How to obtain the equation of a plane which intersects with a given plane in a given line

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Given the plane $2x + y - 3z = 1$.
How to obtain the equation of a plane which intersects this plane in the line
$$r=left(
beginarrayc1\2\1\endarrayright)+tleft(
beginarrayc1\-2\0\endarrayright)?$$

I have assumed that the other plane is $ax+by+cz=d$. Then the line satisfies this plane $$a(1+t)+b(2-2t)+c=d.$$
But I can't proceed from here.

calculus vector-analysis

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asked Aug 1 at 13:43

Sulayman

1807

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up vote
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down vote

favorite

Given the plane $2x + y - 3z = 1$.
How to obtain the equation of a plane which intersects this plane in the line
$$r=left(
beginarrayc1\2\1\endarrayright)+tleft(
beginarrayc1\-2\0\endarrayright)?$$

I have assumed that the other plane is $ax+by+cz=d$. Then the line satisfies this plane $$a(1+t)+b(2-2t)+c=d.$$
But I can't proceed from here.

calculus vector-analysis

share|cite|improve this question

asked Aug 1 at 13:43

Sulayman

1807

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Given the plane $2x + y - 3z = 1$.
How to obtain the equation of a plane which intersects this plane in the line
$$r=left(
beginarrayc1\2\1\endarrayright)+tleft(
beginarrayc1\-2\0\endarrayright)?$$

I have assumed that the other plane is $ax+by+cz=d$. Then the line satisfies this plane $$a(1+t)+b(2-2t)+c=d.$$
But I can't proceed from here.

calculus vector-analysis

share|cite|improve this question

asked Aug 1 at 13:43

Sulayman

1807

Given the plane $2x + y - 3z = 1$.
How to obtain the equation of a plane which intersects this plane in the line
$$r=left(
beginarrayc1\2\1\endarrayright)+tleft(
beginarrayc1\-2\0\endarrayright)?$$

I have assumed that the other plane is $ax+by+cz=d$. Then the line satisfies this plane $$a(1+t)+b(2-2t)+c=d.$$
But I can't proceed from here.

calculus vector-analysis

share|cite|improve this question

asked Aug 1 at 13:43

Sulayman

1807

share|cite|improve this question

share|cite|improve this question

asked Aug 1 at 13:43

Sulayman

1807

asked Aug 1 at 13:43

Sulayman

1807

asked Aug 1 at 13:43

Sulayman

1807

1807

add a comment | 

add a comment | 

7 Answers
7

active

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up vote
3
down vote



accepted

I will follow your idea. From $$a(1+t)+b(2-2t)+c=d$$ you get

$$at-2bt+a+2b+c=d.$$ To have the equality it is obvious that $a=2b.$ So you have the relation

$$4b+c=d.$$

Thus, any plane of the form

$$2bx+by+cz=4b+c$$ solves the problem. (Well, you have to exclude $2x+y-3z=1$.)

share|cite|improve this answer

edited Aug 2 at 7:58

answered Aug 1 at 14:20

mfl

22.9k11837

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I got hung up on “To have equality it is obvious that $a=2b$.” But now I see why: the equation $at-2bt+a+2b+c=d$ must be satisfied for all $t$. The only way this can happen is if the coefficient of $t$ is zero.
  – Matthew Leingang
  Aug 1 at 14:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MatthewLeingang You are right. That's the reason.
  – mfl
  Aug 1 at 14:39
  

  
  
  
  

add a comment | 

up vote
1
down vote

If all you know is that the line $r$ lies in the desired plane, then the plane is not unique. If we find one such plane, we can rotate it about the line $r$ to obtain another plane in which $r$ lies. This means that there is a one-parameter family of planes that could solve the problem.

The fact that the line $r$ lies in another plane is immaterial to the question, unless for some reason $r$ doesn't actually lie in the other plane (in which case there is no plane which intersects the other plane in $r$.)

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answered Aug 1 at 13:54

Michael Seifert

4,449623

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up vote
1
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A plane $P$ is an answer two your question if and only if:

1. $P$ isn't the plane $2x+y-3y=1$;


2. $rsubset P$.

Now, a plane $P$ defined by $ax+by+cz=d$ contains $r$ if and only if $(1,2,1),(2,0,1)in P$. This means that$$left{beginarrayla+2b+c=d\2a+c=d,endarrayright.$$which is equivalent to $a=2b$ and $c=-4b+d$. So, the answer is: the planes of the form$$2bx+by+(-4b+d)z=d,$$for which there's no $lambdainmathbb R$ such that $lambda(2,1,-3,1)=(2b,b,-4b+d,d)$.

