Plane-Plane Intersection
How do you find the equation of the line that is given by the intersection of these planes:
x+y=2
z=3
(Keep in mind that im a beginner in this subject, but i'm looking forward to learn)
algebra-precalculus
edited Nov 28 '18 at 20:58
KReiser
9,31211435
asked Nov 28 '18 at 20:57
Tomas Vortali
33
add a comment |
How do you find the equation of the line that is given by the intersection of these planes:
x+y=2
z=3
(Keep in mind that im a beginner in this subject, but i'm looking forward to learn)
algebra-precalculus
edited Nov 28 '18 at 20:58
KReiser
9,31211435
asked Nov 28 '18 at 20:57
Tomas Vortali
33
I don't know how far along you are in your studies - but if you think about the geometry you can see that the direction of the line is perpendicular to the normals to each plane. Find the two normal vectors. Then find their cross product to find a vector perpendicular to both.
– Ethan Bolker
Nov 28 '18 at 21:00
What do you mean by “the equation of the line?” What form are you looking for?
– amd
Nov 28 '18 at 21:32
Thank you Ethan, that helped a lot! On the other hand, sorry if i didn't express myself correctly. I was looking for the vector form.
– Tomas Vortali
Nov 28 '18 at 21:48
add a comment |
How do you find the equation of the line that is given by the intersection of these planes:
x+y=2
z=3
(Keep in mind that im a beginner in this subject, but i'm looking forward to learn)
algebra-precalculus
edited Nov 28 '18 at 20:58
KReiser
9,31211435
asked Nov 28 '18 at 20:57
Tomas Vortali
33
How do you find the equation of the line that is given by the intersection of these planes:
x+y=2
z=3
(Keep in mind that im a beginner in this subject, but i'm looking forward to learn)
algebra-precalculus
algebra-precalculus
edited Nov 28 '18 at 20:58
KReiser
9,31211435
asked Nov 28 '18 at 20:57
Tomas Vortali
33
edited Nov 28 '18 at 20:58
KReiser
9,31211435
asked Nov 28 '18 at 20:57
Tomas Vortali
33
edited Nov 28 '18 at 20:58
KReiser
9,31211435
edited Nov 28 '18 at 20:58
KReiser
9,31211435
edited Nov 28 '18 at 20:58
KReiser
9,31211435
9,31211435
asked Nov 28 '18 at 20:57
Tomas Vortali
33
asked Nov 28 '18 at 20:57
Tomas Vortali
33
asked Nov 28 '18 at 20:57
Tomas Vortali
33
33
I don't know how far along you are in your studies - but if you think about the geometry you can see that the direction of the line is perpendicular to the normals to each plane. Find the two normal vectors. Then find their cross product to find a vector perpendicular to both.
– Ethan Bolker
Nov 28 '18 at 21:00
What do you mean by “the equation of the line?” What form are you looking for?
– amd
Nov 28 '18 at 21:32
Thank you Ethan, that helped a lot! On the other hand, sorry if i didn't express myself correctly. I was looking for the vector form.
– Tomas Vortali
Nov 28 '18 at 21:48
add a comment |
I don't know how far along you are in your studies - but if you think about the geometry you can see that the direction of the line is perpendicular to the normals to each plane. Find the two normal vectors. Then find their cross product to find a vector perpendicular to both.
– Ethan Bolker
Nov 28 '18 at 21:00
What do you mean by “the equation of the line?” What form are you looking for?
– amd
Nov 28 '18 at 21:32
Thank you Ethan, that helped a lot! On the other hand, sorry if i didn't express myself correctly. I was looking for the vector form.
– Tomas Vortali
Nov 28 '18 at 21:48
I don't know how far along you are in your studies - but if you think about the geometry you can see that the direction of the line is perpendicular to the normals to each plane. Find the two normal vectors. Then find their cross product to find a vector perpendicular to both.
– Ethan Bolker
Nov 28 '18 at 21:00
I don't know how far along you are in your studies - but if you think about the geometry you can see that the direction of the line is perpendicular to the normals to each plane. Find the two normal vectors. Then find their cross product to find a vector perpendicular to both.
– Ethan Bolker
Nov 28 '18 at 21:00
What do you mean by “the equation of the line?” What form are you looking for?
– amd
Nov 28 '18 at 21:32
What do you mean by “the equation of the line?” What form are you looking for?
– amd
Nov 28 '18 at 21:32
Thank you Ethan, that helped a lot! On the other hand, sorry if i didn't express myself correctly. I was looking for the vector form.
– Tomas Vortali
Nov 28 '18 at 21:48
Thank you Ethan, that helped a lot! On the other hand, sorry if i didn't express myself correctly. I was looking for the vector form.
