Candescence
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#1
Report Thread starter 11 years ago
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The plane \Pi_1 passes through the P, with position vector i+2j-k and is perpendicular to the line L with equation:

 3i-2k +\lambda(-i+2j+3k)

a) show the cartesian equation of \Pi_1 is x-5y-3z=-6

Can anyone explain where the 5 comes from? chances are i'll get stuck on the rest of the question after this, but for now i just don't know where the 5 comes from.
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silent ninja
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#2
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I think that's a typo. I remember changing one of my question papers last week when someone spotted a type and this seems to be it. The 2 should be a 5 in the original question I believe.
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Candescence
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(Original post by silent ninja)
I think that's a typo. I remember changing one of my question papers last week when someone spotted a type and this seems to be it. The 2 should be a 5 in the original question I believe.
Oh thank god, i thought that the 2 was probably a typo because it worked fine if i put 5 in. But since i lost my textbook i don't have anything to check my method against >.<

Thanks
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Candescence
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#4
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On the same question:

The plane \Pi_2 contains the line L and passes through Q,i+2j+2k

c)find the equation of \Pi_2 in the form r=a +sb +tc

i get a is 3i-2k and b is -i+5j+3k but the answer scheme says c is 2i-2j-4k which i don't understand =/

Thanks in advance.
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Pixie-Goblin
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#5
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You have to find two non-parallel vectors from a starting point, in this case 3i-2k. So, we know a line which lies on the plane (line L) as spotted. Now the next step is to use the other piece of information given. We know that Q lies on the plane. So how can you link Q in to get the second vector?
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Candescence
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Report Thread starter 11 years ago
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(Original post by Pixie-Goblin)
You have to find two non-parallel vectors from a starting point, in this case 3i-2k. So, we know a line which lies on the plane (line L) as spotted. Now the next step is to use the other piece of information given. We know that Q lies on the plane. So how can you link Q in to get the second vector?
Oh right, that's really easy :p: thanks.

It would be just as valid to use -2i +2j +4k wouldn't it? since the direction of the line doesn't matter, only that it's parallel.
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Pixie-Goblin
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#7
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indeed, in that case the equivalent t value would just be a negative but it still works out no problem
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Candescence
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#8
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Great, thanks.
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