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answered Aug 1 at 14:04

José Carlos Santos

112k1696172

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For the new plane we need that the normal vector $vec n=(a,b,c)$ is orthogonal to the given line that is

$$(a,b,c)cdot (1,-2,0)=0 implies (a,b,c)=(2,1,0)$$

and we obtain

$$2x+y+d=0$$

then we can find $d$ by the condition that $P=(1,2,1)$ belongs to the plane, that is

$$2+2+d=0implies d=-4$$

therefore

$$2x+y-4=0$$

is a possible solution.

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edited Aug 1 at 14:32

answered Aug 1 at 13:54

gimusi

63.9k73480

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For example, take
$$(x,y,z)=(1,2,1)+t(1,-2,0)+s(1,0,1),$$
which gives $$2x+y-2z-2=0.$$

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answered Aug 1 at 13:46

Michael Rozenberg

87.4k1577179

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0
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Try to think geometrically rather than algebraically. You're given a plane $Pi$; think about this plane as determined by a reference point and a normal vector $mathbfn$. You're also given a line $ell$ within that plane; that's determined by a reference point and a direction vector $mathbfv$. Since $ell$ is contained in $Pi$, we can choose the reference points of each to be the same point $P$, and we know that $mathbfn$ and $mathbfv$ are orthogonal.

We want another plane $Pi'$; again, it's determined by a reference point and normal vector $mathbfn'$. It's required to include $ell$, so it must contain $P$, and $mathbfn'$ must be orthogonal to $mathbfv$ too. We might as well use $P$ as the reference point for $Pi'$. As long as $mathbfn'$ is not parallel to $mathbfn$ (otherwise the planes coincide), their intersection will be $ell$ alone.

It remains to find a vector $mathbfn'$ that is perpendicular to $mathbfv$ and not parallel to $mathbfn$. Now it's an algebra problem, but let's keep it in terms of vector algebra. There are infinitely many choices, but one convenient one is $mathbfn' = mathbfn times mathbfv$. This is nonzero because $mathbfn$ and $mathbfv$ are nonzero orthogonal vectors, and it's orthogonal to both $mathbfn$ and $mathbfv$. In particular, it's not parallel to $mathbfn$.

Therefore a solution to your problem is the plane through $P$ normal to $mathbfn times mathbfv$, where $P$ is a point on $ell$, $mathbfv$ is a direction vector of $ell$, and $mathbfn$ is a normal vector to $Pi$. And we did this without any messy algebra!

But you probably need an equation for this plane. Remember $P$ can be any point on $ell$. Since $ell$ is parametrized by a vector function $mathbfr(t)$, a convenient point to call $P$ is the tip of $mathbfr(0)$. That is, let $P = (1,2,1)$. Since $Pi$ has equation $2x + y - 3z = 1$, the vector of coefficients $mathbfn = left<2,1,-3right>$ is a normal vector to $Pi$. Finally, a direction vector $mathbfv$ to $ell$ can be found by taking $mathbfr'(t) = left<1,-2,0right>$. Therefore,
$$
mathbfn' = mathbfn times mathbfv = left<-6,-3,-5right>
$$
And $Pi'$ has equation
beginalign*
-6(x-1) - 3(y-2) - 5(z-1) &= 0
\implies
-6x -3y - 5z &= -17
\implies
6x + 3y + 5z &= 17
endalign*

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answered Aug 1 at 14:30

Matthew Leingang

14.9k12042

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There are infinitely many such planes e.g. one is the plane itself and the other is the perpendicular plane. If we denote the characteristic vector of the unknown plane by $(a,b,c)$ it's obvious that it should be normal on the characteristic vector of the given line, (1,-2,0) which means that $$a-2b=0$$so the vector is $$(2b,b,c)$$ but the plane also includes (1,2,1) which means that $$2bx+by+cz=4b+c$$if $bne0$ we have $$2x+y+lambda z=4+lambdaquad,quad lambdainBbb R$$which is one set of planes. The other plane is $z=1$

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answered Aug 2 at 10:02

Mostafa Ayaz

8,5183630

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7 Answers
7

active

oldest

votes

7 Answers
7

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

I will follow your idea. From $$a(1+t)+b(2-2t)+c=d$$ you get

$$at-2bt+a+2b+c=d.$$ To have the equality it is obvious that $a=2b.$ So you have the relation

$$4b+c=d.$$

Thus, any plane of the form

$$2bx+by+cz=4b+c$$ solves the problem. (Well, you have to exclude $2x+y-3z=1$.)

