Distance of a Point from a Plane

Distance of a Point from a Plane (Cartesian Form)

[Image of perpendicular distance from a point to a plane] To find the perpendicular distance from a specific point to a given plane, we use a standard formula based on the plane’s normal vector.

Let the equation of the plane be $$Ax + By + Cz + D = 0$$ and let the given point be $$P(x_1, y_1, z_1).$$

Then the perpendicular distance $d$ from the point $P$ to the plane is $$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}.$$

Here:

• $Ax_1 + By_1 + Cz_1 + D$ is obtained by substituting the coordinates of the point into the plane equation.
• $\sqrt{A^2 + B^2 + C^2}$ is the magnitude of the normal vector $(A, B, C)$ of the plane.
• Absolute value is taken because distance is always non-negative.

Special Case

For the origin $(0,0,0)$, the formula becomes $$d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}.$$

Distance of a Point from a Plane (Vector Form)

Let the plane be written in vector form as $$\vec{r} \cdot \vec{N} = D,$$ where $\vec{N}$ is a normal vector to the plane. Let the point have position vector $\vec{a}$.

Then the perpendicular distance from the point to the plane is $$d = \frac{|\vec{a} \cdot \vec{N} - D|}{|\vec{N}|}.$$

This is exactly the vector analogue of the Cartesian formula.

Normal Form

If the plane is given in normal form $$\vec{r} \cdot \hat{n} = p,$$ where $\hat{n}$ is a unit normal vector, then since $|\hat{n}| = 1$, the formula simplifies to $$d = |\vec{a} \cdot \hat{n} - p|.$$

Distance Between Two Parallel Planes

Two parallel planes have the same normal direction. So their normal coefficients must first be made identical before applying the formula.

Let the two parallel planes be $$\pi_1: Ax + By + Cz + d_1 = 0$$ $$\pi_2: Ax + By + Cz + d_2 = 0$$ Then the perpendicular distance between them is $$d = \frac{|d_1 - d_2|}{\sqrt{A^2 + B^2 + C^2}}.$$

This works because the numerator measures the difference along the normal direction, and the denominator normalizes by the length of the common normal vector.

Vector Form

If the planes are written as $$\vec{r} \cdot \vec{N} = d_1 \quad \text{and} \quad \vec{r} \cdot \vec{N} = d_2,$$ then the distance between them is $$d = \frac{|d_1 - d_2|}{|\vec{N}|}.$$

Foot of the Perpendicular and Image of a Point

A very important application of point-plane distance is finding the foot of the perpendicular and the image of a point in a plane.

Let the plane be $$Ax + By + Cz + D = 0$$ and the point be $$P(x_1, y_1, z_1).$$

Foot of the Perpendicular

The foot of the perpendicular from $P$ to the plane lies on the line through $P$ in the direction of the plane’s normal vector $(A, B, C)$. So the normal line is $$\frac{x-x_1}{A} = \frac{y-y_1}{B} = \frac{z-z_1}{C} = \lambda.$$ A general point on this line is $$(x_1 + A\lambda,\ y_1 + B\lambda,\ z_1 + C\lambda).$$ Since the foot lies on the plane, we substitute these coordinates into the plane equation and solve for $\lambda$.

Image of a Point in a Plane

If $Q$ is the foot of the perpendicular from $P$ to the plane, and $P'$ is the image of $P$ in the plane, then $Q$ is the midpoint of $PP'$. Hence, $$Q_x = \frac{x_1 + x'}{2}, \quad Q_y = \frac{y_1 + y'}{2}, \quad Q_z = \frac{z_1 + z'}{2}.$$ This gives the coordinates of the image point.

Example 1: Basic Cartesian Distance

Find the distance of the point $(2, 3, -5)$ from the plane $x + 2y - 2z = 9$.

Solution: Step 1: Rewrite the plane in the standard form $Ax + By + Cz + D = 0$. $$x + 2y - 2z - 9 = 0$$ So, $$A = 1, \quad B = 2, \quad C = -2, \quad D = -9.$$ The point is $$(x_1, y_1, z_1) = (2, 3, -5).$$

Step 2: Use the distance formula: $$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}.$$ Substitute the values: $$d = \frac{|1(2) + 2(3) + (-2)(-5) - 9|}{\sqrt{1^2 + 2^2 + (-2)^2}}.$$

Step 3: Simplify the numerator: $$|2 + 6 + 10 - 9| = |9| = 9.$$

Step 4: Simplify the denominator: $$\sqrt{1 + 4 + 4} = \sqrt{9} = 3.$$

Step 5: Compute the distance: $$d = \frac{9}{3} = 3.$$

Answer: 3 units.

Example 2: Vector Form Distance

Find the distance of a point having position vector $2\hat{i} - \hat{j} - 4\hat{k}$ from the plane $\vec{r} \cdot (3\hat{i} - 4\hat{j} + 12\hat{k}) = 9$.

