How to find equation of plane given one vector and two points? The Next CEO of Stack OverflowFind equation of planeFind the equation of a plane when it passes through two points and parallel to two vectorsEquation of a plane, given two points and a perpendicular planeFind an equation of the plane.How to find an equation for the plane that is perpendicular to a line and passes through a point?Equation of a plane perpendicular to two planes and passes through a pointVector equation that passes through a point which is parallel to two planeFind a point on the plane where there are two given points and a orientation vector to that plane is given.Given a plane and a line, find the equation of another plane that has an angle 30 of degree to the given plane and contains the given line.Cartesian equation of plane through $3$ points
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How to find equation of plane given one vector and two points?
The Next CEO of Stack OverflowFind equation of planeFind the equation of a plane when it passes through two points and parallel to two vectorsEquation of a plane, given two points and a perpendicular planeFind an equation of the plane.How to find an equation for the plane that is perpendicular to a line and passes through a point?Equation of a plane perpendicular to two planes and passes through a pointVector equation that passes through a point which is parallel to two planeFind a point on the plane where there are two given points and a orientation vector to that plane is given.Given a plane and a line, find the equation of another plane that has an angle 30 of degree to the given plane and contains the given line.Cartesian equation of plane through $3$ points
$begingroup$
Find an equation of the plane that is perpendicular to $x-2y+z=5$ and passes through $P_1(-2,4,3)$ and $P_2(3,-5,0)$?
How to solve this question? It is needed to find another vector using cross product or just use $(1,-2,1)$ and choose any point?
vectors 3d
edited May 26 '17 at 7:42
Parcly Taxel
44.7k1376109
asked May 26 '17 at 6:38
Mohd AzimuddinMohd Azimuddin
164
$endgroup$
add a comment |
$begingroup$
Find an equation of the plane that is perpendicular to $x-2y+z=5$ and passes through $P_1(-2,4,3)$ and $P_2(3,-5,0)$?
How to solve this question? It is needed to find another vector using cross product or just use $(1,-2,1)$ and choose any point?
vectors 3d
edited May 26 '17 at 7:42
Parcly Taxel
44.7k1376109
asked May 26 '17 at 6:38
Mohd AzimuddinMohd Azimuddin
164
$endgroup$
add a comment |
$begingroup$
Find an equation of the plane that is perpendicular to $x-2y+z=5$ and passes through $P_1(-2,4,3)$ and $P_2(3,-5,0)$?
How to solve this question? It is needed to find another vector using cross product or just use $(1,-2,1)$ and choose any point?
vectors 3d
edited May 26 '17 at 7:42
Parcly Taxel
44.7k1376109
asked May 26 '17 at 6:38
Mohd AzimuddinMohd Azimuddin
164
$endgroup$
Find an equation of the plane that is perpendicular to $x-2y+z=5$ and passes through $P_1(-2,4,3)$ and $P_2(3,-5,0)$?
How to solve this question? It is needed to find another vector using cross product or just use $(1,-2,1)$ and choose any point?
vectors 3d
vectors 3d
edited May 26 '17 at 7:42
Parcly Taxel
44.7k1376109
asked May 26 '17 at 6:38
Mohd AzimuddinMohd Azimuddin
164
edited May 26 '17 at 7:42
Parcly Taxel
44.7k1376109
asked May 26 '17 at 6:38
Mohd AzimuddinMohd Azimuddin
164
edited May 26 '17 at 7:42
Parcly Taxel
44.7k1376109
edited May 26 '17 at 7:42
Parcly Taxel
44.7k1376109
edited May 26 '17 at 7:42
Parcly Taxel
44.7k1376109
44.7k1376109
asked May 26 '17 at 6:38
Mohd AzimuddinMohd Azimuddin
164
asked May 26 '17 at 6:38
Mohd AzimuddinMohd Azimuddin
164
asked May 26 '17 at 6:38
Mohd AzimuddinMohd Azimuddin
164
164
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Two things first:
- If two planes are perpendicular, their normal vectors will also be perpendicular.
