Equation of a Plane Through a Point and Perpendicular to a Vector

A plane in 3D space is uniquely determined by a point \( P(x_0, y_0, z_0) \) and a normal vector \( \vec{n} = \langle a, b, c \rangle \) perpendicular to the plane.

For any point \( M(x,y,z) \) on the plane, the vector \( \overrightarrow{PM} = \langle x-x_0, y-y_0, z-z_0 \rangle \) satisfies the orthogonality condition:

\( \vec{n} \cdot \overrightarrow{PM} = a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \)

This expands to the Cartesian equation \( ax + by + cz = d \), where \( d = a x_0 + b y_0 + c z_0 \).

Plane Equation Solver

Enter point coordinates and normal vector components

Point \( P(x_0, y_0, z_0) \)

Normal Vector \( \vec{n} = \langle n_x, n_y, n_z \rangle \)

⚠️ Normal vector cannot be zero

Plane Equation (Cartesian Form)

\( 3x - y + 2z = 19 \)

Step-by-Step Solution

STEP 1: Components of vector \( \overrightarrow{PM} \)

\( \overrightarrow{PM} = \langle x - 2,\; y + 3,\; z - 5 \rangle \)

STEP 2: Orthogonality condition \( \vec{n} \cdot \overrightarrow{PM} = 0 \)

\( \langle 3,-1,2 \rangle \cdot \langle x-2,\; y+3,\; z-5 \rangle = 0 \)

STEP 3: Expand and simplify → \( ax + by + cz = d \)

\( 3(x-2) -1(y+3) + 2(z-5) = 0 \) → \( 3x - y + 2z = 19 \)

Geometric Interpretation

The normal vector \( \vec{n} \) is perpendicular to every direction vector lying in the plane. The dot product condition ensures all points \( M \) satisfy orthogonality.