Equation of a plane, given two points and a perpendicular plane
The plane passes through the points $(3, 4, 1)$ and $(3, 1, -6)$ and is perpendicular to the plane $7x + 9y + 4z = 17$. Find the equation of the plane.
What I was thinking was to take the cross product of the normal $(7, 9, 4)$ and the line $(3-3, 4-1, 1-(-6)) = (0, 3, 7)$. However, when I get the answer of $51x - 49y + 21z = 0$, it is not accepted as the right answer. Can anyone point out what I'm doing wrong?
vectors analytic-geometry
asked Sep 18 '16 at 14:38
Arghya
135
add a comment |
The plane passes through the points $(3, 4, 1)$ and $(3, 1, -6)$ and is perpendicular to the plane $7x + 9y + 4z = 17$. Find the equation of the plane.
What I was thinking was to take the cross product of the normal $(7, 9, 4)$ and the line $(3-3, 4-1, 1-(-6)) = (0, 3, 7)$. However, when I get the answer of $51x - 49y + 21z = 0$, it is not accepted as the right answer. Can anyone point out what I'm doing wrong?
vectors analytic-geometry
asked Sep 18 '16 at 14:38
Arghya
135
1
Change "$=0$" to something else in order to have the plane pass trough $(3,4,1)$
– Hagen von Eitzen
Sep 18 '16 at 14:42
$(7,9,4)$ is a vector, not a plane.
– anderstood
Sep 18 '16 at 14:42
1
Also, the cross product is $(51,-49,21)$. The 49 is negative.
– Nick
Sep 18 '16 at 14:43
You did good (apart from computing errors: I did not check) but you found just the normal to the plane: then you have to take a plane with that normal and passing through one of the points given (not through the origin).
– G Cab
Sep 18 '16 at 14:44
Thank you all! The biggest mistake was that the cross product was indeed (51, -49, 21) and I needed to solve using (3, 4, 1) which made the equation equal to -22.
– Arghya
Sep 18 '16 at 16:18
add a comment |
The plane passes through the points $(3, 4, 1)$ and $(3, 1, -6)$ and is perpendicular to the plane $7x + 9y + 4z = 17$. Find the equation of the plane.
What I was thinking was to take the cross product of the normal $(7, 9, 4)$ and the line $(3-3, 4-1, 1-(-6)) = (0, 3, 7)$. However, when I get the answer of $51x - 49y + 21z = 0$, it is not accepted as the right answer. Can anyone point out what I'm doing wrong?
vectors analytic-geometry
asked Sep 18 '16 at 14:38
Arghya
135
The plane passes through the points $(3, 4, 1)$ and $(3, 1, -6)$ and is perpendicular to the plane $7x + 9y + 4z = 17$. Find the equation of the plane.
What I was thinking was to take the cross product of the normal $(7, 9, 4)$ and the line $(3-3, 4-1, 1-(-6)) = (0, 3, 7)$. However, when I get the answer of $51x - 49y + 21z = 0$, it is not accepted as the right answer. Can anyone point out what I'm doing wrong?
vectors analytic-geometry
vectors analytic-geometry
asked Sep 18 '16 at 14:38
Arghya
135
asked Sep 18 '16 at 14:38
Arghya
135
edited Sep 18 '16 at 15:03
asked Sep 18 '16 at 14:38
Arghya
135
asked Sep 18 '16 at 14:38
Arghya
135
asked Sep 18 '16 at 14:38
Arghya
135
135
1
Change "$=0$" to something else in order to have the plane pass trough $(3,4,1)$
– Hagen von Eitzen
Sep 18 '16 at 14:42
$(7,9,4)$ is a vector, not a plane.
– anderstood
Sep 18 '16 at 14:42
1
Also, the cross product is $(51,-49,21)$. The 49 is negative.
– Nick
Sep 18 '16 at 14:43
You did good (apart from computing errors: I did not check) but you found just the normal to the plane: then you have to take a plane with that normal and passing through one of the points given (not through the origin).
– G Cab
Sep 18 '16 at 14:44
Thank you all! The biggest mistake was that the cross product was indeed (51, -49, 21) and I needed to solve using (3, 4, 1) which made the equation equal to -22.
– Arghya
Sep 18 '16 at 16:18
add a comment |
1
Change "$=0$" to something else in order to have the plane pass trough $(3,4,1)$
– Hagen von Eitzen
Sep 18 '16 at 14:42
$(7,9,4)$ is a vector, not a plane.
– anderstood
Sep 18 '16 at 14:42
1
Also, the cross product is $(51,-49,21)$. The 49 is negative.
– Nick
Sep 18 '16 at 14:43
You did good (apart from computing errors: I did not check) but you found just the normal to the plane: then you have to take a plane with that normal and passing through one of the points given (not through the origin).
