= Orthogonality =

'''Orthogonality''' is a generalization of perpendicularity.

<<TableOfContents>>

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== Description ==

The notation for orthogonality is ''⊥'', as in ''a⃗ ⊥ b⃗''.

In ''R^3^'' space, a vector is orthogonal to a plane.

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== Dot Product ==

If two vectors are perpendicular, they must satisfy the Pythagorean theorem. It follows that the [[Calculus/VectorOperations#Dot_Product|dot product]] must be 0.

''a^T^a + b^T^b = (a+b)^T^(a+b)''

''a^T^a + b^T^b = a^T^a + b^T^b + a^T^b + b^T^a''

''0 = a^T^b + b^T^a''

''0 = 2(a^T^b)''

''0 = a^T^b''

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== Normal Vectors ==

A frequent operation in vector calculus is identifying a normal vector. These are usually notated as ''N'', or ''n'' if a unit normal vector.

To identify a vector normal to a plane formed by two vectors, use the [[Calculus/VectorOperations#Cross_Product|cross product]] on those vectors.

To identify a vector normal to a plane formed by three points (''A'', ''B'', and ''C''), calculate the tangent vectors as ''B-A'' and ''C-A''. Then use the above strategy.

For a given plane (''ax + by + cz = d''), the trivial normal vector is ''[a b c]''.

In all cases, it should be understood that there are infinitely many normal vectors. For any given vector is normal to a given plane, it must be a similar to the trivial normal vector. That is, the given vector must be a scaled version of the trivial normal vector.

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== Normal Planes ==

To identify a plane normal 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.



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CategoryRicottone