Mixing
Equation of plane parallel to $x$-axis.
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Problem in several editions of Larsen calculus texts:
Plane passes through $(1,-2,-1) $ and $(2,5,6)$ and is parallel to $x$- axis.
Solution given takes cross-product of vector between the points and vector $u = i$, claiming both lie in plane. How could unit vector $ (1,0,0) $ lie in plane parallel to $x$-axis?
Thank you.
vectors
edited Feb 23 '16 at 18:12
MickG
4,28731753
asked Sep 3 '15 at 20:32
Mark
612
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
up vote
1
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Problem in several editions of Larsen calculus texts:
Plane passes through $(1,-2,-1) $ and $(2,5,6)$ and is parallel to $x$- axis.
Solution given takes cross-product of vector between the points and vector $u = i$, claiming both lie in plane. How could unit vector $ (1,0,0) $ lie in plane parallel to $x$-axis?
Thank you.
vectors
edited Feb 23 '16 at 18:12
MickG
4,28731753
asked Sep 3 '15 at 20:32
Mark
612
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
1
It has to do with translation invariance of vectors. The vectors have the same properties at any position in space, so a vector travelling "along" the axis is always parallel to it, even if it is placed somewhere else in space.
– Terra Hyde
Sep 3 '15 at 20:35
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Problem in several editions of Larsen calculus texts:
Plane passes through $(1,-2,-1) $ and $(2,5,6)$ and is parallel to $x$- axis.
Solution given takes cross-product of vector between the points and vector $u = i$, claiming both lie in plane. How could unit vector $ (1,0,0) $ lie in plane parallel to $x$-axis?
Thank you.
vectors
edited Feb 23 '16 at 18:12
MickG
4,28731753
asked Sep 3 '15 at 20:32
Mark
612
Problem in several editions of Larsen calculus texts:
Plane passes through $(1,-2,-1) $ and $(2,5,6)$ and is parallel to $x$- axis.
Solution given takes cross-product of vector between the points and vector $u = i$, claiming both lie in plane. How could unit vector $ (1,0,0) $ lie in plane parallel to $x$-axis?
Thank you.
vectors
vectors
edited Feb 23 '16 at 18:12
MickG
4,28731753
asked Sep 3 '15 at 20:32
Mark
612
edited Feb 23 '16 at 18:12
MickG
4,28731753
asked Sep 3 '15 at 20:32
Mark
612
edited Feb 23 '16 at 18:12
MickG
4,28731753
edited Feb 23 '16 at 18:12
MickG
4,28731753
edited Feb 23 '16 at 18:12
MickG
4,28731753
4,28731753
asked Sep 3 '15 at 20:32
Mark
612
asked Sep 3 '15 at 20:32
Mark
612
asked Sep 3 '15 at 20:32
Mark
612
612
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
1
It has to do with translation invariance of vectors. The vectors have the same properties at any position in space, so a vector travelling "along" the axis is always parallel to it, even if it is placed somewhere else in space.
– Terra Hyde
Sep 3 '15 at 20:35
add a comment |
1
It has to do with translation invariance of vectors. The vectors have the same properties at any position in space, so a vector travelling "along" the axis is always parallel to it, even if it is placed somewhere else in space.
– Terra Hyde
Sep 3 '15 at 20:35
1
1
It has to do with translation invariance of vectors. The vectors have the same properties at any position in space, so a vector travelling "along" the axis is always parallel to it, even if it is placed somewhere else in space.
– Terra Hyde
Sep 3 '15 at 20:35
It has to do with translation invariance of vectors. The vectors have the same properties at any position in space, so a vector travelling "along" the axis is always parallel to it, even if it is placed somewhere else in space.
– Terra Hyde
Sep 3 '15 at 20:35
add a comment |
3 Answers
3
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oldest
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0
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By definition,the vector $i$ is parallel to all planes parallel to $ XY$ and $XZ$ planes, or any other plane rotated about x-axis and translated arbitrarily thereafter.
answered Sep 3 '15 at 20:44
Narasimham
20.4k52158
add a comment |
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0
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Probably you are confusing an Orientated Segment with a Vector.
An orientated segment (sometimes called applied vector in physics) is defined by the ordered couple of two points (A,B), which are the ends of the segment, and in which the direction from $A$ to $B$ is chosen.