share|cite|improve this answer

edited Aug 2 at 7:58

answered Aug 1 at 14:20

mfl

22.9k11837

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I got hung up on “To have equality it is obvious that $a=2b$.” But now I see why: the equation $at-2bt+a+2b+c=d$ must be satisfied for all $t$. The only way this can happen is if the coefficient of $t$ is zero.
  – Matthew Leingang
  Aug 1 at 14:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MatthewLeingang You are right. That's the reason.
  – mfl
  Aug 1 at 14:39
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

I will follow your idea. From $$a(1+t)+b(2-2t)+c=d$$ you get

$$at-2bt+a+2b+c=d.$$ To have the equality it is obvious that $a=2b.$ So you have the relation

$$4b+c=d.$$

Thus, any plane of the form

$$2bx+by+cz=4b+c$$ solves the problem. (Well, you have to exclude $2x+y-3z=1$.)

share|cite|improve this answer

edited Aug 2 at 7:58

answered Aug 1 at 14:20

mfl

22.9k11837

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I got hung up on “To have equality it is obvious that $a=2b$.” But now I see why: the equation $at-2bt+a+2b+c=d$ must be satisfied for all $t$. The only way this can happen is if the coefficient of $t$ is zero.
  – Matthew Leingang
  Aug 1 at 14:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MatthewLeingang You are right. That's the reason.
  – mfl
  Aug 1 at 14:39
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

I will follow your idea. From $$a(1+t)+b(2-2t)+c=d$$ you get

$$at-2bt+a+2b+c=d.$$ To have the equality it is obvious that $a=2b.$ So you have the relation

$$4b+c=d.$$

Thus, any plane of the form

$$2bx+by+cz=4b+c$$ solves the problem. (Well, you have to exclude $2x+y-3z=1$.)

share|cite|improve this answer

edited Aug 2 at 7:58

answered Aug 1 at 14:20

mfl

22.9k11837

I will follow your idea. From $$a(1+t)+b(2-2t)+c=d$$ you get

$$at-2bt+a+2b+c=d.$$ To have the equality it is obvious that $a=2b.$ So you have the relation

$$4b+c=d.$$

Thus, any plane of the form

$$2bx+by+cz=4b+c$$ solves the problem. (Well, you have to exclude $2x+y-3z=1$.)

share|cite|improve this answer

edited Aug 2 at 7:58

answered Aug 1 at 14:20

mfl

22.9k11837

share|cite|improve this answer

share|cite|improve this answer

edited Aug 2 at 7:58

edited Aug 2 at 7:58

edited Aug 2 at 7:58

answered Aug 1 at 14:20

mfl

22.9k11837

answered Aug 1 at 14:20

mfl

22.9k11837

answered Aug 1 at 14:20

mfl

22.9k11837

22.9k11837

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I got hung up on “To have equality it is obvious that $a=2b$.” But now I see why: the equation $at-2bt+a+2b+c=d$ must be satisfied for all $t$. The only way this can happen is if the coefficient of $t$ is zero.
  – Matthew Leingang
  Aug 1 at 14:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MatthewLeingang You are right. That's the reason.
  – mfl
  Aug 1 at 14:39
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I got hung up on “To have equality it is obvious that $a=2b$.” But now I see why: the equation $at-2bt+a+2b+c=d$ must be satisfied for all $t$. The only way this can happen is if the coefficient of $t$ is zero.
  – Matthew Leingang
  Aug 1 at 14:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MatthewLeingang You are right. That's the reason.
  – mfl
  Aug 1 at 14:39
  

  
  
  
  

I got hung up on “To have equality it is obvious that $a=2b$.” But now I see why: the equation $at-2bt+a+2b+c=d$ must be satisfied for all $t$. The only way this can happen is if the coefficient of $t$ is zero.
– Matthew Leingang
Aug 1 at 14:36

I got hung up on “To have equality it is obvious that $a=2b$.” But now I see why: the equation $at-2bt+a+2b+c=d$ must be satisfied for all $t$. The only way this can happen is if the coefficient of $t$ is zero.
– Matthew Leingang
Aug 1 at 14:36

@MatthewLeingang You are right. That's the reason.
– mfl
Aug 1 at 14:39

@MatthewLeingang You are right. That's the reason.
– mfl
Aug 1 at 14:39

add a comment | 

up vote
1
down vote

If all you know is that the line $r$ lies in the desired plane, then the plane is not unique. If we find one such plane, we can rotate it about the line $r$ to obtain another plane in which $r$ lies. This means that there is a one-parameter family of planes that could solve the problem.

