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How do you prove two planes are coincident? - Answers

By "coincident", do you mean that they lie on the same plane (i.e. they are instances of the same plane)? Assuming that's what you're after: To begin with, you should have a plane equation defining each plane in the following form: ax + by + cz + d = 0 This form can be re-written in vector form as: (a,b,c) dot (x,y,z) + d = 0 Written in this form, (a,b,c) is a vector that defines the normal of the plane (a vector perpendicular to the plane), Which determines the orientation of the plane around the origin. The scalar value d defines a distance that the plane is offset from the origin, times the length of vector (a,b,c).In case you aren't familiar with the dot product, it is a way of multiplying two vectors resulting in a scalar value that represents the angle and the magnitude of the two vectors. Specifically: v1 dot v2 = |v1| |v2| cos(theta)where: * theta is the angle between the vectors * |v| is the length of a vector To compare the two planes, the first thing that we will want to do is "normalize" the plane equations, that is, scale them so that the length of (a,b,c) is 1.0. To do this, divide each constant in the equation by the length of the vector (a,b,c): (ax + by + cz + d) / |(a, b, c)| = 0 The normalized version of the plane equation defines the same plane as the original plane equation. The difference is that now we can readily use the dot product to determine the angle between the two plane's normals, as "v1 dot v2 = cos(theta)" when |v1| and |v2| are both equal to 1.0. The cosine of zero degrees is 1.0, so if we calculate the dot product of the normal vectors of the two planes we can determine if the normals, and thus the planes, are parallel: cos(theta) = (a1,b1,c1) dot (a2,b2,c2)IF cos(theta) < 1.0the planes are not parallel, and thus cannot be coincidental.IF cos(theta) = 1.0the planes are parallel, and may be coincidentalIF cos(theta) = -1.0the planes are parallel, with normals facing opposite directionsthe planes may be coincidental If the planes are not parallel, you don't have to test any further, they are not coincidental. If they are parallel, then you can determine if they are coincidental by comparing the normalized d from each equation to see if they are equal, flipping the sign if the normals are facing opposite directions: IF d1 = cos(theta) d2the planes are coincident



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How do you prove two planes are coincident? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_prove_two_planes_are_coincident

By "coincident", do you mean that they lie on the same plane (i.e. they are instances of the same plane)? Assuming that's what you're after: To begin with, you should have a plane equation defining each plane in the following form: ax + by + cz + d = 0 This form can be re-written in vector form as: (a,b,c) dot (x,y,z) + d = 0 Written in this form, (a,b,c) is a vector that defines the normal of the plane (a vector perpendicular to the plane), Which determines the orientation of the plane around the origin. The scalar value d defines a distance that the plane is offset from the origin, times the length of vector (a,b,c).In case you aren't familiar with the dot product, it is a way of multiplying two vectors resulting in a scalar value that represents the angle and the magnitude of the two vectors. Specifically: v1 dot v2 = |v1| |v2| cos(theta)where: * theta is the angle between the vectors * |v| is the length of a vector To compare the two planes, the first thing that we will want to do is "normalize" the plane equations, that is, scale them so that the length of (a,b,c) is 1.0. To do this, divide each constant in the equation by the length of the vector (a,b,c): (ax + by + cz + d) / |(a, b, c)| = 0 The normalized version of the plane equation defines the same plane as the original plane equation. The difference is that now we can readily use the dot product to determine the angle between the two plane's normals, as "v1 dot v2 = cos(theta)" when |v1| and |v2| are both equal to 1.0. The cosine of zero degrees is 1.0, so if we calculate the dot product of the normal vectors of the two planes we can determine if the normals, and thus the planes, are parallel: cos(theta) = (a1,b1,c1) dot (a2,b2,c2)IF cos(theta) < 1.0the planes are not parallel, and thus cannot be coincidental.IF cos(theta) = 1.0the planes are parallel, and may be coincidentalIF cos(theta) = -1.0the planes are parallel, with normals facing opposite directionsthe planes may be coincidental If the planes are not parallel, you don't have to test any further, they are not coincidental. If they are parallel, then you can determine if they are coincidental by comparing the normalized d from each equation to see if they are equal, flipping the sign if the normals are facing opposite directions: IF d1 = cos(theta) d2the planes are coincident



