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# Planes question watch

1. http://www.ocr.org.uk/Images/57750-q...hematics-3.pdf

Question 6 part 2).

So I thought that since it's the line parallel to L1 but contains L2 you do

L2 x L1 to get the vector i-2k as the normal, but apparently the markscheme says it's the same normal as for part 1 I.E -i+2k why is that?
2. (Original post by Music99)
http://www.ocr.org.uk/Images/57750-q...hematics-3.pdf

Question 6 part 2).

So I thought that since it's the line parallel to L1 but contains L2 you do

L2 x L1 to get the vector i-2k as the normal, but apparently the markscheme says it's the same normal as for part 1 I.E -i+2k why is that?
Well if a plane exists such that it contains line l1 and that it is parallel to l2, then l1 and l2 must be (I can't recall the right word...not parallel - kind of parallel but they don't have the same direction - I think it's skew)? Basically l1 and l2 never intersect in 3d. So if you imagine that then the planes containing l1 and l2 must be parallel, so the normal vector of both planes must be the same. It doesn't matter which way you do your cross product - either A x B or B x A, you still get an equivalent answer - the only thing is if you do A x B or B x A, it'll change your corresponding 'p' value in the required form of from positive to negative
3. (Original post by Music99)
http://www.ocr.org.uk/Images/57750-q...hematics-3.pdf

Question 6 part 2).

So I thought that since it's the line parallel to L1 but contains L2 you do

L2 x L1 to get the vector i-2k as the normal, but apparently the markscheme says it's the same normal as for part 1 I.E -i+2k why is that?
There is no THE normal, as there are many possibilities, each a scalar multiple of the other.

Both i-2k and -i+2k are normals for both of the two planes.

And i-2k = -1 (-i+2k)

Following from L2 x L1 = - L1 x L2, refering to the direction vectors of each line.

Edit: Too slow.
4. (Original post by Music99)
http://www.ocr.org.uk/Images/57750-q...hematics-3.pdf

Question 6 part 2).

So I thought that since it's the line parallel to L1 but contains L2 you do

L2 x L1 to get the vector i-2k as the normal, but apparently the markscheme says it's the same normal as for part 1 I.E -i+2k why is that?
L2 X L1 = -(L1 X L2)

The normal vector you've calculated is the 'same' in that it's a scalar of the previous one (factor of -1 due to the way the cross product was calculated).

Edit: Deleted the previous post due to mark scheme confusing me, my bad.
5. (Original post by Felix Felicis)
Well if a plane exists such that it contains line l1 and that it is parallel to l2, then l1 and l2 must be (I can't recall the right word...not parallel - kind of parallel but they don't have the same direction - I think it's skew)? Basically l1 and l2 never intersect in 3d. So if you imagine that then the planes containing l1 and l2 must be parallel, so the normal vector of both planes must be the same. It doesn't matter which way you do your cross product - either A x B or B x A, you still get an equivalent answer - the only thing is if you do A x B or B x A, it'll change your corresponding 'p' value in the required form of from positive to negative
Ah okay makes sense, thank you.
6. (Original post by Blazy)
L2 X L1 = -(L1 X L2)

The normal vector you've calculated is the 'same' in that it's a scalar of the previous one (factor of -1 due to the way the cross product was calculated).

Edit: Deleted the previous post due to various reasons.
Yeah, but when you use that value you end up with a different answer from the markscheme for the rest of the question.
7. (Original post by Music99)
Yeah, but when you use that value you end up with a different answer from the markscheme for the rest of the question.
Surely the mark scheme should have something saying accept any equivalent answers?
8. (Original post by Felix Felicis)
Surely the mark scheme should have something saying accept any equivalent answers?
Nah it didn't - you get 3 instead of -3 which can be fixed by swapping the signs I guess, but the mark scheme doesn't say anything about it.
9. (Original post by Blazy)
Nah it didn't - you get 3 instead of -3 which can be fixed by swapping the signs I guess, but the mark scheme doesn't say anything about it.
Seriously? Seems harsh penalising candidates who may have done the cross product the other way around...it doesn't even state that the equation has to be given strictly in a required form - just
10. (Original post by Felix Felicis)
Seriously? Seems harsh penalising candidates who may have done the cross product the other way around...it doesn't even state that the equation has to be given strictly in a required form - just
I suppose the fact that it was 2 marks whilst the previous was 5 could be an indicator to the amount of work you had to do. It shouldn't be too much of a problem in the exam though.
11. (Original post by Felix Felicis)
Seriously? Seems harsh penalising candidates who may have done the cross product the other way around...it doesn't even state that the equation has to be given strictly in a required form - just
Yeah pretty annoying, so then I got the next bit wrong :/. Also on the next bit why do you not have to work out both distances and add them if you want the distance between both planes. You can do the distance between a point on one line and the other plane and that's the overall distance, bit unsure on why...
12. (Original post by Music99)
Yeah pretty annoying, so then I got the next bit wrong :/. Also on the next bit why do you not have to work out both distances and add them if you want the distance between both planes. You can do the distance between a point on one line and the other plane and that's the overall distance, bit unsure on why...
Sorry, not quite sure on what you're asking

If I were to calculate the distance between two planes, I'd usually calculate the distance from the origin of each plane - the if you have a plane in scalar product form then the distance from the origin is and just subtract the distances from the origin of each plane

Or, l1 is contained in plane 1 and l2 is contained in plane 2, you could calculate the distance between the two lines
13. (Original post by Felix Felicis)
Sorry, not quite sure on what you're asking

If I were to calculate the distance between two planes, I'd usually calculate the distance from the origin of each plane - the if you have a plane in scalar product form then the distance from the origin is and just subtract the distances from the origin of each plane

Or, l1 is contained in plane 1 and l2 is contained in plane 2, you could calculate the distance between the two lines
Don't worry I added instead of subtracting -.-. Would you mind if I quote you a bit later with another question I'm stuck on?
14. (Original post by Music99)
Don't worry I added instead of subtracting -.-. Would you mind if I quote you a bit later with another question I'm stuck on?
Depends on the topic and if I've covered it yet :P But yeah, feel free

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