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| ---- == Orthogonal Vector of a Plane == To identify a vector that is orthogonal to a plane formed by two vectors, use the [[Calculus/VectorOperations#Cross_Product|cross product]] on those vectors. ---- == Orthogonal Vector of a Given Plane == To identify a vector that is orthogonal to a given plane, by which I mean as a system of equations like ''x + 2y + 3z = 0'', simply extract the coefficients. The vector ''[1 2 3]'' passes through the origin and is orthogonal to that example plane. A test for orthogonality between a vector and a plane follows from this. A given vector is only orthogonal to a given plane if it is parallel to that trivial orthogonal vector. Two vectors are parallel if they are similar, i.e. they are equal given some scaling factor. By the same logic, a test for parallelism between a vector and plane is to test for a dot product of 0 between the trivial orthogonal vector and the given vector. ---- == Orthogonal Plane of a Given Vector == To identify a plane that is orthogonal to a given vector a⃗, use the dot product to design a system of equations. If the origin should be included in the solution, then the plane must be composed of a vector b⃗ from the origin to a point at ''(x, y, z)''. b⃗ is then characterized by ''[(x-0) (y-0) (z-0)]'' or ''[x y z]''. The dot product of this and a⃗ must be 0. If the given vector a⃗ is ''[1 5 10]'' then: ''a⃗ · b⃗ = 0'' ''[1 5 10] · [x y z] = 0'' ''x + 5y + 10z = 0'' The plane is characterized by this equation. If a given point P should be included in the solution, as opposed to the origin, simply update with ''[(x-p,,1,,) (y-p,,2,,) (z-p,,3,,)]''. If the given point P is ''(2, 1, -1)'' then: ''a⃗ · b⃗' = 0'' ''[1 5 10] · [(x-2) (y-1) (z+1)] = 0'' ''(x-2) + 5(y-1) + 10(z+1) = 0'' ''x + 5y + 10z = -3'' This system reveals a parallel plane with a constant offset of -3. |
Orthogonality
Orthogonality is a generalization of perpendicularity.
Contents
Description
The notation for orthogonality is ⊥, as in a⃗ ⊥ b⃗.
In R3 space, a vector is orthogonal to a plane.
Dot Product
If two vectors are perpendicular, they must satisfy the Pythagorean theorem. It follows that the dot product must be 0.
aTa + bTb = (a+b)T(a+b)
aTa + bTb = aTa + bTb + aTb + bTa
0 = aTb + bTa
0 = 2(aTb)
0 = aTb
Orthogonal Vector of a Plane
To identify a vector that is orthogonal to a plane formed by two vectors, use the cross product on those vectors.
Orthogonal Vector of a Given Plane
To identify a vector that is orthogonal to a given plane, by which I mean as a system of equations like x + 2y + 3z = 0, simply extract the coefficients. The vector [1 2 3] passes through the origin and is orthogonal to that example plane.
A test for orthogonality between a vector and a plane follows from this. A given vector is only orthogonal to a given plane if it is parallel to that trivial orthogonal vector. Two vectors are parallel if they are similar, i.e. they are equal given some scaling factor.
By the same logic, a test for parallelism between a vector and plane is to test for a dot product of 0 between the trivial orthogonal vector and the given vector.
Orthogonal Plane of a Given Vector
To identify a plane that is orthogonal to a given vector a⃗, use the dot product to design a system of equations.
If the origin should be included in the solution, then the plane must be composed of a vector b⃗ from the origin to a point at (x, y, z). b⃗ is then characterized by [(x-0) (y-0) (z-0)] or [x y z]. The dot product of this and a⃗ must be 0. If the given vector a⃗ is [1 5 10] then:
a⃗ · b⃗ = 0
[1 5 10] · [x y z] = 0
x + 5y + 10z = 0
The plane is characterized by this equation.
If a given point P should be included in the solution, as opposed to the origin, simply update with [(x-p1) (y-p2) (z-p3)]. If the given point P is (2, 1, -1) then:
a⃗ · b⃗' = 0
[1 5 10] · [(x-2) (y-1) (z+1)] = 0
(x-2) + 5(y-1) + 10(z+1) = 0
x + 5y + 10z = -3
This system reveals a parallel plane with a constant offset of -3.