Question about vector equations of lines and planes
Multi tool use
$begingroup$
Find the equation of the line going through the point $(2,-3,4)$ ,and which is perpendicular to the plane $ x+2y + 2z = 13$
So I tried this: the normal of the plane is $(1,2,2)$, random point on the line is $(x,y,z)$, so direction vector is $(x-2,y+3,z-4)$. We'll require $(x-2,y+3,z-4)cdot (1,2,2) = 0$. This yields $x-2y+2z -4 =0$
let us take $z=y=1$. that makes $x=0$ and so $(0,1,1)$ is a random point on the plane; and so we get direction vector $(-2,4,-3)$. This gives a line equation of: $(2,-3,4) + t(-2,4,-3)$ ;
general point on such a line is: $(2-2t,4t-3,-3t+4)$.
We also need to find it's point of intersection with the plane $ x+2y + 2z = 13$, so we substitute values of the
general point to the plane $ x+2y + 2z = 13$. this however yields $0=9$ which is a little unsatisfying.
vector-spaces vector-analysis
asked Apr 14 '14 at 19:18
Bak1139Bak1139
1,39121232
$endgroup$
add a comment |
$begingroup$
Find the equation of the line going through the point $(2,-3,4)$ ,and which is perpendicular to the plane $ x+2y + 2z = 13$
So I tried this: the normal of the plane is $(1,2,2)$, random point on the line is $(x,y,z)$, so direction vector is $(x-2,y+3,z-4)$. We'll require $(x-2,y+3,z-4)cdot (1,2,2) = 0$. This yields $x-2y+2z -4 =0$
let us take $z=y=1$. that makes $x=0$ and so $(0,1,1)$ is a random point on the plane; and so we get direction vector $(-2,4,-3)$. This gives a line equation of: $(2,-3,4) + t(-2,4,-3)$ ;
general point on such a line is: $(2-2t,4t-3,-3t+4)$.
We also need to find it's point of intersection with the plane $ x+2y + 2z = 13$, so we substitute values of the
general point to the plane $ x+2y + 2z = 13$. this however yields $0=9$ which is a little unsatisfying.
vector-spaces vector-analysis
asked Apr 14 '14 at 19:18
Bak1139Bak1139
1,39121232
$endgroup$
add a comment |
$begingroup$
Find the equation of the line going through the point $(2,-3,4)$ ,and which is perpendicular to the plane $ x+2y + 2z = 13$
So I tried this: the normal of the plane is $(1,2,2)$, random point on the line is $(x,y,z)$, so direction vector is $(x-2,y+3,z-4)$. We'll require $(x-2,y+3,z-4)cdot (1,2,2) = 0$. This yields $x-2y+2z -4 =0$
let us take $z=y=1$. that makes $x=0$ and so $(0,1,1)$ is a random point on the plane; and so we get direction vector $(-2,4,-3)$. This gives a line equation of: $(2,-3,4) + t(-2,4,-3)$ ;
general point on such a line is: $(2-2t,4t-3,-3t+4)$.
We also need to find it's point of intersection with the plane $ x+2y + 2z = 13$, so we substitute values of the
general point to the plane $ x+2y + 2z = 13$. this however yields $0=9$ which is a little unsatisfying.
vector-spaces vector-analysis
asked Apr 14 '14 at 19:18
Bak1139Bak1139
1,39121232
$endgroup$
Find the equation of the line going through the point $(2,-3,4)$ ,and which is perpendicular to the plane $ x+2y + 2z = 13$
So I tried this: the normal of the plane is $(1,2,2)$, random point on the line is $(x,y,z)$, so direction vector is $(x-2,y+3,z-4)$. We'll require $(x-2,y+3,z-4)cdot (1,2,2) = 0$. This yields $x-2y+2z -4 =0$
let us take $z=y=1$. that makes $x=0$ and so $(0,1,1)$ is a random point on the plane; and so we get direction vector $(-2,4,-3)$. This gives a line equation of: $(2,-3,4) + t(-2,4,-3)$ ;
general point on such a line is: $(2-2t,4t-3,-3t+4)$.