The fact that the line $r$ lies in another plane is immaterial to the question, unless for some reason $r$ doesn't actually lie in the other plane (in which case there is no plane which intersects the other plane in $r$.)

share|cite|improve this answer

answered Aug 1 at 13:54

Michael Seifert

4,449623

add a comment | 

up vote
1
down vote

If all you know is that the line $r$ lies in the desired plane, then the plane is not unique. If we find one such plane, we can rotate it about the line $r$ to obtain another plane in which $r$ lies. This means that there is a one-parameter family of planes that could solve the problem.

The fact that the line $r$ lies in another plane is immaterial to the question, unless for some reason $r$ doesn't actually lie in the other plane (in which case there is no plane which intersects the other plane in $r$.)

share|cite|improve this answer

answered Aug 1 at 13:54

Michael Seifert

4,449623

add a comment | 

up vote
1
down vote

up vote
1
down vote

If all you know is that the line $r$ lies in the desired plane, then the plane is not unique. If we find one such plane, we can rotate it about the line $r$ to obtain another plane in which $r$ lies. This means that there is a one-parameter family of planes that could solve the problem.

The fact that the line $r$ lies in another plane is immaterial to the question, unless for some reason $r$ doesn't actually lie in the other plane (in which case there is no plane which intersects the other plane in $r$.)

share|cite|improve this answer

answered Aug 1 at 13:54

Michael Seifert

4,449623

If all you know is that the line $r$ lies in the desired plane, then the plane is not unique. If we find one such plane, we can rotate it about the line $r$ to obtain another plane in which $r$ lies. This means that there is a one-parameter family of planes that could solve the problem.

The fact that the line $r$ lies in another plane is immaterial to the question, unless for some reason $r$ doesn't actually lie in the other plane (in which case there is no plane which intersects the other plane in $r$.)

share|cite|improve this answer

answered Aug 1 at 13:54

Michael Seifert

4,449623

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share|cite|improve this answer

answered Aug 1 at 13:54

Michael Seifert

4,449623

answered Aug 1 at 13:54

Michael Seifert

4,449623

answered Aug 1 at 13:54

Michael Seifert

4,449623

4,449623

add a comment | 

add a comment | 

up vote
1
down vote

A plane $P$ is an answer two your question if and only if:

1. $P$ isn't the plane $2x+y-3y=1$;


2. $rsubset P$.

Now, a plane $P$ defined by $ax+by+cz=d$ contains $r$ if and only if $(1,2,1),(2,0,1)in P$. This means that$$left{beginarrayla+2b+c=d\2a+c=d,endarrayright.$$which is equivalent to $a=2b$ and $c=-4b+d$. So, the answer is: the planes of the form$$2bx+by+(-4b+d)z=d,$$for which there's no $lambdainmathbb R$ such that $lambda(2,1,-3,1)=(2b,b,-4b+d,d)$.

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answered Aug 1 at 14:04

José Carlos Santos

112k1696172

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up vote
1
down vote

A plane $P$ is an answer two your question if and only if:

1. $P$ isn't the plane $2x+y-3y=1$;


2. $rsubset P$.

Now, a plane $P$ defined by $ax+by+cz=d$ contains $r$ if and only if $(1,2,1),(2,0,1)in P$. This means that$$left{beginarrayla+2b+c=d\2a+c=d,endarrayright.$$which is equivalent to $a=2b$ and $c=-4b+d$. So, the answer is: the planes of the form$$2bx+by+(-4b+d)z=d,$$for which there's no $lambdainmathbb R$ such that $lambda(2,1,-3,1)=(2b,b,-4b+d,d)$.

share|cite|improve this answer

answered Aug 1 at 14:04

José Carlos Santos

112k1696172

add a comment | 

up vote
1
down vote

up vote
1
down vote

A plane $P$ is an answer two your question if and only if:

1. $P$ isn't the plane $2x+y-3y=1$;


2. $rsubset P$.

Now, a plane $P$ defined by $ax+by+cz=d$ contains $r$ if and only if $(1,2,1),(2,0,1)in P$. This means that$$left{beginarrayla+2b+c=d\2a+c=d,endarrayright.$$which is equivalent to $a=2b$ and $c=-4b+d$. So, the answer is: the planes of the form$$2bx+by+(-4b+d)z=d,$$for which there's no $lambdainmathbb R$ such that $lambda(2,1,-3,1)=(2b,b,-4b+d,d)$.

share|cite|improve this answer

answered Aug 1 at 14:04

José Carlos Santos

112k1696172

A plane $P$ is an answer two your question if and only if:

1. $P$ isn't the plane $2x+y-3y=1$;


2. $rsubset P$.