- If a plane contains two points, it contains the line through those points; in particular its normal vector is perpendicular to the direction vector of the line.
So the desired plane's normal vector is perpendicular to $(1,-2,1)$ and $(3,-5,0)-(-2,4,3)=(5,-9,-3)$. The cross product yields such a perpendicular vector: $(15,8,1)$. So the equation of the plane is $15x+8y+z=k$, where $k$ is found by substituting either given point in: $15cdot(-2)+8cdot4+1cdot3=5$. The final answer is $15x+8y+z=5$.
answered May 26 '17 at 7:41
Parcly TaxelParcly Taxel
44.7k1376109
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
active
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votes
$begingroup$
Two things first:
- If two planes are perpendicular, their normal vectors will also be perpendicular.
- If a plane contains two points, it contains the line through those points; in particular its normal vector is perpendicular to the direction vector of the line.
So the desired plane's normal vector is perpendicular to $(1,-2,1)$ and $(3,-5,0)-(-2,4,3)=(5,-9,-3)$. The cross product yields such a perpendicular vector: $(15,8,1)$. So the equation of the plane is $15x+8y+z=k$, where $k$ is found by substituting either given point in: $15cdot(-2)+8cdot4+1cdot3=5$. The final answer is $15x+8y+z=5$.
answered May 26 '17 at 7:41
Parcly TaxelParcly Taxel
44.7k1376109
$endgroup$
add a comment |
$begingroup$
Two things first:
- If two planes are perpendicular, their normal vectors will also be perpendicular.
- If a plane contains two points, it contains the line through those points; in particular its normal vector is perpendicular to the direction vector of the line.
So the desired plane's normal vector is perpendicular to $(1,-2,1)$ and $(3,-5,0)-(-2,4,3)=(5,-9,-3)$. The cross product yields such a perpendicular vector: $(15,8,1)$. So the equation of the plane is $15x+8y+z=k$, where $k$ is found by substituting either given point in: $15cdot(-2)+8cdot4+1cdot3=5$. The final answer is $15x+8y+z=5$.
answered May 26 '17 at 7:41
Parcly TaxelParcly Taxel
44.7k1376109
$endgroup$
add a comment |
$begingroup$
Two things first:
- If two planes are perpendicular, their normal vectors will also be perpendicular.
- If a plane contains two points, it contains the line through those points; in particular its normal vector is perpendicular to the direction vector of the line.
So the desired plane's normal vector is perpendicular to $(1,-2,1)$ and $(3,-5,0)-(-2,4,3)=(5,-9,-3)$. The cross product yields such a perpendicular vector: $(15,8,1)$. So the equation of the plane is $15x+8y+z=k$, where $k$ is found by substituting either given point in: $15cdot(-2)+8cdot4+1cdot3=5$. The final answer is $15x+8y+z=5$.
answered May 26 '17 at 7:41
Parcly TaxelParcly Taxel
44.7k1376109
$endgroup$
Two things first:
- If two planes are perpendicular, their normal vectors will also be perpendicular.
- If a plane contains two points, it contains the line through those points; in particular its normal vector is perpendicular to the direction vector of the line.
So the desired plane's normal vector is perpendicular to $(1,-2,1)$ and $(3,-5,0)-(-2,4,3)=(5,-9,-3)$. The cross product yields such a perpendicular vector: $(15,8,1)$. So the equation of the plane is $15x+8y+z=k$, where $k$ is found by substituting either given point in: $15cdot(-2)+8cdot4+1cdot3=5$. The final answer is $15x+8y+z=5$.
answered May 26 '17 at 7:41
Parcly TaxelParcly Taxel
44.7k1376109
answered May 26 '17 at 7:41
Parcly TaxelParcly Taxel
44.7k1376109
answered May 26 '17 at 7:41
Parcly TaxelParcly Taxel
44.7k1376109
answered May 26 '17 at 7:41
Parcly TaxelParcly Taxel
44.7k1376109
44.7k1376109
add a comment |
add a comment |
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