Main Concept

A plane can be defined by five different methods:

The Cartesian equation of a plane π is $a\cdot x\+ b\cdot y \+ c\cdot z \+ d \= 0$, where $⟨a\,b\,c⟩$is the vector normal to the plane.

How to find the equation of the plane through a point with a given normal vector

Let $P \left(x_p\, y_p\, z_p\right)$ be the point and $A \left(x_{a\,}{y}_a\, z_a\right)$ be the normal vector. 1.  Substitute$x_{a\,}{y}_a\, z_a$into $a\, b\, c$ respectively. 2.  Plug in point P and solve for the last unknown variable d. Example: Find the equation of the plane that passes through the point $p \= \left(1\,1\,1\right)$ with a normal vector $\stackrel{\rightharpoonup }{A} \= \left(2\,3\,4\right)$ 1.  Substitute $\mathit{ }{x}_a\mathit{\,}\mathit{ }{y}_a\mathit{\,}\mathit{ }\mathit{ }{z}_a$ into $a\mathit{\,}\mathit{ }\mathit{ }b\mathit{\,}\mathit{ }\mathit{ }c$ respectively $\mathrm{\π}\: 2\cdot x \+ 3\cdot y \+ 4\cdot z \+ d \= 0$ 2.  Plug in point P and solve for the last unknown variable d. ${\mathrm{\π}}_{}\:$ $2\cdot x \+ 3\cdot y \+ 4\cdot z \+ d$ $\=$ $0$ $2\cdot \left(1\right) \+ 3\cdot \left(1\right)\+ 4\cdot \left(1\right) \+ d$ $\=$ $0$ $d$ $\=$ $-9$ 3.  The equation of the plane is $\mathit{2}\mathit{\cdot }x\mathit{ }\mathit{\+}\mathit{ }\mathit{3}\mathit{\cdot }y\mathit{ }\mathit{\+}\mathit{ }\mathit{4}\mathit{\cdot }z\mathit{ }\mathit{-}\mathit{9}\mathit{ }\mathit{\=}\mathit{0}$

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