– Tomas Vortali
Nov 28 '18 at 21:48
add a comment |
1 Answer
1
active
oldest
votes
The line is the solution of the system
$$
begin{cases}
x+y=2\
z=3
end{cases}
$$
that we can write in the form:
$$
begin{cases}
x=2-y\
z=3\
y=t
end{cases}
$$
where $y=t in mathbb{R}$ means that $y$ can have any real value.
reordering the system we have:
$$
begin{cases}
x=2-t\
y=t\
z=3
end{cases}
$$
that, in vector form, becomes the equation of the line in vector form :
$$
begin{bmatrix}
x\y\z
end{bmatrix}
=
begin{bmatrix}
2\0\3
end{bmatrix}
+t
begin{bmatrix}
-1\1\0
end{bmatrix}
$$
answered Nov 28 '18 at 21:37
Emilio Novati
51.5k43472
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The line is the solution of the system
$$
begin{cases}
x+y=2\
z=3
end{cases}
$$
that we can write in the form:
$$
begin{cases}
x=2-y\
z=3\
y=t
end{cases}
$$
where $y=t in mathbb{R}$ means that $y$ can have any real value.
reordering the system we have:
$$
begin{cases}
x=2-t\
y=t\
z=3
end{cases}
$$
that, in vector form, becomes the equation of the line in vector form :
$$
begin{bmatrix}
x\y\z
end{bmatrix}
=
begin{bmatrix}
2\0\3
end{bmatrix}
+t
begin{bmatrix}
-1\1\0
end{bmatrix}
$$
answered Nov 28 '18 at 21:37
Emilio Novati
51.5k43472
add a comment |
The line is the solution of the system
$$
begin{cases}
x+y=2\
z=3
end{cases}
$$
that we can write in the form:
$$
begin{cases}
x=2-y\
z=3\
y=t
end{cases}
$$
where $y=t in mathbb{R}$ means that $y$ can have any real value.
reordering the system we have:
$$
begin{cases}
x=2-t\
y=t\
z=3
end{cases}
$$
that, in vector form, becomes the equation of the line in vector form :
$$
begin{bmatrix}
x\y\z
end{bmatrix}
=
begin{bmatrix}
2\0\3
end{bmatrix}
+t
begin{bmatrix}
-1\1\0
end{bmatrix}
$$
answered Nov 28 '18 at 21:37
Emilio Novati
51.5k43472
add a comment |
The line is the solution of the system
$$
begin{cases}
x+y=2\
z=3
end{cases}
$$
that we can write in the form:
$$
begin{cases}
x=2-y\
z=3\
y=t
end{cases}
$$
where $y=t in mathbb{R}$ means that $y$ can have any real value.
reordering the system we have:
$$
begin{cases}
x=2-t\
y=t\
z=3
end{cases}
$$
that, in vector form, becomes the equation of the line in vector form :
$$
begin{bmatrix}
x\y\z
end{bmatrix}
=
begin{bmatrix}
2\0\3
end{bmatrix}
+t
begin{bmatrix}
-1\1\0
end{bmatrix}
$$
answered Nov 28 '18 at 21:37
Emilio Novati
51.5k43472
The line is the solution of the system
$$
begin{cases}
x+y=2\
z=3
end{cases}
$$
that we can write in the form:
$$
begin{cases}
x=2-y\
z=3\
y=t
end{cases}
$$
where $y=t in mathbb{R}$ means that $y$ can have any real value.
reordering the system we have:
$$
begin{cases}
x=2-t\
y=t\
z=3
end{cases}
$$
that, in vector form, becomes the equation of the line in vector form :
$$
begin{bmatrix}
x\y\z
end{bmatrix}
=
begin{bmatrix}
2\0\3
end{bmatrix}
+t
begin{bmatrix}
-1\1\0
end{bmatrix}
$$
answered Nov 28 '18 at 21:37
Emilio Novati
51.5k43472
edited Nov 28 '18 at 21:49
answered Nov 28 '18 at 21:37
Emilio Novati
51.5k43472
answered Nov 28 '18 at 21:37
Emilio Novati
51.5k43472
answered Nov 28 '18 at 21:37
Emilio Novati
51.5k43472
51.5k43472
add a comment |
add a comment |
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I don't know how far along you are in your studies - but if you think about the geometry you can see that the direction of the line is perpendicular to the normals to each plane. Find the two normal vectors. Then find their cross product to find a vector perpendicular to both.
– Ethan Bolker
Nov 28 '18 at 21:00
What do you mean by “the equation of the line?” What form are you looking for?
– amd
Nov 28 '18 at 21:32
Thank you Ethan, that helped a lot! On the other hand, sorry if i didn't express myself correctly. I was looking for the vector form.
– Tomas Vortali
Nov 28 '18 at 21:48