# Thornton & Marion, Classical Dynamics, Fifth Edition

## Chapter 1. Matrices, Vectors, and Vector Calculus

### Problem 08. An equation of a plane in vector form

##### The problem asks you to
show that the given equation is the equation of a plane.
##### This problem assumes
1) $\vec{A}$ $\vec{A}$ be a vector from the origin to a fixed point $P$ $P$

2) $\vec{r}$ $\vec{r}$ be a vector from the origin to a variabel point $Q(x_1,x_2,x_3)$ $Q(x_1,x_2,x_3)$
the equation of a plane, which is the form of $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$ $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$ 
##### Solution
Let the vector $\vec{A}$ $\vec{A}$ be $\vec{A} = (A_1, A_2, A_3)$ $\vec{A} = (A_1, A_2, A_3)$

and the vector $\vec{r}$ $\vec{r}$ be $\vec{r} = (x_1, x_2, x_3)$ $\vec{r} = (x_1, x_2, x_3)$

Then, $\begin{array}{rcl} \vec{A} \cdot \vec{r} & = & A_1 x_1 + A_2 x_2 + A_3 x_3 \\ & = & A^2 \\ & = & A^2_1 + A^2_2 + A^2_3 \end{array}$ $\begin{array}{rcl} \vec{A} \cdot \vec{r} & = & A_1 x_1 + A_2 x_2 + A_3 x_3 \\ & = & A^2 \\ & = & A^2_1 + A^2_2 + A^2_3 \end{array}$

Thus, $A_1 x_1 + A_2 x_2 + A_3 x_3 = A^2_1 + A^2_2 + A^2_3$ $A_1 x_1 + A_2 x_2 + A_3 x_3 = A^2_1 + A^2_2 + A^2_3$

and it becomes $A_1(x_1 - A_1) + A_2(x_2 - A_2) + A_3(x_3 - A_3) = 0$ $A_1(x_1 - A_1) + A_2(x_2 - A_2) + A_3(x_3 - A_3) = 0$

It is the equation of a plane perpendicular to $\vec{A}$ $\vec{A}$ and passing through the point $P$ $P$.