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Lesson /10
Equation of a Plane

0.1.  Scalar equation of a plane in $\mathbb{R}^3$ is described by the general equation,

n_1x_1 + n_2x_2 +n_3x_3 = d

where $(n_1,n_2,n_3)$ are the coordinates of the vector that is normal to the plane.
This equation can be derived if we know the normal vector, and a known point on the plane, $P(p_1,p_2,p_3)$. Then, we define any general point on plane, $Q(x_1,x_2,x_3)$.
First, we create $\vec{PQ}$.

\vec{PQ} = \begin{bmatrix}x_1-p_1\\x_2-p_2\\x_3-p_3\end{bmatrix}

We know that $\vec{PQ}$ is on the plane, and will be orthogonal to the normal vector. Which means,
$\vec{n} \cdot \vec{PQ} = 0$
Expanding it out,

\begin{split}
n_1(x_1-p_1)+n_2(x_2-p_2)+n_3(x_3-p_3)=0
n_1x_1 + n_2x_2 +n_3x_3 – (n_1p_1+n_2p_2+n_3p_3)= 0
\end{split}

Since $(n_1p_1+n_2p_2+n_3p_3)$ are given numbers, it is possible to group them as another real number, $d$.

\begin{split}
n_1x_1 + n_2x_2 +n_3x_3 – d = 0
n_1x_1 + n_2x_2 +n_3x_3 = d
\end{split}

where $d = n\cdot p$.
Expanding this same concept to $\mathbb{R}^n$,

n_1x_1 + n_2x_2 +n_3x_3…n_nx_n = d

In this case, it is not called scalar equation of a plane, but scalar equation of a hyperplane.

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