Proving that the solution set of $ax+by+cz=0$ and $dx+ey+fz=0$ is a line in $mathbb{R}^3$
I have a question here which says:
"The solution set of:
$$ax+by+cz=0$$
$$dx+ey+fz=0$$
is a line in $mathbb{R^3}$"
I am supposed to show that this is true or false. If it's true, I have to explain why and if it's false, I have to explain why it's false.
I don't think it's true. We COULD have a line if the planes intersected but if $a,b,c$ are scalar multiples of $d,e,f$, then we would just have the same plane since there are infinitely many solutions. Is that the right should process or did I mess up my reasoning?
linear-algebra geometry systems-of-equations
edited Nov 30 at 5:08
Blue
47.6k870151
asked Nov 30 at 3:43
Future Math person
972717
|
show 2 more comments
I have a question here which says:
"The solution set of:
$$ax+by+cz=0$$
$$dx+ey+fz=0$$
is a line in $mathbb{R^3}$"
I am supposed to show that this is true or false. If it's true, I have to explain why and if it's false, I have to explain why it's false.
I don't think it's true. We COULD have a line if the planes intersected but if $a,b,c$ are scalar multiples of $d,e,f$, then we would just have the same plane since there are infinitely many solutions. Is that the right should process or did I mess up my reasoning?
linear-algebra geometry systems-of-equations
edited Nov 30 at 5:08
Blue
47.6k870151
asked Nov 30 at 3:43
Future Math person
972717
Awesome. Thanks :) .
– Future Math person
Nov 30 at 3:45
3
There's also the possibility that the planes are parallel and therefore never intersect, in which case there are no solutions.
– Robert Howard
Nov 30 at 3:45
2
@RobertHoward actually no solution is NOT a possibility here because of the homogeneous system, so either a line or a plane are the only possible solutions.
– Anurag A
Nov 30 at 3:48
1
I caught it though so no worries!
– Future Math person
Nov 30 at 4:03
1
@RobertHoward I’m astonished that your comment got three upvotes!
– amd
Nov 30 at 5:06
|
show 2 more comments
I have a question here which says:
"The solution set of:
$$ax+by+cz=0$$
$$dx+ey+fz=0$$
is a line in $mathbb{R^3}$"
I am supposed to show that this is true or false. If it's true, I have to explain why and if it's false, I have to explain why it's false.
I don't think it's true. We COULD have a line if the planes intersected but if $a,b,c$ are scalar multiples of $d,e,f$, then we would just have the same plane since there are infinitely many solutions. Is that the right should process or did I mess up my reasoning?
linear-algebra geometry systems-of-equations
edited Nov 30 at 5:08
Blue
47.6k870151
asked Nov 30 at 3:43
Future Math person
972717
I have a question here which says:
"The solution set of:
$$ax+by+cz=0$$
$$dx+ey+fz=0$$
is a line in $mathbb{R^3}$"
I am supposed to show that this is true or false. If it's true, I have to explain why and if it's false, I have to explain why it's false.
I don't think it's true. We COULD have a line if the planes intersected but if $a,b,c$ are scalar multiples of $d,e,f$, then we would just have the same plane since there are infinitely many solutions. Is that the right should process or did I mess up my reasoning?
linear-algebra geometry systems-of-equations
linear-algebra geometry systems-of-equations
edited Nov 30 at 5:08
Blue
47.6k870151
asked Nov 30 at 3:43
Future Math person
972717
edited Nov 30 at 5:08
Blue
47.6k870151
asked Nov 30 at 3:43
Future Math person
972717
edited Nov 30 at 5:08
Blue
47.6k870151
edited Nov 30 at 5:08
Blue
47.6k870151
edited Nov 30 at 5:08
Blue
47.6k870151
47.6k870151
asked Nov 30 at 3:43
Future Math person
972717
asked Nov 30 at 3:43
Future Math person
972717
asked Nov 30 at 3:43
Future Math person
972717
972717
Awesome. Thanks :) .
– Future Math person
Nov 30 at 3:45
3
There's also the possibility that the planes are parallel and therefore never intersect, in which case there are no solutions.
– Robert Howard
Nov 30 at 3:45
2
@RobertHoward actually no solution is NOT a possibility here because of the homogeneous system, so either a line or a plane are the only possible solutions.
