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Edexcel FP3 June 2015 - Official Thread

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Then expand the bracket of (1+x)^1 so the integral could be expressed in terms of

I_n and I_{n-1}

Thanks a lot!! That was really frustrating me. I'm worried I wouldn't see something like that in the exam.

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Reply 41

simonli2575

Did the 2014 paper, forgot what orthogonal matrix meant and q6 was insane.

Reply 42

BP_Tranquility

How do you find the line of intersection between 2 planes?

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Reply 43

Elcor

Original post by BP_Tranquility

How do you find the line of intersection between 2 planes?

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You know the line is perpendicular to both of the planes' normal vectors, so d=n1xn2. Find a point on the line by converting the planes to cartesian form, letting any of x, y or z=0 and solving simultaneously for the other two ordinates (this is basically finding x and y when z=0 where the planes intersect, for example)

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Reply 44

H0PEL3SS

Hmm, only just found this thread. Done all but vectors, and the only real problem seems to be the more complex integrations and loci. Hopefully vectors aren't too difficult.

Reply 45

BP_Tranquility

Original post by Elcor

Do you have to let z=0 or can you let it equal any other constant as well?

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Reply 46

Elcor

Original post by BP_Tranquility

Do you have to let z=0 or can you let it equal any other constant as well?

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Any

You'd end up with a different point but it's still on the line

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Reply 47

Elcor

For the plane equation r.n=a.n, can a be ANY point on the plane?

Reply 48

simonli2575

Original post by Elcor

For the plane equation r.n=a.n, can a be ANY point on the plane?

I'm pretty sure yes.

Reply 49

Elcor

Original post by simonli2575

I'm pretty sure yes.

I just can't see how a.n is a constant

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Reply 50

simonli2575

Original post by Elcor

I just can't see how a.n is a constant

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I think about it like this:
r.n=a.n comes from (r-a).n=0
And if you choose another point on the plane for a, r-a is still going to be perpendicular to n
thus it makes no change.

Reply 51

Elcor

Original post by simonli2575

I think about it like this:
r.n=a.n comes from (r-a).n=0
And if you choose another point on the plane for a, r-a is still going to be perpendicular to n
thus it makes no change.

Ah good point

Reply 52

Elcor

When proving a line does not meet a plane I imagine it's not enough to show that it's perpendicular to the normal vector, as the line could still lie on the plane.

Reply 53

Elcor

I proved r(L).n=/=a.n which I think is sufficient. Will it always be that the lambdas cancel out in r(L).n (if they do not meet), since no lambda satisfies the equation?

Reply 54

simonli2575

Original post by Elcor

I proved r(L).n=/=a.n which I think is sufficient. Will it always be that the lambdas cancel out in r(L).n (if they do not meet), since no lambda satisfies the equation?

If it's parallel to but does not intersect the plane, then r(L).n will result in an impossible equation, like 2=5.

Reply 55

alexmufc1995

Original post by simonli2575

Did the 2014 paper, forgot what orthogonal matrix meant and q6 was insane.

Assuming q6 was the loci question; I didn't answer anything beyond part 1 and still got 100 UMS

Reply 56

BP_Tranquility

Can someone help me with 14f please? I found a vector which was perpendicular to both planes (as I thought it'd give a parallel vector to plane 1) and then set that equal to the dot product between the new normal vector I found and point A. However, the solution bank instead uses the normal vector for plane A but surely that's a vector which is perpendicular to plane 1 and not parallel?

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Reply 57

Elcor

Original post by BP_Tranquility

There does not exist a vector perpendicular to both planes, as they themselves are perpendicular. Try visualising it.

The plane will have the same normal vector as plane 1, as it is parallel. Then just use OA.n

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