– G Cab
Sep 18 '16 at 14:44
Thank you all! The biggest mistake was that the cross product was indeed (51, -49, 21) and I needed to solve using (3, 4, 1) which made the equation equal to -22.
– Arghya
Sep 18 '16 at 16:18
1
1
Change "$=0$" to something else in order to have the plane pass trough $(3,4,1)$
– Hagen von Eitzen
Sep 18 '16 at 14:42
Change "$=0$" to something else in order to have the plane pass trough $(3,4,1)$
– Hagen von Eitzen
Sep 18 '16 at 14:42
$(7,9,4)$ is a vector, not a plane.
– anderstood
Sep 18 '16 at 14:42
$(7,9,4)$ is a vector, not a plane.
– anderstood
Sep 18 '16 at 14:42
1
1
Also, the cross product is $(51,-49,21)$. The 49 is negative.
– Nick
Sep 18 '16 at 14:43
Also, the cross product is $(51,-49,21)$. The 49 is negative.
– Nick
Sep 18 '16 at 14:43
You did good (apart from computing errors: I did not check) but you found just the normal to the plane: then you have to take a plane with that normal and passing through one of the points given (not through the origin).
– G Cab
Sep 18 '16 at 14:44
You did good (apart from computing errors: I did not check) but you found just the normal to the plane: then you have to take a plane with that normal and passing through one of the points given (not through the origin).
– G Cab
Sep 18 '16 at 14:44
Thank you all! The biggest mistake was that the cross product was indeed (51, -49, 21) and I needed to solve using (3, 4, 1) which made the equation equal to -22.
– Arghya
Sep 18 '16 at 16:18
Thank you all! The biggest mistake was that the cross product was indeed (51, -49, 21) and I needed to solve using (3, 4, 1) which made the equation equal to -22.
– Arghya
Sep 18 '16 at 16:18
add a comment |
1 Answer
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make the ansatz the plane has the equation
$$ax+by+cz+d=0$$ then we have
$$3a+4b+c+d=0$$(1)
$$3a+b-6c+d=0$$ (2)
and
$$[a,b,c]cdot[7,9,4]=0$$ (3)
can you proceed?
answered Sep 18 '16 at 15:07
Dr. Sonnhard Graubner
72.7k32865
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
make the ansatz the plane has the equation
$$ax+by+cz+d=0$$ then we have
$$3a+4b+c+d=0$$(1)
$$3a+b-6c+d=0$$ (2)
and
$$[a,b,c]cdot[7,9,4]=0$$ (3)
can you proceed?
answered Sep 18 '16 at 15:07
Dr. Sonnhard Graubner
72.7k32865
add a comment |
make the ansatz the plane has the equation
$$ax+by+cz+d=0$$ then we have
$$3a+4b+c+d=0$$(1)
$$3a+b-6c+d=0$$ (2)
and
$$[a,b,c]cdot[7,9,4]=0$$ (3)
can you proceed?
answered Sep 18 '16 at 15:07
Dr. Sonnhard Graubner
72.7k32865
add a comment |
make the ansatz the plane has the equation
$$ax+by+cz+d=0$$ then we have
$$3a+4b+c+d=0$$(1)
$$3a+b-6c+d=0$$ (2)
and
$$[a,b,c]cdot[7,9,4]=0$$ (3)
can you proceed?
answered Sep 18 '16 at 15:07
Dr. Sonnhard Graubner
72.7k32865
make the ansatz the plane has the equation
$$ax+by+cz+d=0$$ then we have
$$3a+4b+c+d=0$$(1)
$$3a+b-6c+d=0$$ (2)
and
$$[a,b,c]cdot[7,9,4]=0$$ (3)
can you proceed?
answered Sep 18 '16 at 15:07
Dr. Sonnhard Graubner
72.7k32865
answered Sep 18 '16 at 15:07
Dr. Sonnhard Graubner
72.7k32865
answered Sep 18 '16 at 15:07
Dr. Sonnhard Graubner
72.7k32865
answered Sep 18 '16 at 15:07
Dr. Sonnhard Graubner
72.7k32865
72.7k32865
add a comment |
add a comment |
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1
Change "$=0$" to something else in order to have the plane pass trough $(3,4,1)$
– Hagen von Eitzen
Sep 18 '16 at 14:42
$(7,9,4)$ is a vector, not a plane.
– anderstood
Sep 18 '16 at 14:42
1
Also, the cross product is $(51,-49,21)$. The 49 is negative.
– Nick
Sep 18 '16 at 14:43
You did good (apart from computing errors: I did not check) but you found just the normal to the plane: then you have to take a plane with that normal and passing through one of the points given (not through the origin).
– G Cab
Sep 18 '16 at 14:44
Thank you all! The biggest mistake was that the cross product was indeed (51, -49, 21) and I needed to solve using (3, 4, 1) which made the equation equal to -22.
– Arghya
Sep 18 '16 at 16:18