# FP3: intersection of planesWatch

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Thread starter 1 week ago
#1
are r equal to each other...as in thats the equation we're looking for and what unites both planes... as in, have I just answered my own question?! 0
1 week ago
#2
(Original post by Maths&physics)
are r equal to each other...as in thats the equation we're looking for and what unites both planes... as in, have I just answered my own question?! Well r (=(x,y,z)) represents a point on the plane in each case, and where the two planes intersect both equations will be satisfied simultanesously, i.e. by the same value of r in each.

Your working looks good up until you try to convert the cartesian form of the line back into the vector form - then it goes all over.
i,j,k have changed to j,k,l.

Don't forget the form of r will be that of a line: x1,y1,z1 refer to your vector a, and l,m,n refer to the vector b.

PS:

You can check if your value of r is correct: Does it have the right form and does it satisfy the equations of the planes?
Last edited by ghostwalker; 1 week ago
Thread starter 1 week ago
#3
(Original post by ghostwalker)
Well r (=(x,y,z)) represents a point on the plane in each case, and where the two planes intersect both equations will be satisfied simultanesously, i.e. by the same value of r in each.

Your working looks good up until you try to convert the cartesian form of the line back into the vector form - then it goes all over.
i,j,k have changed to j,k,l.

Don't forget the form of r will be that of a line: x1,y1,z1 refer to your vector a, and l,m,n refer to the vector b.

PS:

You can check if your value of r is correct: Does it have the right form and does it satisfy the equations of the planes?
thanks that's a tutorial I'm watching.

in this question: (1i + -1j +2z) and (ai + 3j + 2z) and both positions on the plane?

the (1i + -1j +2z) - (ai + 3j + 2z) = a vector parallel to the pane and the for/scaler product with give us the normal.

and to get the answer (r.n=D), I can use the direction vector of either line and dot it with n, and get the value for D (d in the equation)?
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1 week ago
#4
(Original post by Maths&physics)
thanks that's a tutorial I'm watching.

in this question: (1i + -1j +2z) and (ai + 3j + 2z) and both positions on the plane?
So, this is part b). And you should have a value for alpha from part a).

the (1i + -1j +2z) - (ai + 3j + 2z) = a vector parallel to the pane and the for/scaler product with give us the normal.
In red, will be parallel to the plane, as will the direction vectors of each of the lines.

I don't know what a "for/scalar product" is.

and to get the answer (r.n=D), I can use the direction vector of either line and dot it with n, and get the value for D (d in the equation)?
If you know n, then you can use any point on the plane to find D. The direction vector of the lines dotted with n, will be 0.
Thread starter 1 week ago
#5
(Original post by ghostwalker)
So, this is part b). And you should have a value for alpha from part a).

In red, will be parallel to the plane, as will the direction vectors of each of the lines.

I don't know what a "for/scalar product" is.

If you know n, then you can use any point on the plane to find D. The direction vector of the lines dotted with n, will be 0.
yes, I have a = 1.

ok, in red is the vector parallel to the plane, so that's the point both vectors meet/cross the plane?

sorry, I meant to dot product of the vector to the plane and the vector normal to the plane (n) will = 0?

so, is either or both (1i + -1j +2z) and (ai + 3j + 2z) a point on the line?
0
1 week ago
#6
(Original post by Maths&physics)
yes, I have a = 1.

ok, in red is the vector parallel to the plane, so that's the point both vectors meet/cross the plane?
In red is a vector parallel to the plane.

The rest of the line doesn't make sense in relation to that.

sorry, I meant to dot product of the vector to the plane and the vector normal to the plane (n) will = 0?
What do you mean by "the vector to the plane".

The dot product of n with any vector in the plane or parallel to a vector in the plane will be zero.

so, is either or both (1i + -1j +2z) and (ai + 3j + 2z) a point on the line?
First one defines a point on l1 and second one is a point on l2.
Correction: Second is the direction vector for the line l2, and not a point on l2

Please think about what you're trying to say - I am really struggling to make sense of your posts relating to this question.
Last edited by ghostwalker; 1 week ago
Thread starter 1 week ago
#7
(Original post by ghostwalker)
In red is a vector parallel to the plane.

The rest of the line doesn't make sense in relation to that.

What do you mean by "the vector to the plane".

The dot product of n with any vector in the plane or parallel to a vector in the plane will be zero.

First one defines a point on l1 and second one is a point on l2.

Please think about what you're trying to say - I am really struggling to make sense of your posts relating to this question.
the red vector is parallel to the plane/the vector between the points where L1 and L2 meet the plane [ (1i + -1j +2z) and (ai + 3j + 2z)] - represented by the blue and green circles respectively.

vector n is parallel to both the plane and the red vector - the dot product of vector n and the red vector = 0 => this is how we find n.

the orange dot is where they interest but now I'm confused because this is also suppose to be the place they intersect the plane.
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Thread starter 1 week ago
#8
bump
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Thread starter 1 week ago
#9
(Original post by ghostwalker)
In red is a vector parallel to the plane.