Now, a plane $P$ defined by $ax+by+cz=d$ contains $r$ if and only if $(1,2,1),(2,0,1)in P$. This means that$$left{beginarrayla+2b+c=d\2a+c=d,endarrayright.$$which is equivalent to $a=2b$ and $c=-4b+d$. So, the answer is: the planes of the form$$2bx+by+(-4b+d)z=d,$$for which there's no $lambdainmathbb R$ such that $lambda(2,1,-3,1)=(2b,b,-4b+d,d)$.

share|cite|improve this answer

answered Aug 1 at 14:04

José Carlos Santos

112k1696172

share|cite|improve this answer

share|cite|improve this answer

answered Aug 1 at 14:04

José Carlos Santos

112k1696172

answered Aug 1 at 14:04

José Carlos Santos

112k1696172

answered Aug 1 at 14:04

José Carlos Santos

112k1696172

112k1696172

add a comment | 

add a comment | 

up vote
1
down vote

For the new plane we need that the normal vector $vec n=(a,b,c)$ is orthogonal to the given line that is

$$(a,b,c)cdot (1,-2,0)=0 implies (a,b,c)=(2,1,0)$$

and we obtain

$$2x+y+d=0$$

then we can find $d$ by the condition that $P=(1,2,1)$ belongs to the plane, that is

$$2+2+d=0implies d=-4$$

therefore

$$2x+y-4=0$$

is a possible solution.

share|cite|improve this answer

edited Aug 1 at 14:32

answered Aug 1 at 13:54

gimusi

63.9k73480

add a comment | 

up vote
1
down vote

For the new plane we need that the normal vector $vec n=(a,b,c)$ is orthogonal to the given line that is

$$(a,b,c)cdot (1,-2,0)=0 implies (a,b,c)=(2,1,0)$$

and we obtain

$$2x+y+d=0$$

then we can find $d$ by the condition that $P=(1,2,1)$ belongs to the plane, that is

$$2+2+d=0implies d=-4$$

therefore

$$2x+y-4=0$$

is a possible solution.

share|cite|improve this answer

edited Aug 1 at 14:32

answered Aug 1 at 13:54

gimusi

63.9k73480

add a comment | 

up vote
1
down vote

up vote
1
down vote

For the new plane we need that the normal vector $vec n=(a,b,c)$ is orthogonal to the given line that is

$$(a,b,c)cdot (1,-2,0)=0 implies (a,b,c)=(2,1,0)$$

and we obtain

$$2x+y+d=0$$

then we can find $d$ by the condition that $P=(1,2,1)$ belongs to the plane, that is

$$2+2+d=0implies d=-4$$

therefore

$$2x+y-4=0$$

is a possible solution.

share|cite|improve this answer

edited Aug 1 at 14:32

answered Aug 1 at 13:54

gimusi

63.9k73480

For the new plane we need that the normal vector $vec n=(a,b,c)$ is orthogonal to the given line that is

$$(a,b,c)cdot (1,-2,0)=0 implies (a,b,c)=(2,1,0)$$

and we obtain

$$2x+y+d=0$$

then we can find $d$ by the condition that $P=(1,2,1)$ belongs to the plane, that is

$$2+2+d=0implies d=-4$$

therefore

$$2x+y-4=0$$

is a possible solution.

share|cite|improve this answer

edited Aug 1 at 14:32

answered Aug 1 at 13:54

gimusi

63.9k73480

share|cite|improve this answer

share|cite|improve this answer

edited Aug 1 at 14:32

edited Aug 1 at 14:32

edited Aug 1 at 14:32

answered Aug 1 at 13:54

gimusi

63.9k73480

answered Aug 1 at 13:54

gimusi

63.9k73480

answered Aug 1 at 13:54

gimusi

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63.9k73480

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For example, take
$$(x,y,z)=(1,2,1)+t(1,-2,0)+s(1,0,1),$$
which gives $$2x+y-2z-2=0.$$

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answered Aug 1 at 13:46

Michael Rozenberg

87.4k1577179

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For example, take
$$(x,y,z)=(1,2,1)+t(1,-2,0)+s(1,0,1),$$
which gives $$2x+y-2z-2=0.$$