We also need to find it's point of intersection with the plane $ x+2y + 2z = 13$, so we substitute values of the
general point to the plane $ x+2y + 2z = 13$. this however yields $0=9$ which is a little unsatisfying.
vector-spaces vector-analysis
vector-spaces vector-analysis
asked Apr 14 '14 at 19:18
Bak1139Bak1139
1,39121232
asked Apr 14 '14 at 19:18
Bak1139Bak1139
1,39121232
edited Apr 14 '14 at 19:32
Bak1139
asked Apr 14 '14 at 19:18
Bak1139Bak1139
1,39121232
asked Apr 14 '14 at 19:18
Bak1139Bak1139
1,39121232
asked Apr 14 '14 at 19:18
Bak1139Bak1139
1,39121232
1,39121232
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The reason why you get $0=9$ is that there is no intersection; the intersection is empty. Notice that the plane $P: x+2y+2z=13$ is parallel to the original plane, and so is the line. This means that either the line lies on the plane $P$ , or the line is parallel to it--and the second choice here is the case. If you start with a wrong assumption you may not get a good result (i.e., you are not guaranteed to, which you would be if you started with a true premise and argued correctly) , even if you argue correctly .
EDIT: Since the point lies on a line perpendicular to the plane thru $(2, -3,4)$, the line must be of the type : $L: (2,-3,4)+ t(1,2,2)=(2+t, -3+2t, 4+2t)=(x,y,z)$ . We then substitute this form into $x+2y+2z=13$, and find that $t=1$ is the solution ( makes sense, since both must intersect at a single point by a (co)dimension argument ). From $t=1$, we get that the intersection point is $(2+1, -3+2, 4+2)=(3,-1,6)$
answered Apr 14 '14 at 19:24
gueroguero
362
$endgroup$
$begingroup$
Well, according to the answers such an intersection exists; it is $(3,-1,6)$
$endgroup$
– Bak1139
Apr 14 '14 at 19:26
$begingroup$
@Bak1139 If you find a line parallel to a plane, then the plane and line will never intersect (unless the line is part of the plane). Since they point in the same direction.
$endgroup$
– George1811
Apr 14 '14 at 19:29
$begingroup$
you all are absolutely right, there is a typo...I'll fix
$endgroup$
– Bak1139
Apr 14 '14 at 19:31
$begingroup$
But the line is parallel to the original plane, and the second plane you're given; $x+2y+2z=13$ is parallel to the original plane, since both have the same ( up to scaling) normal vector $(1,2,2)$ . So the line and the two planes are either parallel, or the line lies on the plane. The answer is that the line is parallel to the second plane. I think the answer $(3,-1,6)$ is a mistake; notice that $(3,-1,6)$ is not on the line.
$endgroup$
– guero
Apr 14 '14 at 19:34
$begingroup$
perpendicularity is indeed the correct given data, at the problem beginning
$endgroup$
– Bak1139
Apr 14 '14 at 19:36
|
show 4 more comments
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1 Answer
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$begingroup$
The reason why you get $0=9$ is that there is no intersection; the intersection is empty. Notice that the plane $P: x+2y+2z=13$ is parallel to the original plane, and so is the line. This means that either the line lies on the plane $P$ , or the line is parallel to it--and the second choice here is the case. If you start with a wrong assumption you may not get a good result (i.e., you are not guaranteed to, which you would be if you started with a true premise and argued correctly) , even if you argue correctly .
EDIT: Since the point lies on a line perpendicular to the plane thru $(2, -3,4)$, the line must be of the type : $L: (2,-3,4)+ t(1,2,2)=(2+t, -3+2t, 4+2t)=(x,y,z)$ . We then substitute this form into $x+2y+2z=13$, and find that $t=1$ is the solution ( makes sense, since both must intersect at a single point by a (co)dimension argument ). From $t=1$, we get that the intersection point is $(2+1, -3+2, 4+2)=(3,-1,6)$
answered Apr 14 '14 at 19:24
gueroguero
362
$endgroup$
$begingroup$
Well, according to the answers such an intersection exists; it is $(3,-1,6)$
$endgroup$
– Bak1139
Apr 14 '14 at 19:26
$begingroup$
@Bak1139 If you find a line parallel to a plane, then the plane and line will never intersect (unless the line is part of the plane). Since they point in the same direction.
$endgroup$
– George1811
Apr 14 '14 at 19:29
$begingroup$
you all are absolutely right, there is a typo...I'll fix
$endgroup$
– Bak1139
Apr 14 '14 at 19:31
$begingroup$
But the line is parallel to the original plane, and the second plane you're given; $x+2y+2z=13$ is parallel to the original plane, since both have the same ( up to scaling) normal vector $(1,2,2)$ . So the line and the two planes are either parallel, or the line lies on the plane. The answer is that the line is parallel to the second plane. I think the answer $(3,-1,6)$ is a mistake; notice that $(3,-1,6)$ is not on the line.