The rest of the line doesn't make sense in relation to that.

What do you mean by "the vector to the plane".

The dot product of n with any vector in the plane or parallel to a vector in the plane will be zero.

First one defines a point on l1 and second one is a point on l2.

Please think about what you're trying to say - I am really struggling to make sense of your posts relating to this question.
are (1i + -1j +2z) and (ai + 3j + 2z) points on the plane?
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1 week ago
#10
(Original post by Maths&physics)
are (1i + -1j +2z) and (ai + 3j + 2z) points on the plane?
A couple of things to note:
1) You need to go back and reconsider your diagram, as the question clearly states that the plane contains L1 and L2 whereas you have shown the lines passing through the plane.
2) You seem to think that the known point on L2 is (ai + 3j + 2z) whereas it is in fact (ai - 4j). Furthermore the value of a is known from earlier in the question, so you can and should use the known value.
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Thread starter 1 week ago
#11
(Original post by old_engineer)
A couple of things to note:
1) You need to go back and reconsider your diagram, as the question clearly states that the plane contains L1 and L2 whereas you have shown the lines passing through the plane.
2) You seem to think that the known point on L2 is (ai + 3j + 2z) whereas it is in fact (ai - 4j). Furthermore the value of a is known from earlier in the question, so you can and should use the known value.
ok, so, that makes sense - they are on the plane and not passing through. so would both direction vectors of the lines, each be perpendicular to n?

sorry, I realised that mistake. yeah, the value for a = 1. so, the vector between both of the known points on the lines, would also be parallel to the plane?
Last edited by Maths&physics; 1 week ago
0
1 week ago
#12
(Original post by Maths&physics)
ok, so, that makes sense - they are on the plane and not passing through. so would both direction vectors of the lines, each be parallel to n?

sorry, I realised that mistake. yeah, the value for a = 1. so, the vector between both of the known points on the lines, would also be parallel to the plane?
Direction vectors of lines in a plane are perpendicular to the normal n, not parallel.
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Thread starter 1 week ago
#13
(Original post by DFranklin)
Direction vectors of lines in a plane are perpendicular to the normal n, not parallel.
sorry, thats what I meant. So could I use either direction vector from both lines to find n?

and would they give me the same answer because n is traveling in the same direction?
Last edited by Maths&physics; 1 week ago
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1 week ago
#14
(Original post by Maths&physics)
sorry, thats what I meant. So could I use either direction vector from both lines to find n?

and would they give me the same answer because n is traveling in the same direction?
You need to use both direction vectors. n is not traveling in the same direction - it's perpendicular.

You don't seem to understand the difference between parallel and perpendicular.
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1 week ago
#15
(Original post by Maths&physics)
sorry, thats what I meant. So could I use either direction vector from both lines to find n?

and would they give me the same answer because n is traveling in the same direction?
The most direct way of finding n, the direction vector perpendicular to the plane, is to use the direction vectors of L1 and L2. n must be perpendicular to both the line direction vectors.
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Thread starter 1 week ago
#16
(Original post by old_engineer)
The most direct way of finding n, the direction vector perpendicular to the plane, is to use the direction vectors of L1 and L2. n must be perpendicular to both the line direction vectors.
ok, for L1: (-1.x) + (3.y) + (4.z) = 0

for L2: (0.x) + (3.y) + (2.z) = 0

and solve for x y and z?
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1 week ago
#17
(Original post by Maths&physics)
ok, for L1: (-1.x) + (3.y) + (4.z) = 0

for L2: (0.x) + (3.y) + (2.z) = 0

and solve for x y and z?
Yes (noting I haven't actually checked those are the correct direction vectors). And you need to find a solution *other* than x=y=z=0.

Do you know what the vector cross-product is?
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Thread starter 1 week ago
#18
(Original post by old_engineer)
The most direct way of finding n, the direction vector perpendicular to the plane, is to use the direction vectors of L1 and L2. n must be perpendicular to both the line direction vectors.
ok, for L1: (-1.x) + (3.y) + (4.z) = 0

for L2: (0.x) + (3.y) + (2.z) = 0

and solve for x y and z?

ok: (-1.x) + (3.y) + (4.z) = 0 => x = 3y + 4z

(0.x) + (3.y) + (2.z) = 0 => y = -(2/3)z

sub L2 into L1

2z = x

therefore, x is double z. vectors have direction, so, I chose x = 6, therefore, y = -2 and z = 3 ???
Last edited by Maths&physics; 1 week ago
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Thread starter 1 week ago
#19
(Original post by DFranklin)
Yes (noting I haven't actually checked those are the correct direction vectors). And you need to find a solution *other* than x=y=z=0.

Do you know what the vector cross-product is?
yes, I know and thats where I got the dot product from.
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1 week ago
#20
(Original post by Maths&physics)
.....

therefore, x is double z. vectors have direction, so, I chose x = 6, therefore, y = -2 and z = 3 ???
Yes n = (6i - 2j + 3k) or any multiple of it.
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