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answered Aug 1 at 13:46

Michael Rozenberg

87.4k1577179

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up vote
0
down vote

up vote
0
down vote

For example, take
$$(x,y,z)=(1,2,1)+t(1,-2,0)+s(1,0,1),$$
which gives $$2x+y-2z-2=0.$$

share|cite|improve this answer

answered Aug 1 at 13:46

Michael Rozenberg

87.4k1577179

For example, take
$$(x,y,z)=(1,2,1)+t(1,-2,0)+s(1,0,1),$$
which gives $$2x+y-2z-2=0.$$

share|cite|improve this answer

answered Aug 1 at 13:46

Michael Rozenberg

87.4k1577179

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share|cite|improve this answer

answered Aug 1 at 13:46

Michael Rozenberg

87.4k1577179

answered Aug 1 at 13:46

Michael Rozenberg

87.4k1577179

answered Aug 1 at 13:46

Michael Rozenberg

87.4k1577179

87.4k1577179

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add a comment | 

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Try to think geometrically rather than algebraically. You're given a plane $Pi$; think about this plane as determined by a reference point and a normal vector $mathbfn$. You're also given a line $ell$ within that plane; that's determined by a reference point and a direction vector $mathbfv$. Since $ell$ is contained in $Pi$, we can choose the reference points of each to be the same point $P$, and we know that $mathbfn$ and $mathbfv$ are orthogonal.

We want another plane $Pi'$; again, it's determined by a reference point and normal vector $mathbfn'$. It's required to include $ell$, so it must contain $P$, and $mathbfn'$ must be orthogonal to $mathbfv$ too. We might as well use $P$ as the reference point for $Pi'$. As long as $mathbfn'$ is not parallel to $mathbfn$ (otherwise the planes coincide), their intersection will be $ell$ alone.

It remains to find a vector $mathbfn'$ that is perpendicular to $mathbfv$ and not parallel to $mathbfn$. Now it's an algebra problem, but let's keep it in terms of vector algebra. There are infinitely many choices, but one convenient one is $mathbfn' = mathbfn times mathbfv$. This is nonzero because $mathbfn$ and $mathbfv$ are nonzero orthogonal vectors, and it's orthogonal to both $mathbfn$ and $mathbfv$. In particular, it's not parallel to $mathbfn$.

Therefore a solution to your problem is the plane through $P$ normal to $mathbfn times mathbfv$, where $P$ is a point on $ell$, $mathbfv$ is a direction vector of $ell$, and $mathbfn$ is a normal vector to $Pi$. And we did this without any messy algebra!

But you probably need an equation for this plane. Remember $P$ can be any point on $ell$. Since $ell$ is parametrized by a vector function $mathbfr(t)$, a convenient point to call $P$ is the tip of $mathbfr(0)$. That is, let $P = (1,2,1)$. Since $Pi$ has equation $2x + y - 3z = 1$, the vector of coefficients $mathbfn = left<2,1,-3right>$ is a normal vector to $Pi$. Finally, a direction vector $mathbfv$ to $ell$ can be found by taking $mathbfr'(t) = left<1,-2,0right>$. Therefore,
$$
mathbfn' = mathbfn times mathbfv = left<-6,-3,-5right>
$$
And $Pi'$ has equation
beginalign*
-6(x-1) - 3(y-2) - 5(z-1) &= 0
\implies
-6x -3y - 5z &= -17
\implies
6x + 3y + 5z &= 17
endalign*

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answered Aug 1 at 14:30

Matthew Leingang

14.9k12042

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up vote
0
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Try to think geometrically rather than algebraically. You're given a plane $Pi$; think about this plane as determined by a reference point and a normal vector $mathbfn$. You're also given a line $ell$ within that plane; that's determined by a reference point and a direction vector $mathbfv$. Since $ell$ is contained in $Pi$, we can choose the reference points of each to be the same point $P$, and we know that $mathbfn$ and $mathbfv$ are orthogonal.

We want another plane $Pi'$; again, it's determined by a reference point and normal vector $mathbfn'$. It's required to include $ell$, so it must contain $P$, and $mathbfn'$ must be orthogonal to $mathbfv$ too. We might as well use $P$ as the reference point for $Pi'$. As long as $mathbfn'$ is not parallel to $mathbfn$ (otherwise the planes coincide), their intersection will be $ell$ alone.

It remains to find a vector $mathbfn'$ that is perpendicular to $mathbfv$ and not parallel to $mathbfn$. Now it's an algebra problem, but let's keep it in terms of vector algebra. There are infinitely many choices, but one convenient one is $mathbfn' = mathbfn times mathbfv$. This is nonzero because $mathbfn$ and $mathbfv$ are nonzero orthogonal vectors, and it's orthogonal to both $mathbfn$ and $mathbfv$. In particular, it's not parallel to $mathbfn$.

Therefore a solution to your problem is the plane through $P$ normal to $mathbfn times mathbfv$, where $P$ is a point on $ell$, $mathbfv$ is a direction vector of $ell$, and $mathbfn$ is a normal vector to $Pi$. And we did this without any messy algebra!

