direction of the line of intersection between 2 planes ALWAYS going to be perpendicular to both planes?

Look at the slopes of the edges of each plane if you were
to do the cross product between those you'd get a normal(the line of intersection)

Think if you put a plus sign on the floor and the the middle of the plus sign
you'd have a line coming straight up towards your face. Idk how else to describe

(edited 7 years ago)

Reply 87

pic

Original post by B_9710

Notice how the line of intersection lies in

\Pi_1

and

\Pi_2

so is parallel to both?

Thanks this makes perfect sense.

However, I'm still confused in the context of jan 2011 5d. (1 -12 2) is the direction of the line of intersection and so is parallel to both planes, and perpendicular to both normals. But the question wants the direction perpenciular to both planes not their normals, so I don't get why (1 -12 2) is the normal direction...

edit: got it! dw

(edited 7 years ago)

Reply 88

B_9710

Original post by pic

I wouldn't overcomplicate this question. You want to find a pl;ane whcih is perpendicular to both

\Pi_1

and

\Pi_2

so you want the plane to be parallel to each of the nromal vectors to

\Pi_1

and

\Pi_2

so I would just give the euqation fo the plane as

\displaystyle \mathbf{r}=\begin{bmatrix} 4 \\ 1 \\9 \end{bmatrix} +\lambda \begin{bmatrix} 6 \\ 2 \\9 \end{bmatrix} + \mu \begin{bmatrix} 10 \\ -1 \\ -11 \end{bmatrix}

Reply 89

B_9710

Original post by Argylesocksrox

You gonna post those answers? :biggrin:

For the first part I get

9x^2+4y^2+14xy+36x+10y-10=0

.
And for the second part I get

188x^2-216xy+62y^2+3=0

(edited 7 years ago)

Reply 90

MahirYuksel

Can someone explain how to do the right hand rule for cross product?

Reply 91

B_9710

Original post by MahirYuksel

Can someone explain how to do the right hand rule for cross product?

Why do you need the rule?

Reply 92

MahirYuksel

Original post by B_9710

Why do you need the rule?

I guess to aid understanding of normals.

Reply 93

01binary

When finding a point of intersection between two planes, why can you just set say X=0?
if you had found a eq of say 15x-3y=45 by eliminating z

Reply 94

C0balt

I'm feeling comfortable with the module now but I bet I'll take my words back when they pull an FP3 tomorrow

Reply 95

Argylesocksrox

This is either going to be my best or worst module, depending on how my algebra is tommorow.

Reply 96

pic

Original post by B_9710

I wouldn't overcomplicate this question. You want to find a pl;ane whcih is perpendicular to both

\Pi_1

and

\Pi_2

so you want the plane to be parallel to each of the nromal vectors to

\Pi_1

and

\Pi_2

so I would just give the euqation fo the plane as

\displaystyle \mathbf{r}=\begin{bmatrix} 4 \\ 1 \\9 \end{bmatrix} +\lambda \begin{bmatrix} 6 \\ 2 \\9 \end{bmatrix} + \mu \begin{bmatrix} 10 \\ -1 \\ -11 \end{bmatrix}

Good shout, good luck tomorrow mate!

Reply 97

B_9710

Original post by pic

Good shout, good luck tomorrow mate!

Good luck to you too.

Reply 98

Qcomber

Original post by B_9710

I always thought a repeated eigenvalue meant invariant plane.

Reply 99

B_9710

Original post by Qcomber

I always thought a repeated eigenvalue meant invariant plane.

I don't think so. It's when all 3 equations are the same. Whether this always happens with a repeated eigenvalue or not, I'm not sure.
Perhaps you're right, that's what I used to think.
Just from thinking about it, when you have been doing questions have you found that a repeated eigenvalue leads to all 3 equations being identical, hence an invariant plane. I think it may be the case.
Since searching, I have found that a repeated eigenvalues do NOT imply an invariant plane.

(edited 7 years ago)