– Anurag A
Nov 30 at 3:48
1
I caught it though so no worries!
– Future Math person
Nov 30 at 4:03
1
@RobertHoward I’m astonished that your comment got three upvotes!
– amd
Nov 30 at 5:06
|
show 2 more comments
Awesome. Thanks :) .
– Future Math person
Nov 30 at 3:45
3
There's also the possibility that the planes are parallel and therefore never intersect, in which case there are no solutions.
– Robert Howard
Nov 30 at 3:45
2
@RobertHoward actually no solution is NOT a possibility here because of the homogeneous system, so either a line or a plane are the only possible solutions.
– Anurag A
Nov 30 at 3:48
1
I caught it though so no worries!
– Future Math person
Nov 30 at 4:03
1
@RobertHoward I’m astonished that your comment got three upvotes!
– amd
Nov 30 at 5:06
Awesome. Thanks :) .
– Future Math person
Nov 30 at 3:45
Awesome. Thanks :) .
– Future Math person
Nov 30 at 3:45
3
3
There's also the possibility that the planes are parallel and therefore never intersect, in which case there are no solutions.
– Robert Howard
Nov 30 at 3:45
There's also the possibility that the planes are parallel and therefore never intersect, in which case there are no solutions.
– Robert Howard
Nov 30 at 3:45
2
2
@RobertHoward actually no solution is NOT a possibility here because of the homogeneous system, so either a line or a plane are the only possible solutions.
– Anurag A
Nov 30 at 3:48
@RobertHoward actually no solution is NOT a possibility here because of the homogeneous system, so either a line or a plane are the only possible solutions.
– Anurag A
Nov 30 at 3:48
1
1
I caught it though so no worries!
– Future Math person
Nov 30 at 4:03
I caught it though so no worries!
– Future Math person
Nov 30 at 4:03
1
1
@RobertHoward I’m astonished that your comment got three upvotes!
– amd
Nov 30 at 5:06
@RobertHoward I’m astonished that your comment got three upvotes!
– amd
Nov 30 at 5:06
|
show 2 more comments
1 Answer
1
active
oldest
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When absolutely nothing is provided about $a,b,c,d,e,f$ then for one you may even just take say all of them to be equal to zero, and then the system of solutions is the whole of $mathbb R^3$, which is not a line.
Of course, a plane is also obtainable : it is obtained exactly when the ratios $d:a,e:b,f:c$ are the same (else, the planes would not be parallel and hence would intersect at a line passing through zero). Then, the equations $ax+by+cz = 0$ and $dx+ey+fz = 0$ describe the same plane. This is clearly seen for example with $a,b,c,d,e,f$ all equal to $1$, or say $a,b,c=1$, $d,e,f = 2$. So, obviously the statement given can be false.
Note , however, that a solution must exist. This is because of the rank-nullity theorem, applied on the linear transformation $(x,y,z) to (ax+by+cz,dx+ey+fz)$. This is a map from $mathbb R^3$ to $mathbb R^2$. It has rank atmost $2$, therefore nullity at least one i.e. the null space has at least one line.
answered Nov 30 at 3:53
астон вілла олоф мэллбэрг
37.2k33376
add a comment |
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When absolutely nothing is provided about $a,b,c,d,e,f$ then for one you may even just take say all of them to be equal to zero, and then the system of solutions is the whole of $mathbb R^3$, which is not a line.
Of course, a plane is also obtainable : it is obtained exactly when the ratios $d:a,e:b,f:c$ are the same (else, the planes would not be parallel and hence would intersect at a line passing through zero). Then, the equations $ax+by+cz = 0$ and $dx+ey+fz = 0$ describe the same plane. This is clearly seen for example with $a,b,c,d,e,f$ all equal to $1$, or say $a,b,c=1$, $d,e,f = 2$. So, obviously the statement given can be false.
Note , however, that a solution must exist. This is because of the rank-nullity theorem, applied on the linear transformation $(x,y,z) to (ax+by+cz,dx+ey+fz)$. This is a map from $mathbb R^3$ to $mathbb R^2$. It has rank atmost $2$, therefore nullity at least one i.e. the null space has at least one line.
answered Nov 30 at 3:53
астон вілла олоф мэллбэрг
37.2k33376
add a comment |
When absolutely nothing is provided about $a,b,c,d,e,f$ then for one you may even just take say all of them to be equal to zero, and then the system of solutions is the whole of $mathbb R^3$, which is not a line.
