# Couple of Further Maths Vector Qs

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#1
Hey guys, I'm a bit stuck on a couple of Qs.

1) P1: r.(1,2,-1) = 7 and P2: r.(2,1,1) = -1
Find P3 which contains y-axis and intersects P1 & P2 to form prism.

2) R1: r.(2,-3,6) = 0, p = (7,6,5) and q=(1,3,-1)
Find R2 which contains line PQ and is perpendic. to R1

With 1) I'm really struggling to visualise the problem, thus confused.

Any help appreciated.

EDIT: I've done 2). We need a vector perpendic. to both line PQ and (2, -3, 6) for n_R2. Then sub P(7,6,5) into r.(n_R2) = d to find d.
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10 years ago
#2
I thought you were like a 3rd year imperial physicist , whyre you doing FP3 stuff ?
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#3
(Original post by rbnphlp)
I thought you were like a 3rd year imperial physicist , whyre you doing FP3 stuff ?
Because I'm not at Imperial at the mo.

I've done Q 2) and feel like I'm along the right lines in Q 1). The answer is x + z = 0. Sounds plausible, but how do I get that?
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10 years ago
#4
(Original post by Physics Enemy)
Q 1). The answer is x + z = 0. Sounds plausible, but how do I get that?
Since P3 forms a prism with P1 and P2, what do you know about the lines of pairwise intersection? What do they have in common?
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#5
(Original post by ghostwalker)
Since P3 forms a prism with P1 and P2, what do you know about the lines of pairwise intersection? What do they have in common?
Not sure. They form the same angle with P3? So this is replicated when looking at the dot product between their normals? What about containing y=0?

Another Q: If I add 2 planes together that have a common line of intersection, will my resultant plane contain this line? I did a Q and it seemed to imply this. But is it true?
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10 years ago
#6
(Original post by Physics Enemy)
Not sure. They form the same angle with P3? So this is replicated when looking at the dot product between their normals? What about containing y=0?
The direction vector for that line will lie in the plane P3 (in fact in all three planes), and you already have another vector lying in the plane.
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#7
(Original post by ghostwalker)
The direction vector for that line will lie in the plane P3 (in fact in all three planes), and you already have another vector lying in the plane.
Sorry I'm lost. I don't get which line you're refering to and why it would lie in all 3 planes. And which vector do I have lying in P3?

Also, do you know the answer to my other Q - If you add 2 planes together, do you get a new plane that contains their line of intersection?
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10 years ago
#8
(Original post by Physics Enemy)
Sorry I'm lost. I don't get which line you're refering to and why it would lie in all 3 planes. And which vector do I have lying in P3?
Three planes forming a prism will intersect in three lines. Those lines are all parallel to each other; they have the same direction vector. That direction vector lies in all three planes, specifically it lies in P3.

P3 contains the y-axis, and hence contains the direction vector (0,1,0)

You now have two vectors in the plane and forming their cross product will give you the normal for the plane.

You'll need to work out the direction vector of the line of intersection of P1 and P2 initially.

Also, do you know the answer to my other Q - If you add 2 planes together, do you get a new plane that contains their line of intersection?
In terms of co-ordinate geometry.

P1: ax + by+ cz=d

P2: a'x+b'y+c'z=d'

all points on their line if intersection will satisfy both equations.

I assume by adding that you just add one equation to the other to get:

P3: (a+a')x + (b+b')y + (c+c')z = d+d'

Any point that satisfies both P1 and P2, will satisfy P3.
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#9
(Original post by ghostwalker)
Three planes forming a prism will intersect in three lines. Those lines are all parallel to each other; they have the same direction vector. That direction vector lies in all three planes, specifically it lies in P3.
I get the 3 lines thing. I don't see how they're all parallel to each other; how do you make a prism with three parallel lines? I get that parallel lines have the same DV. I get that each plane contains this parallel line and hence this DV.

So my sticking point here is the parallel bit / my visualisation of it.

(Original post by ghostwalker)
P3 contains the y-axis, and hence contains the direction vector (0,1,0)
Fair enough.

(Original post by ghostwalker)
You now have two vectors in the plane and forming their cross product will give you the normal for the plane.
Yes, I get that.

(Original post by ghostwalker)
You'll need to work out the direction vector of the line of intersection of P1 and P2 initially.
That's fine. But why would I need to do this?

(Original post by ghostwalker)
In terms of co-ordinate geometry.

P1: ax + by+ cz=d

P2: a'x+b'y+c'z=d'

all points on their line if intersection will satisfy both equations.

I assume by adding that you just add one equation to the other to get:

P3: (a+a')x + (b+b')y + (c+c')z = d+d'

Any point that satisfies both P1 and P2, will satisfy P3.
Yes, thanks.
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10 years ago
#10
(Original post by Physics Enemy)
...
Someone posted a good diagram on one of the other threads in the last couple of days, unfortunately I haven't got one, and you can find it as easily as I can.

The three planes do not contain the same line; the points of intersection of the three planes form three lines and each line has the same direction vector.

Imagine the packet a bar of toblerone comes in. Its cross section is triangular. The three edges that run from one end to the other are the three lines I'm talking about. The faces/sides are parts of the three planes.

Hope that clarifies.
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#11
(Original post by ghostwalker)
Someone posted a good diagram on one of the other threads in the last couple of days, unfortunately I haven't got one, and you can find it as easily as I can.

The three planes do not contain the same line; the points of intersection of the three planes form three lines and each line has the same direction vector.

Imagine the packet a bar of toblerone comes in. Its cross section is triangular. The three edges that run from one end to the other are the three lines I'm talking about. The faces/sides are parts of the three planes.

Hope that clarifies.
I GET IT NOW! I was thinking of the lines (sides) of the triangular cross-section, and was like 'they can't be parallel?!'. Haha. Thanks so much, you were of great help! Really really clear and concise with your responses too, great help.

I think I get the set up now and which 2 vectors I need to cross multiply. Final thing; don't I need a point on the plane? I can use (0, 0, 0) right? Or (0, k, 0) generally.
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10 years ago
#12
(Original post by Physics Enemy)
Final thing; don't I need a point on the plane? That would be (0, 0, 0) right?
That's the one I'd go for, the more zeros the better.
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