$endgroup$
– guero
Apr 14 '14 at 19:34
$begingroup$
perpendicularity is indeed the correct given data, at the problem beginning
$endgroup$
– Bak1139
Apr 14 '14 at 19:36
|
show 4 more comments
$begingroup$
The reason why you get $0=9$ is that there is no intersection; the intersection is empty. Notice that the plane $P: x+2y+2z=13$ is parallel to the original plane, and so is the line. This means that either the line lies on the plane $P$ , or the line is parallel to it--and the second choice here is the case. If you start with a wrong assumption you may not get a good result (i.e., you are not guaranteed to, which you would be if you started with a true premise and argued correctly) , even if you argue correctly .
EDIT: Since the point lies on a line perpendicular to the plane thru $(2, -3,4)$, the line must be of the type : $L: (2,-3,4)+ t(1,2,2)=(2+t, -3+2t, 4+2t)=(x,y,z)$ . We then substitute this form into $x+2y+2z=13$, and find that $t=1$ is the solution ( makes sense, since both must intersect at a single point by a (co)dimension argument ). From $t=1$, we get that the intersection point is $(2+1, -3+2, 4+2)=(3,-1,6)$
answered Apr 14 '14 at 19:24
gueroguero
362
$endgroup$
$begingroup$
Well, according to the answers such an intersection exists; it is $(3,-1,6)$
$endgroup$
– Bak1139
Apr 14 '14 at 19:26
$begingroup$
@Bak1139 If you find a line parallel to a plane, then the plane and line will never intersect (unless the line is part of the plane). Since they point in the same direction.
$endgroup$
– George1811
Apr 14 '14 at 19:29
$begingroup$
you all are absolutely right, there is a typo...I'll fix
$endgroup$
– Bak1139
Apr 14 '14 at 19:31
$begingroup$
But the line is parallel to the original plane, and the second plane you're given; $x+2y+2z=13$ is parallel to the original plane, since both have the same ( up to scaling) normal vector $(1,2,2)$ . So the line and the two planes are either parallel, or the line lies on the plane. The answer is that the line is parallel to the second plane. I think the answer $(3,-1,6)$ is a mistake; notice that $(3,-1,6)$ is not on the line.
$endgroup$
– guero
Apr 14 '14 at 19:34
$begingroup$
perpendicularity is indeed the correct given data, at the problem beginning
$endgroup$
– Bak1139
Apr 14 '14 at 19:36
|
show 4 more comments
$begingroup$
The reason why you get $0=9$ is that there is no intersection; the intersection is empty. Notice that the plane $P: x+2y+2z=13$ is parallel to the original plane, and so is the line. This means that either the line lies on the plane $P$ , or the line is parallel to it--and the second choice here is the case. If you start with a wrong assumption you may not get a good result (i.e., you are not guaranteed to, which you would be if you started with a true premise and argued correctly) , even if you argue correctly .
EDIT: Since the point lies on a line perpendicular to the plane thru $(2, -3,4)$, the line must be of the type : $L: (2,-3,4)+ t(1,2,2)=(2+t, -3+2t, 4+2t)=(x,y,z)$ . We then substitute this form into $x+2y+2z=13$, and find that $t=1$ is the solution ( makes sense, since both must intersect at a single point by a (co)dimension argument ). From $t=1$, we get that the intersection point is $(2+1, -3+2, 4+2)=(3,-1,6)$
answered Apr 14 '14 at 19:24
gueroguero
362
$endgroup$
The reason why you get $0=9$ is that there is no intersection; the intersection is empty. Notice that the plane $P: x+2y+2z=13$ is parallel to the original plane, and so is the line. This means that either the line lies on the plane $P$ , or the line is parallel to it--and the second choice here is the case. If you start with a wrong assumption you may not get a good result (i.e., you are not guaranteed to, which you would be if you started with a true premise and argued correctly) , even if you argue correctly .
EDIT: Since the point lies on a line perpendicular to the plane thru $(2, -3,4)$, the line must be of the type : $L: (2,-3,4)+ t(1,2,2)=(2+t, -3+2t, 4+2t)=(x,y,z)$ . We then substitute this form into $x+2y+2z=13$, and find that $t=1$ is the solution ( makes sense, since both must intersect at a single point by a (co)dimension argument ). From $t=1$, we get that the intersection point is $(2+1, -3+2, 4+2)=(3,-1,6)$
answered Apr 14 '14 at 19:24
gueroguero
362
edited Apr 14 '14 at 20:21
answered Apr 14 '14 at 19:24
gueroguero
362
answered Apr 14 '14 at 19:24
gueroguero
362
answered Apr 14 '14 at 19:24
gueroguero
362
362
$begingroup$
Well, according to the answers such an intersection exists; it is $(3,-1,6)$
$endgroup$
– Bak1139
Apr 14 '14 at 19:26
$begingroup$
@Bak1139 If you find a line parallel to a plane, then the plane and line will never intersect (unless the line is part of the plane). Since they point in the same direction.