Solution: Step 1: Identify the data. Point position vector: $$\vec{a} = 2\hat{i} - \hat{j} - 4\hat{k}.$$ Plane normal vector: $$\vec{N} = 3\hat{i} - 4\hat{j} + 12\hat{k}.$$ Constant: $$D = 9.$$

Step 2: Use the vector formula: $$d = \frac{|\vec{a} \cdot \vec{N} - D|}{|\vec{N}|}.$$

Step 3: Compute the dot product: $$\vec{a} \cdot \vec{N} = (2)(3) + (-1)(-4) + (-4)(12) = 6 + 4 - 48 = -38.$$

Step 4: Compute the magnitude of $\vec{N}$: $$|\vec{N}| = \sqrt{3^2 + (-4)^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13.$$

Step 5: Substitute: $$d = \frac{|-38 - 9|}{13} = \frac{|-47|}{13} = \frac{47}{13}.$$

Answer: $47/13$ units.

Example 3: Distance from the Origin

Find the distance of the plane $2x - 3y + 6z + 14 = 0$ from the origin.

Solution: Step 1: The point is the origin $(0,0,0)$. For a plane $$Ax + By + Cz + D = 0,$$ the distance of the origin is $$d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}.$$

Step 2: Identify the coefficients: $$A = 2, \quad B = -3, \quad C = 6, \quad D = 14.$$

Step 3: Compute the denominator: $$\sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7.$$

Step 4: Compute the distance: $$d = \frac{|14|}{7} = 2.$$

Answer: 2 units.

Example 4: Distance Between Parallel Planes

Find the distance between the parallel planes $2x - y + 2z + 3 = 0$ and $4x - 2y + 4z + 5 = 0$.

Solution: Step 1: The normals must be made identical before using the formula. Divide the second plane by 2: $$4x - 2y + 4z + 5 = 0 \implies 2x - y + 2z + \frac{5}{2} = 0.$$

Step 2: Now the planes are $$\pi_1: 2x - y + 2z + 3 = 0,$$ $$\pi_2: 2x - y + 2z + \frac{5}{2} = 0.$$

Step 3: Use the formula for distance between parallel planes: $$d = \frac{|d_1 - d_2|}{\sqrt{A^2 + B^2 + C^2}}.$$ Here, $$d_1 = 3, \quad d_2 = \frac{5}{2}, \quad A = 2, \quad B = -1, \quad C = 2.$$ So, $$d = \frac{\left|3 - \frac{5}{2}\right|}{\sqrt{2^2 + (-1)^2 + 2^2}}.$$

Step 4: Simplify: $$\left|3 - \frac{5}{2}\right| = \frac{1}{2},$$ $$\sqrt{4 + 1 + 4} = \sqrt{9} = 3.$$ Thus, $$d = \frac{1/2}{3} = \frac{1}{6}.$$

Answer: $1/6$ units.

Example 5: Finding a Plane at a Given Distance

Find the equation of the plane(s) parallel to the plane $x - 2y + 2z - 3 = 0$ which are at a unit distance from the point $(1, 2, 3)$.

Solution: Step 1: Any plane parallel to $$x - 2y + 2z - 3 = 0$$ has the same normal coefficients. So its equation is of the form $$x - 2y + 2z + d = 0,$$ where $d$ is to be found.

Step 2: The distance from $(1,2,3)$ to this plane is 1. Use the point-plane distance formula: $$1 = \frac{|1(1) - 2(2) + 2(3) + d|}{\sqrt{1^2 + (-2)^2 + 2^2}}.$$

Step 3: Simplify: $$1 = \frac{|1 - 4 + 6 + d|}{\sqrt{1 + 4 + 4}} = \frac{|3 + d|}{3}.$$ So, $$|3 + d| = 3.$$

Step 4: Solve the absolute value equation: $$3 + d = 3 \quad \text{or} \quad 3 + d = -3.$$ Hence, $$d = 0 \quad \text{or} \quad d = -6.$$

Step 5: Therefore the required planes are $$x - 2y + 2z = 0$$ and $$x - 2y + 2z - 6 = 0.$$

Answer: $x - 2y + 2z = 0$ and $x - 2y + 2z - 6 = 0$

Example 6: Foot of the Perpendicular

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $2x - 3y + 4z - 6 = 0$.

Solution: Step 1: The normal vector to the plane is $$(2, -3, 4).$$ So the perpendicular from the origin to the plane lies along the line with direction ratios $2, -3, 4$.