A vector $mathbf v$, i.e. an ordered D-uple, can be thought as individuating the "class" of all the orientated segments such that
$$
{bf v} = mathop {OB}limits^ to - mathop {OA}limits^ to
$$
which can be otherwise written as
$$
mathop {OB}limits^ to = mathop {OA}limits^ to + {bf v}
$$
(vector $bf v$ "applied" in $A$).
Thus:
- the or. segment $((0,0,0),(1,0,0))$ lies on the $x$ axis but does not lie on a plane parallel (and distant) from it;
- the vector $(1,0,0)$ is parallel to the $x$ axis and to any a plane parallel from it.
answered Aug 7 '17 at 9:56
G Cab
16.9k31237
add a comment |
up vote
-1
down vote
Well just think, point $(1,0,0)$ lies on the $x$-axis, not on $y$ or $z$-axis. Hence your vector is parallel to $x$-axis and therefore lies in an arbitrary plane parallel to $x$-axis. 3D is just the working of your brain :)
edited Oct 19 '16 at 16:21
zar
2,74631837
answered Oct 19 '16 at 14:48
Kismat
1
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
By definition,the vector $i$ is parallel to all planes parallel to $ XY$ and $XZ$ planes, or any other plane rotated about x-axis and translated arbitrarily thereafter.
answered Sep 3 '15 at 20:44
Narasimham
20.4k52158
add a comment |
up vote
0
down vote
By definition,the vector $i$ is parallel to all planes parallel to $ XY$ and $XZ$ planes, or any other plane rotated about x-axis and translated arbitrarily thereafter.
answered Sep 3 '15 at 20:44
Narasimham
20.4k52158
add a comment |
up vote
0
down vote
up vote
0
down vote
By definition,the vector $i$ is parallel to all planes parallel to $ XY$ and $XZ$ planes, or any other plane rotated about x-axis and translated arbitrarily thereafter.
answered Sep 3 '15 at 20:44
Narasimham
20.4k52158
By definition,the vector $i$ is parallel to all planes parallel to $ XY$ and $XZ$ planes, or any other plane rotated about x-axis and translated arbitrarily thereafter.
answered Sep 3 '15 at 20:44
Narasimham
20.4k52158
answered Sep 3 '15 at 20:44
Narasimham
20.4k52158
answered Sep 3 '15 at 20:44
Narasimham
20.4k52158
answered Sep 3 '15 at 20:44
Narasimham
20.4k52158
20.4k52158
add a comment |
add a comment |
up vote
0
down vote
Probably you are confusing an Orientated Segment with a Vector.
An orientated segment (sometimes called applied vector in physics) is defined by the ordered couple of two points (A,B), which are the ends of the segment, and in which the direction from $A$ to $B$ is chosen.
A vector $mathbf v$, i.e. an ordered D-uple, can be thought as individuating the "class" of all the orientated segments such that
$$
{bf v} = mathop {OB}limits^ to - mathop {OA}limits^ to
$$
which can be otherwise written as
$$
mathop {OB}limits^ to = mathop {OA}limits^ to + {bf v}
$$
(vector $bf v$ "applied" in $A$).
Thus:
- the or. segment $((0,0,0),(1,0,0))$ lies on the $x$ axis but does not lie on a plane parallel (and distant) from it;
- the vector $(1,0,0)$ is parallel to the $x$ axis and to any a plane parallel from it.
answered Aug 7 '17 at 9:56
G Cab
16.9k31237
add a comment |
up vote
0
down vote
Probably you are confusing an Orientated Segment with a Vector.
An orientated segment (sometimes called applied vector in physics) is defined by the ordered couple of two points (A,B), which are the ends of the segment, and in which the direction from $A$ to $B$ is chosen.
A vector $mathbf v$, i.e. an ordered D-uple, can be thought as individuating the "class" of all the orientated segments such that
$$
{bf v} = mathop {OB}limits^ to - mathop {OA}limits^ to
$$
which can be otherwise written as
$$
mathop {OB}limits^ to = mathop {OA}limits^ to + {bf v}
$$
(vector $bf v$ "applied" in $A$).
Thus:
- the or. segment $((0,0,0),(1,0,0))$ lies on the $x$ axis but does not lie on a plane parallel (and distant) from it;
- the vector $(1,0,0)$ is parallel to the $x$ axis and to any a plane parallel from it.
answered Aug 7 '17 at 9:56
G Cab
16.9k31237
add a comment |
up vote
0
down vote
up vote
0
down vote
Probably you are confusing an Orientated Segment with a Vector.
