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| ---- == Orthogonal Vector of a Given Plan == To identify a vector that is orthogonal to a given plane, use the [[Calculus/VectorOperations#Cross_Product|cross product]] on the vectors that make up that plane. ---- == 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 Given Plan
To identify a vector that is orthogonal to a given plane, use the cross product on the vectors that make up that plane.
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.