# FP3: intersection of planes

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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?!

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#2

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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?!

**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?!

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; 2 years ago

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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.

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?

**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?

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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#4

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thanks that's a tutorial I'm watching.

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

**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?

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

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)?

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(Original post by

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.

**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.

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?

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#6

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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?

**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?

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?

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?

**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; 2 years ago

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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

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.

**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.

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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**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.

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#10

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are (1i + -1j +2z) and (ai + 3j + 2z) points on the plane?

**Maths&physics**)are (1i + -1j +2z) and (ai + 3j + 2z) points on the plane?

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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(Original post by

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.

**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.

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; 2 years ago

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#12

(Original post by

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?

**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?

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(Original post by

Direction vectors of lines in a plane are perpendicular to the normal n, not parallel.

**DFranklin**)Direction vectors of lines in a plane are perpendicular to the normal n, not parallel.

and would they give me the same answer because n is traveling in the same direction?

Last edited by Maths&physics; 2 years ago

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#14

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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?

**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 don't seem to understand the difference between parallel and perpendicular.

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#15

**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?

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(Original post by

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.

**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.

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

and solve for x y and z?

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#17

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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?

**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?

Do you know what the vector cross-product is?

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**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.

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; 2 years ago

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(Original post by

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?

**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?

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#20

(Original post by

.....

therefore, x is double z. vectors have direction, so, I chose x = 6, therefore, y = -2 and z = 3 ???

**Maths&physics**).....

therefore, x is double z. vectors have direction, so, I chose x = 6, therefore, y = -2 and z = 3 ???

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