But you probably need an equation for this plane. Remember $P$ can be any point on $ell$. Since $ell$ is parametrized by a vector function $mathbfr(t)$, a convenient point to call $P$ is the tip of $mathbfr(0)$. That is, let $P = (1,2,1)$. Since $Pi$ has equation $2x + y - 3z = 1$, the vector of coefficients $mathbfn = left<2,1,-3right>$ is a normal vector to $Pi$. Finally, a direction vector $mathbfv$ to $ell$ can be found by taking $mathbfr'(t) = left<1,-2,0right>$. Therefore,
$$
mathbfn' = mathbfn times mathbfv = left<-6,-3,-5right>
$$
And $Pi'$ has equation
beginalign*
-6(x-1) - 3(y-2) - 5(z-1) &= 0
\implies
-6x -3y - 5z &= -17
\implies
6x + 3y + 5z &= 17
endalign*

share|cite|improve this answer

answered Aug 1 at 14:30

Matthew Leingang

14.9k12042

add a comment | 

up vote
0
down vote

up vote
0
down vote

Try to think geometrically rather than algebraically. You're given a plane $Pi$; think about this plane as determined by a reference point and a normal vector $mathbfn$. You're also given a line $ell$ within that plane; that's determined by a reference point and a direction vector $mathbfv$. Since $ell$ is contained in $Pi$, we can choose the reference points of each to be the same point $P$, and we know that $mathbfn$ and $mathbfv$ are orthogonal.

We want another plane $Pi'$; again, it's determined by a reference point and normal vector $mathbfn'$. It's required to include $ell$, so it must contain $P$, and $mathbfn'$ must be orthogonal to $mathbfv$ too. We might as well use $P$ as the reference point for $Pi'$. As long as $mathbfn'$ is not parallel to $mathbfn$ (otherwise the planes coincide), their intersection will be $ell$ alone.

It remains to find a vector $mathbfn'$ that is perpendicular to $mathbfv$ and not parallel to $mathbfn$. Now it's an algebra problem, but let's keep it in terms of vector algebra. There are infinitely many choices, but one convenient one is $mathbfn' = mathbfn times mathbfv$. This is nonzero because $mathbfn$ and $mathbfv$ are nonzero orthogonal vectors, and it's orthogonal to both $mathbfn$ and $mathbfv$. In particular, it's not parallel to $mathbfn$.

Therefore a solution to your problem is the plane through $P$ normal to $mathbfn times mathbfv$, where $P$ is a point on $ell$, $mathbfv$ is a direction vector of $ell$, and $mathbfn$ is a normal vector to $Pi$. And we did this without any messy algebra!

But you probably need an equation for this plane. Remember $P$ can be any point on $ell$. Since $ell$ is parametrized by a vector function $mathbfr(t)$, a convenient point to call $P$ is the tip of $mathbfr(0)$. That is, let $P = (1,2,1)$. Since $Pi$ has equation $2x + y - 3z = 1$, the vector of coefficients $mathbfn = left<2,1,-3right>$ is a normal vector to $Pi$. Finally, a direction vector $mathbfv$ to $ell$ can be found by taking $mathbfr'(t) = left<1,-2,0right>$. Therefore,
$$
mathbfn' = mathbfn times mathbfv = left<-6,-3,-5right>
$$
And $Pi'$ has equation
beginalign*
-6(x-1) - 3(y-2) - 5(z-1) &= 0
\implies
-6x -3y - 5z &= -17
\implies
6x + 3y + 5z &= 17
endalign*

share|cite|improve this answer

answered Aug 1 at 14:30

Matthew Leingang

14.9k12042

Try to think geometrically rather than algebraically. You're given a plane $Pi$; think about this plane as determined by a reference point and a normal vector $mathbfn$. You're also given a line $ell$ within that plane; that's determined by a reference point and a direction vector $mathbfv$. Since $ell$ is contained in $Pi$, we can choose the reference points of each to be the same point $P$, and we know that $mathbfn$ and $mathbfv$ are orthogonal.

We want another plane $Pi'$; again, it's determined by a reference point and normal vector $mathbfn'$. It's required to include $ell$, so it must contain $P$, and $mathbfn'$ must be orthogonal to $mathbfv$ too. We might as well use $P$ as the reference point for $Pi'$. As long as $mathbfn'$ is not parallel to $mathbfn$ (otherwise the planes coincide), their intersection will be $ell$ alone.

It remains to find a vector $mathbfn'$ that is perpendicular to $mathbfv$ and not parallel to $mathbfn$. Now it's an algebra problem, but let's keep it in terms of vector algebra. There are infinitely many choices, but one convenient one is $mathbfn' = mathbfn times mathbfv$. This is nonzero because $mathbfn$ and $mathbfv$ are nonzero orthogonal vectors, and it's orthogonal to both $mathbfn$ and $mathbfv$. In particular, it's not parallel to $mathbfn$.