$endgroup$
– George1811
Apr 14 '14 at 19:29
$begingroup$
you all are absolutely right, there is a typo...I'll fix
$endgroup$
– Bak1139
Apr 14 '14 at 19:31
$begingroup$
But the line is parallel to the original plane, and the second plane you're given; $x+2y+2z=13$ is parallel to the original plane, since both have the same ( up to scaling) normal vector $(1,2,2)$ . So the line and the two planes are either parallel, or the line lies on the plane. The answer is that the line is parallel to the second plane. I think the answer $(3,-1,6)$ is a mistake; notice that $(3,-1,6)$ is not on the line.
$endgroup$
– guero
Apr 14 '14 at 19:34
$begingroup$
perpendicularity is indeed the correct given data, at the problem beginning
$endgroup$
– Bak1139
Apr 14 '14 at 19:36
|
show 4 more comments
$begingroup$
Well, according to the answers such an intersection exists; it is $(3,-1,6)$
$endgroup$
– Bak1139
Apr 14 '14 at 19:26
$begingroup$
@Bak1139 If you find a line parallel to a plane, then the plane and line will never intersect (unless the line is part of the plane). Since they point in the same direction.
$endgroup$
– George1811
Apr 14 '14 at 19:29
$begingroup$
you all are absolutely right, there is a typo...I'll fix
$endgroup$
– Bak1139
Apr 14 '14 at 19:31
$begingroup$
But the line is parallel to the original plane, and the second plane you're given; $x+2y+2z=13$ is parallel to the original plane, since both have the same ( up to scaling) normal vector $(1,2,2)$ . So the line and the two planes are either parallel, or the line lies on the plane. The answer is that the line is parallel to the second plane. I think the answer $(3,-1,6)$ is a mistake; notice that $(3,-1,6)$ is not on the line.
$endgroup$
– guero
Apr 14 '14 at 19:34
$begingroup$
perpendicularity is indeed the correct given data, at the problem beginning
$endgroup$
– Bak1139
Apr 14 '14 at 19:36
$begingroup$
Well, according to the answers such an intersection exists; it is $(3,-1,6)$
$endgroup$
– Bak1139
Apr 14 '14 at 19:26
$begingroup$
Well, according to the answers such an intersection exists; it is $(3,-1,6)$
$endgroup$
– Bak1139
Apr 14 '14 at 19:26
$begingroup$
@Bak1139 If you find a line parallel to a plane, then the plane and line will never intersect (unless the line is part of the plane). Since they point in the same direction.
$endgroup$
– George1811
Apr 14 '14 at 19:29
$begingroup$
@Bak1139 If you find a line parallel to a plane, then the plane and line will never intersect (unless the line is part of the plane). Since they point in the same direction.
$endgroup$
– George1811
Apr 14 '14 at 19:29
$begingroup$
you all are absolutely right, there is a typo...I'll fix
$endgroup$
– Bak1139
Apr 14 '14 at 19:31
$begingroup$
you all are absolutely right, there is a typo...I'll fix
$endgroup$
– Bak1139
Apr 14 '14 at 19:31
$begingroup$
But the line is parallel to the original plane, and the second plane you're given; $x+2y+2z=13$ is parallel to the original plane, since both have the same ( up to scaling) normal vector $(1,2,2)$ . So the line and the two planes are either parallel, or the line lies on the plane. The answer is that the line is parallel to the second plane. I think the answer $(3,-1,6)$ is a mistake; notice that $(3,-1,6)$ is not on the line.
$endgroup$
– guero
Apr 14 '14 at 19:34
$begingroup$
But the line is parallel to the original plane, and the second plane you're given; $x+2y+2z=13$ is parallel to the original plane, since both have the same ( up to scaling) normal vector $(1,2,2)$ . So the line and the two planes are either parallel, or the line lies on the plane. The answer is that the line is parallel to the second plane. I think the answer $(3,-1,6)$ is a mistake; notice that $(3,-1,6)$ is not on the line.
$endgroup$
– guero
Apr 14 '14 at 19:34
$begingroup$
perpendicularity is indeed the correct given data, at the problem beginning
$endgroup$
– Bak1139
Apr 14 '14 at 19:36
$begingroup$
perpendicularity is indeed the correct given data, at the problem beginning
$endgroup$
– Bak1139
Apr 14 '14 at 19:36
|
show 4 more comments
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