Vectors

Can anybody do this simple question, I'm quite new to vectors.

'Find the vector and cartesian form of the plane that contains the points (1,2,0), (2,2,1) and (1,1,0)'

Thanks.

Reply 1

dvs

Suppose the points are A, B and C. Then:
AB = (2, 2, 1) - (1, 2, 0) = (1, 0, 1)
AC = (1, 1, 0) - (1, 2, 0) = (0 , -1, 0)

So we have that r = s(1, 0, 1) + t(0, -1, 0) determines a bunch of parallel planes. So the plane we want is:
r = (1, 2, 0) + s(1, 0, 1) + t(0, -1, 0)

We could have used (2, 2, 1) or (1, 1, 0) instead of (1, 2, 0).

Reply 2

Dekota

dvs

Thanks. I got that. :smile:

But I'm having difficulty in expressing the vector equation in cartesian form...

Reply 3

dvs

Oh, sorry. I didn't notice that in the question.

Find ABxAC. This vector will be perpendicular to both AB and AC, and hence to the plane. So [(x, y, z) - (1, 2, 0)] . (ABxAC) = 0. Then simplify.

Reply 4

Dekota

dvs

Oh, sorry. I didn't notice that in the question.

Find ABxAC. This vector will be perpendicular to both AB and AC, and hence to the plane. So [(x, y, z) - (1, 2, 0)] . (ABxAC) = 0. Then simplify.

Great help dvs. Thank you.

Thank you again. :biggrin:

Reply 5

Dekota

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What is the quickest way of finding the vector form of an equation like 2x + y + 3z = 6 ?

Thanks.

Reply 6

Dekota

please help.

This is worrying..

Reply 7

bobbob

How you could do it (in theory) is find three points on the plane from the equation and then you can work it the two direction vectors from there. If you need any more help, I'll work the question through for you.

Also, out of interest, what's this for? FP4?

Reply 8

Dekota

bobbob

How you could do it (in theory) is find three points on the plane from the equation and then you can work it the two direction vectors from there.

Also, out of interest, what's this for? FP4?

It is FP3 and some of it is confusing, esp. because I've started studying FP3 vectors before C4 vector chapter!

I knew I can find 3 points and do that, but does the scalar/vector products have no use over here? It's just finding the three points and doing all that is a bit too tedious.

Reply 9

bobbob

Dekota

Ah, didn't realise Edexcel had vectors on FP3. Haven't done C4 either...

You cannot get two points easily from the normal vector on it's own - that's what the cartesian equation is for. I can't think of many practical reasons why you would actually do this. I personally think it's quite quick easy and enjoyable rather than tedious. :P

Not sure how far FP3 goes, but all the cartesian equation is derived from is the normal vector and a point on the plane (n.r = n.a), so just using the normal vector and a point you could go back to the vector equation, and in such a question you would use the dot product. So it is related to scalar/vector products, just not that question.

Reply 10

Dekota

Great. guess i'll have to do it that way.

thanks for the help. I'm doing OCR MEI btw. :smile:

Reply 11

bobbob

Ah! Get confused by all the exam boards.

If you have any other problems post 'em and I'll have a go. (I dug my notes out from a month ago - I had a whole exam just based on vectors and matrices, so feel glad that you're just doing this).

Reply 12

Dekota

gosh! FP3 is quite a mission.

I see that vectors and matrices are very closely similar. Any relationship between the two?

Reply 13

bobbob

Well, a vector is often written as a matrix (whether 2x1 or 3x1) :smile:

The two sections also meet when it comes to eigenvalues and eigenvectors (which allow you to easily work out large matrix powers). Other than that, not really.

My FP3 course contains no matrices/vectors at all, but I did FP4 which was JUST vectors/matrices. FP3 so far has contained mainly more integration and polar co-ordinates, but we didn't start it that long ago (and we're relying on teacher notes since we don't have a textbook) :smile: