Equation of Plane Passing through a Point and Perpendicular to a Vector

Theorem 6.8 (vector form). The plane through the point $A(\vec a)$ that is perpendicular to a fixed non-zero vector $\vec n$ has the equation $\vec r \cdot \vec n = \vec a \cdot \vec n$, where $P(\vec r)$ denotes an arbitrary point of the plane.

Why this is true: if $P(\vec r)$ lies on the plane, the chord $\overrightarrow{AP}$ lies entirely in the plane, so it must be perpendicular to the normal $\vec n$. Writing this perpendicularity as a dot product, $\overrightarrow{AP}\cdot\vec n = 0$. Since $\overrightarrow{AP} = \vec r - \vec a$, this becomes $(\vec r - \vec a)\cdot \vec n = 0$, and expanding the dot product over the subtraction gives $\vec r \cdot \vec n - \vec a \cdot \vec n = 0$, i.e. $\vec r\cdot\vec n = \vec a \cdot \vec n$. This is called the vector equation of a plane in scalar product form (Fig. 6.8 shows the point $A$, a general point $P$ on the plane, the chord $\overrightarrow{AP}$ lying flat in the plane, and $\vec n$ standing perpendicular to it). If the fixed number $\vec a \cdot \vec n$ is called $d$, the equation shortens further to $\vec r \cdot \vec n = d$.

Theorem 6.9 (Cartesian form). If $A(x_1, y_1, z_1)$ is a point of the plane and $a, b, c$ are the direction ratios of the normal, the same idea translated into coordinates gives $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$. Here $P(x, y, z)$ is a general point of the plane, so $x - x_1, y - y_1, z - z_1$ are the direction ratios of $\overrightarrow{AP}$; because $\overrightarrow{AP}$ is perpendicular to the normal, the sum of the products of corresponding direction ratios must vanish, which is exactly this equation. Multiplying out gives the familiar linear form $ax + by + cz + d = 0$.

Ex.(1): the plane through the point with position vector $2\hat i + 3\hat j + 4\hat k$ that is perpendicular to $2\hat i + \hat j - 2\hat k$. Here $\vec a \cdot \vec n = (2)(2) + (3)(1) + (4)(-2) = 4 + 3 - 8 = -1$, so by Theorem 6.8 the required equation is $\vec r \cdot (2\hat i + \hat j - 2\hat k) = -1$.

Ex.(2): the plane through $A(1, 2, 3)$ whose normal has direction ratios $3, 2, 5$. By Theorem 6.9, $3(x-1) + 2(y-2) + 5(z-3) = 0$, which simplifies to $3x + 2y + 5z - 22 = 0$. …