Of course, a plane is also obtainable : it is obtained exactly when the ratios $d:a,e:b,f:c$ are the same (else, the planes would not be parallel and hence would intersect at a line passing through zero). Then, the equations $ax+by+cz = 0$ and $dx+ey+fz = 0$ describe the same plane. This is clearly seen for example with $a,b,c,d,e,f$ all equal to $1$, or say $a,b,c=1$, $d,e,f = 2$. So, obviously the statement given can be false.
Note , however, that a solution must exist. This is because of the rank-nullity theorem, applied on the linear transformation $(x,y,z) to (ax+by+cz,dx+ey+fz)$. This is a map from $mathbb R^3$ to $mathbb R^2$. It has rank atmost $2$, therefore nullity at least one i.e. the null space has at least one line.
answered Nov 30 at 3:53
астон вілла олоф мэллбэрг
37.2k33376
add a comment |
When absolutely nothing is provided about $a,b,c,d,e,f$ then for one you may even just take say all of them to be equal to zero, and then the system of solutions is the whole of $mathbb R^3$, which is not a line.
Of course, a plane is also obtainable : it is obtained exactly when the ratios $d:a,e:b,f:c$ are the same (else, the planes would not be parallel and hence would intersect at a line passing through zero). Then, the equations $ax+by+cz = 0$ and $dx+ey+fz = 0$ describe the same plane. This is clearly seen for example with $a,b,c,d,e,f$ all equal to $1$, or say $a,b,c=1$, $d,e,f = 2$. So, obviously the statement given can be false.
Note , however, that a solution must exist. This is because of the rank-nullity theorem, applied on the linear transformation $(x,y,z) to (ax+by+cz,dx+ey+fz)$. This is a map from $mathbb R^3$ to $mathbb R^2$. It has rank atmost $2$, therefore nullity at least one i.e. the null space has at least one line.
answered Nov 30 at 3:53
астон вілла олоф мэллбэрг
37.2k33376
When absolutely nothing is provided about $a,b,c,d,e,f$ then for one you may even just take say all of them to be equal to zero, and then the system of solutions is the whole of $mathbb R^3$, which is not a line.
Of course, a plane is also obtainable : it is obtained exactly when the ratios $d:a,e:b,f:c$ are the same (else, the planes would not be parallel and hence would intersect at a line passing through zero). Then, the equations $ax+by+cz = 0$ and $dx+ey+fz = 0$ describe the same plane. This is clearly seen for example with $a,b,c,d,e,f$ all equal to $1$, or say $a,b,c=1$, $d,e,f = 2$. So, obviously the statement given can be false.
Note , however, that a solution must exist. This is because of the rank-nullity theorem, applied on the linear transformation $(x,y,z) to (ax+by+cz,dx+ey+fz)$. This is a map from $mathbb R^3$ to $mathbb R^2$. It has rank atmost $2$, therefore nullity at least one i.e. the null space has at least one line.
answered Nov 30 at 3:53
астон вілла олоф мэллбэрг
37.2k33376
answered Nov 30 at 3:53
астон вілла олоф мэллбэрг
37.2k33376
answered Nov 30 at 3:53
астон вілла олоф мэллбэрг
37.2k33376
answered Nov 30 at 3:53
астон вілла олоф мэллбэрг
37.2k33376
37.2k33376
add a comment |
add a comment |
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Awesome. Thanks :) .
– Future Math person
Nov 30 at 3:45
3
There's also the possibility that the planes are parallel and therefore never intersect, in which case there are no solutions.
– Robert Howard
Nov 30 at 3:45
2
@RobertHoward actually no solution is NOT a possibility here because of the homogeneous system, so either a line or a plane are the only possible solutions.
– Anurag A
Nov 30 at 3:48
1
I caught it though so no worries!
– Future Math person
Nov 30 at 4:03
1
@RobertHoward I’m astonished that your comment got three upvotes!
– amd
Nov 30 at 5:06