An orientated segment (sometimes called applied vector in physics) is defined by the ordered couple of two points (A,B), which are the ends of the segment, and in which the direction from $A$ to $B$ is chosen.
A vector $mathbf v$, i.e. an ordered D-uple, can be thought as individuating the "class" of all the orientated segments such that
$$
{bf v} = mathop {OB}limits^ to - mathop {OA}limits^ to
$$
which can be otherwise written as
$$
mathop {OB}limits^ to = mathop {OA}limits^ to + {bf v}
$$
(vector $bf v$ "applied" in $A$).
Thus:
- the or. segment $((0,0,0),(1,0,0))$ lies on the $x$ axis but does not lie on a plane parallel (and distant) from it;
- the vector $(1,0,0)$ is parallel to the $x$ axis and to any a plane parallel from it.
answered Aug 7 '17 at 9:56
G Cab
16.9k31237
Probably you are confusing an Orientated Segment with a Vector.
An orientated segment (sometimes called applied vector in physics) is defined by the ordered couple of two points (A,B), which are the ends of the segment, and in which the direction from $A$ to $B$ is chosen.
A vector $mathbf v$, i.e. an ordered D-uple, can be thought as individuating the "class" of all the orientated segments such that
$$
{bf v} = mathop {OB}limits^ to - mathop {OA}limits^ to
$$
which can be otherwise written as
$$
mathop {OB}limits^ to = mathop {OA}limits^ to + {bf v}
$$
(vector $bf v$ "applied" in $A$).
Thus:
- the or. segment $((0,0,0),(1,0,0))$ lies on the $x$ axis but does not lie on a plane parallel (and distant) from it;
- the vector $(1,0,0)$ is parallel to the $x$ axis and to any a plane parallel from it.
answered Aug 7 '17 at 9:56
G Cab
16.9k31237
answered Aug 7 '17 at 9:56
G Cab
16.9k31237
answered Aug 7 '17 at 9:56
G Cab
16.9k31237
answered Aug 7 '17 at 9:56
G Cab
16.9k31237
16.9k31237
add a comment |
add a comment |
up vote
-1
down vote
Well just think, point $(1,0,0)$ lies on the $x$-axis, not on $y$ or $z$-axis. Hence your vector is parallel to $x$-axis and therefore lies in an arbitrary plane parallel to $x$-axis. 3D is just the working of your brain :)
edited Oct 19 '16 at 16:21
zar
2,74631837
answered Oct 19 '16 at 14:48
Kismat
1
add a comment |
up vote
-1
down vote
Well just think, point $(1,0,0)$ lies on the $x$-axis, not on $y$ or $z$-axis. Hence your vector is parallel to $x$-axis and therefore lies in an arbitrary plane parallel to $x$-axis. 3D is just the working of your brain :)
edited Oct 19 '16 at 16:21
zar
2,74631837
answered Oct 19 '16 at 14:48
Kismat
1
add a comment |
up vote
-1
down vote
up vote
-1
down vote
Well just think, point $(1,0,0)$ lies on the $x$-axis, not on $y$ or $z$-axis. Hence your vector is parallel to $x$-axis and therefore lies in an arbitrary plane parallel to $x$-axis. 3D is just the working of your brain :)
edited Oct 19 '16 at 16:21
zar
2,74631837
answered Oct 19 '16 at 14:48
Kismat
1
Well just think, point $(1,0,0)$ lies on the $x$-axis, not on $y$ or $z$-axis. Hence your vector is parallel to $x$-axis and therefore lies in an arbitrary plane parallel to $x$-axis. 3D is just the working of your brain :)
edited Oct 19 '16 at 16:21
zar
2,74631837
answered Oct 19 '16 at 14:48
Kismat
1
edited Oct 19 '16 at 16:21
zar
2,74631837
edited Oct 19 '16 at 16:21
zar
2,74631837
edited Oct 19 '16 at 16:21
zar
2,74631837
2,74631837
answered Oct 19 '16 at 14:48
Kismat
1
answered Oct 19 '16 at 14:48
Kismat
1
answered Oct 19 '16 at 14:48
Kismat
1
1
add a comment |
add a comment |
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It has to do with translation invariance of vectors. The vectors have the same properties at any position in space, so a vector travelling "along" the axis is always parallel to it, even if it is placed somewhere else in space.
– Terra Hyde
Sep 3 '15 at 20:35