Therefore a solution to your problem is the plane through $P$ normal to $mathbfn times mathbfv$, where $P$ is a point on $ell$, $mathbfv$ is a direction vector of $ell$, and $mathbfn$ is a normal vector to $Pi$. And we did this without any messy algebra!

But you probably need an equation for this plane. Remember $P$ can be any point on $ell$. Since $ell$ is parametrized by a vector function $mathbfr(t)$, a convenient point to call $P$ is the tip of $mathbfr(0)$. That is, let $P = (1,2,1)$. Since $Pi$ has equation $2x + y - 3z = 1$, the vector of coefficients $mathbfn = left<2,1,-3right>$ is a normal vector to $Pi$. Finally, a direction vector $mathbfv$ to $ell$ can be found by taking $mathbfr'(t) = left<1,-2,0right>$. Therefore,
$$
mathbfn' = mathbfn times mathbfv = left<-6,-3,-5right>
$$
And $Pi'$ has equation
beginalign*
-6(x-1) - 3(y-2) - 5(z-1) &= 0
\implies
-6x -3y - 5z &= -17
\implies
6x + 3y + 5z &= 17
endalign*

share|cite|improve this answer

answered Aug 1 at 14:30

Matthew Leingang

14.9k12042

share|cite|improve this answer

share|cite|improve this answer

answered Aug 1 at 14:30

Matthew Leingang

14.9k12042

answered Aug 1 at 14:30

Matthew Leingang

14.9k12042

answered Aug 1 at 14:30

Matthew Leingang

14.9k12042

14.9k12042

add a comment | 

add a comment | 

up vote
0
down vote

There are infinitely many such planes e.g. one is the plane itself and the other is the perpendicular plane. If we denote the characteristic vector of the unknown plane by $(a,b,c)$ it's obvious that it should be normal on the characteristic vector of the given line, (1,-2,0) which means that $$a-2b=0$$so the vector is $$(2b,b,c)$$ but the plane also includes (1,2,1) which means that $$2bx+by+cz=4b+c$$if $bne0$ we have $$2x+y+lambda z=4+lambdaquad,quad lambdainBbb R$$which is one set of planes. The other plane is $z=1$

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answered Aug 2 at 10:02

Mostafa Ayaz

8,5183630

add a comment | 

up vote
0
down vote

There are infinitely many such planes e.g. one is the plane itself and the other is the perpendicular plane. If we denote the characteristic vector of the unknown plane by $(a,b,c)$ it's obvious that it should be normal on the characteristic vector of the given line, (1,-2,0) which means that $$a-2b=0$$so the vector is $$(2b,b,c)$$ but the plane also includes (1,2,1) which means that $$2bx+by+cz=4b+c$$if $bne0$ we have $$2x+y+lambda z=4+lambdaquad,quad lambdainBbb R$$which is one set of planes. The other plane is $z=1$

share|cite|improve this answer

answered Aug 2 at 10:02

Mostafa Ayaz

8,5183630

add a comment | 

up vote
0
down vote

up vote
0
down vote

There are infinitely many such planes e.g. one is the plane itself and the other is the perpendicular plane. If we denote the characteristic vector of the unknown plane by $(a,b,c)$ it's obvious that it should be normal on the characteristic vector of the given line, (1,-2,0) which means that $$a-2b=0$$so the vector is $$(2b,b,c)$$ but the plane also includes (1,2,1) which means that $$2bx+by+cz=4b+c$$if $bne0$ we have $$2x+y+lambda z=4+lambdaquad,quad lambdainBbb R$$which is one set of planes. The other plane is $z=1$

share|cite|improve this answer

answered Aug 2 at 10:02

Mostafa Ayaz

8,5183630

There are infinitely many such planes e.g. one is the plane itself and the other is the perpendicular plane. If we denote the characteristic vector of the unknown plane by $(a,b,c)$ it's obvious that it should be normal on the characteristic vector of the given line, (1,-2,0) which means that $$a-2b=0$$so the vector is $$(2b,b,c)$$ but the plane also includes (1,2,1) which means that $$2bx+by+cz=4b+c$$if $bne0$ we have $$2x+y+lambda z=4+lambdaquad,quad lambdainBbb R$$which is one set of planes. The other plane is $z=1$

share|cite|improve this answer

answered Aug 2 at 10:02

Mostafa Ayaz

8,5183630

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 10:02

Mostafa Ayaz

8,5183630

answered Aug 2 at 10:02

Mostafa Ayaz

8,5183630

answered Aug 2 at 10:02

Mostafa Ayaz

8,5183630

8,5183630

add a comment | 

add a comment | 

 

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