1. **Problem Statement:** Find the equation of a plane given certain conditions (e.g., a point and a normal vector). 2. **Formula:** The general equation of a plane is given by $$Ax + By + Cz = D$$ where $(A, B, C)$ is the normal vector to the plane and $(x, y, z)$ are coordinates of any point on the plane. 3. **Important Rule:** The normal vector is perpendicular to every vector lying on the plane. 4. **Step-by-step:** - Suppose the plane passes through point $P_0(x_0, y_0, z_0)$ and has a normal vector $\vec{n} = (A, B, C)$. - The equation of the plane is derived from the dot product: $$\vec{n} \cdot (\vec{r} - \vec{r_0}) = 0$$ where $\vec{r} = (x, y, z)$ is any point on the plane and $\vec{r_0} = (x_0, y_0, z_0)$. - Expanding the dot product: $$A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$$ - Simplify to get the plane equation: $$Ax + By + Cz = D$$ where $$D = Ax_0 + By_0 + Cz_0$$. 5. **Example:** If the plane passes through $P_0(1, 2, 3)$ and has normal vector $\vec{n} = (4, -5, 6)$, then: - Calculate $D = 4(1) + (-5)(2) + 6(3) = 4 - 10 + 18 = 12$ - Equation: $$4x - 5y + 6z = 12$$ This is the equation of the plane.