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https://math.answers.com/math-and-arithmetic/How_do_you_prove_two_planes_are_coincident

How do you prove two planes are coincident? - Answers

By "coincident", do you mean that they lie on the same plane (i.e. they are instances of the same plane)? Assuming that's what you're after: To begin with, you should have a plane equation defining each plane in the following form: ax + by + cz + d = 0 This form can be re-written in vector form as: (a,b,c) dot (x,y,z) + d = 0 Written in this form, (a,b,c) is a vector that defines the normal of the plane (a vector perpendicular to the plane), Which determines the orientation of the plane around the origin. The scalar value d defines a distance that the plane is offset from the origin, times the length of vector (a,b,c).In case you aren't familiar with the dot product, it is a way of multiplying two vectors resulting in a scalar value that represents the angle and the magnitude of the two vectors. Specifically: v1 dot v2 = |v1| |v2| cos(theta)where: * theta is the angle between the vectors * |v| is the length of a vector To compare the two planes, the first thing that we will want to do is "normalize" the plane equations, that is, scale them so that the length of (a,b,c) is 1.0. To do this, divide each constant in the equation by the length of the vector (a,b,c): (ax + by + cz + d) / |(a, b, c)| = 0 The normalized version of the plane equation defines the same plane as the original plane equation. The difference is that now we can readily use the dot product to determine the angle between the two plane's normals, as "v1 dot v2 = cos(theta)" when |v1| and |v2| are both equal to 1.0. The cosine of zero degrees is 1.0, so if we calculate the dot product of the normal vectors of the two planes we can determine if the normals, and thus the planes, are parallel: cos(theta) = (a1,b1,c1) dot (a2,b2,c2)IF cos(theta) < 1.0the planes are not parallel, and thus cannot be coincidental.IF cos(theta) = 1.0the planes are parallel, and may be coincidentalIF cos(theta) = -1.0the planes are parallel, with normals facing opposite directionsthe planes may be coincidental If the planes are not parallel, you don't have to test any further, they are not coincidental. If they are parallel, then you can determine if they are coincidental by comparing the normalized d from each equation to see if they are equal, flipping the sign if the normals are facing opposite directions: IF d1 = cos(theta) d2the planes are coincident

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      By "coincident", do you mean that they lie on the same plane (i.e. they are instances of the same plane)? Assuming that's what you're after: To begin with, you should have a plane equation defining each plane in the following form: ax + by + cz + d = 0 This form can be re-written in vector form as: (a,b,c) dot (x,y,z) + d = 0 Written in this form, (a,b,c) is a vector that defines the normal of the plane (a vector perpendicular to the plane), Which determines the orientation of the plane around the origin. The scalar value d defines a distance that the plane is offset from the origin, times the length of vector (a,b,c).In case you aren't familiar with the dot product, it is a way of multiplying two vectors resulting in a scalar value that represents the angle and the magnitude of the two vectors. Specifically: v1 dot v2 = |v1| |v2| cos(theta)where: * theta is the angle between the vectors * |v| is the length of a vector To compare the two planes, the first thing that we will want to do is "normalize" the plane equations, that is, scale them so that the length of (a,b,c) is 1.0. To do this, divide each constant in the equation by the length of the vector (a,b,c): (ax + by + cz + d) / |(a, b, c)| = 0 The normalized version of the plane equation defines the same plane as the original plane equation. The difference is that now we can readily use the dot product to determine the angle between the two plane's normals, as "v1 dot v2 = cos(theta)" when |v1| and |v2| are both equal to 1.0. The cosine of zero degrees is 1.0, so if we calculate the dot product of the normal vectors of the two planes we can determine if the normals, and thus the planes, are parallel: cos(theta) = (a1,b1,c1) dot (a2,b2,c2)IF cos(theta) < 1.0the planes are not parallel, and thus cannot be coincidental.IF cos(theta) = 1.0the planes are parallel, and may be coincidentalIF cos(theta) = -1.0the planes are parallel, with normals facing opposite directionsthe planes may be coincidental If the planes are not parallel, you don't have to test any further, they are not coincidental. If they are parallel, then you can determine if they are coincidental by comparing the normalized d from each equation to see if they are equal, flipping the sign if the normals are facing opposite directions: IF d1 = cos(theta) d2the planes are coincident
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