Shortest Distance from Point to Plane Formula

Observation

The provided image displays a geometric diagram (Figure 7.6) illustrating a point $A$ positioned in 3D space above a shaded plane. A line, representing the normal vector's path, passes through point $A$ and intersects the plane perpendicularly at point $P$. Point $P$ is the foot of the perpendicular, and the segment $AP$ represents the shortest distance from the point to the plane.

Answer

The shortest distance from a point $A(x_1, y_1, z_1)$ to a plane defined by $ax + by + cz + d = 0$ is found by calculating the length of the perpendicular segment $AP$. This is analytically determined using the formula $D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$.

Explanation

1. Identify the Normal Vector and Point To find the distance $AP$, we first identify the normal vector $\vec{n}$ of the plane from its Cartesian equation. For a plane $ax + by + cz + d = 0$, the normal vector $\vec{n}$ is: $\vec{n} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ The normal vector is a vector perpendicular to every line lying on the plane surface.

2. Formulate the Vector Line Equation The line passing through $A$ and $P$ is parallel to the normal vector $\vec{n}$ because the shortest distance is along a perpendicular path. We express the line equation in vector form using point $A(\vec{a})$ and direction $\vec{n}$: $\vec{r} = \vec{a} + \lambda \vec{n}$ This equation describes the position of any point on the line through $A$ that is perpendicular to the plane.

3. Find the Intersection Point P We substitute the components of the line equation into the plane equation $ax + by + cz + d = 0$. Solving for the scalar parameter $\lambda$ gives us the specific value at which the line intersects the plane at point $P$. ⚠️ This step is required on exams to find the coordinates of the "foot of the perpendicular."

4. Calculate the Magnitude of AP Once $P$ is found, the distance is the magnitude of the vector $\vec{AP}$. Alternatively, by applying the projection of a vector onto the normal, we derive the general distance formula: $D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$ This formula calculates the perpendicular distance by scaling the "error" of the point in the plane equation by the magnitude of the normal vector.

Final Answer

The shortest distance $D$ from a point $A(x_1, y_1, z_1)$ to a plane $ax + by + cz + d = 0$ is: $\boxed{D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}}$

Common Mistakes

• Forgetting the Absolute Value: Students often forget the modulus $|...|$ in the numerator. Since distance must be a non-negative scalar, the absolute value ensures a positive result even if the point lies "below" the plane relative to the normal direction.
• Incorrect Plane Constant: Ensure the plane equation is in the form $ax + by + cz + d = 0$. If the equation is given as $ax + by + cz = k$, you must rewrite it as $ax + by + cz - k = 0$ before identifying $d$.