Step 2: Equation of the normal line through the origin: $$\frac{x}{2} = \frac{y}{-3} = \frac{z}{4} = \lambda.$$ Hence a general point on the line is $$Q(2\lambda, -3\lambda, 4\lambda).$$

Step 3: Since $Q$ lies on the plane, substitute into the plane equation: $$2(2\lambda) - 3(-3\lambda) + 4(4\lambda) - 6 = 0.$$ This gives $$4\lambda + 9\lambda + 16\lambda - 6 = 0,$$ $$29\lambda = 6,$$ $$\lambda = \frac{6}{29}.$$

Step 4: Substitute back into coordinates: $$x = 2\cdot\frac{6}{29} = \frac{12}{29},$$ $$y = -3\cdot\frac{6}{29} = -\frac{18}{29},$$ $$z = 4\cdot\frac{6}{29} = \frac{24}{29}.$$

Answer: $(12/29, -18/29, 24/29)$

Example 7: Image of a Point in a Plane

Find the image of the point $P(1, 3, 4)$ in the plane $2x - y + z + 3 = 0$.

Solution: Step 1: The normal vector to the plane is $$(2, -1, 1).$$ So the perpendicular line through $P(1,3,4)$ is $$\frac{x-1}{2} = \frac{y-3}{-1} = \frac{z-4}{1} = \lambda.$$ Hence a general point on this line is $$Q(2\lambda + 1, -\lambda + 3, \lambda + 4).$$

Step 2: Since $Q$ is the foot of the perpendicular, it lies on the plane. Substitute into $$2x - y + z + 3 = 0.$$ So, $$2(2\lambda + 1) - (-\lambda + 3) + (\lambda + 4) + 3 = 0.$$ Simplify: $$4\lambda + 2 + \lambda - 3 + \lambda + 4 + 3 = 0,$$ $$6\lambda + 6 = 0,$$ $$\lambda = -1.$$

Step 3: Coordinates of the foot of the perpendicular are $$Q = (2(-1)+1, -(-1)+3, -1+4) = (-1, 4, 3).$$

Step 4: Let the image point be $P'(x',y',z')$. Since $Q$ is the midpoint of $PP'$, use midpoint formula: $$-1 = \frac{1 + x'}{2}, \qquad 4 = \frac{3 + y'}{2}, \qquad 3 = \frac{4 + z'}{2}.$$ Solving: $$x' = -3, \qquad y' = 5, \qquad z' = 2.$$

Answer: $(-3, 5, 2)$

Example 8: Locus based on Distance

Find the locus of a point which moves such that its distance from the plane $x + y + z = 1$ is twice its distance from the plane $x - y + z = 2$.

Solution: Step 1: Let the moving point be $$P(x,y,z).$$

Step 2: Distance from the first plane: $$d_1 = \frac{|x + y + z - 1|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{|x + y + z - 1|}{\sqrt{3}}.$$

Step 3: Distance from the second plane: $$d_2 = \frac{|x - y + z - 2|}{\sqrt{1^2 + (-1)^2 + 1^2}} = \frac{|x - y + z - 2|}{\sqrt{3}}.$$

Step 4: Given condition is $$d_1 = 2d_2.$$ So, $$\frac{|x + y + z - 1|}{\sqrt{3}} = 2\cdot\frac{|x - y + z - 2|}{\sqrt{3}}.$$ Cancel $\sqrt{3}$: $$|x + y + z - 1| = 2|x - y + z - 2|.$$

Step 5: Remove absolute values by taking both signs.

Case 1: $$x + y + z - 1 = 2(x - y + z - 2)$$ $$x + y + z - 1 = 2x - 2y + 2z - 4$$ $$x - 3y + z - 3 = 0.$$

Case 2: $$x + y + z - 1 = -2(x - y + z - 2)$$ $$x + y + z - 1 = -2x + 2y - 2z + 4$$ $$3x - y + 3z - 5 = 0.$$

So the locus is the pair of planes: $$x - 3y + z - 3 = 0$$ or $$3x - y + 3z - 5 = 0.$$

Answer: $x - 3y + z - 3 = 0$ or $3x - y + 3z - 5 = 0$

Example 9: Distance to the Intersection Point of a Line and a Plane

Find the distance of the point $P(-1, -5, -10)$ from the point of intersection of the line $\vec{r} = 2\hat{i} - \hat{j} + 2\hat{k} + \lambda(3\hat{i} + 4\hat{j} + 2\hat{k})$ and the plane $\vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5$.

Solution: Step 1: Write a general point on the line: $$\vec{r} = (2 + 3\lambda)\hat{i} + (-1 + 4\lambda)\hat{j} + (2 + 2\lambda)\hat{k}.$$ So coordinates are $$(2 + 3\lambda,\ -1 + 4\lambda,\ 2 + 2\lambda).$$

Step 2: Substitute into the plane equation $$\vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5.$$ Thus, $$(2 + 3\lambda)(1) + (-1 + 4\lambda)(-1) + (2 + 2\lambda)(1) = 5.$$ Simplify: $$2 + 3\lambda + 1 - 4\lambda + 2 + 2\lambda = 5,$$ $$\lambda + 5 = 5,$$ $$\lambda = 0.$$

Step 3: Therefore the point of intersection is obtained by putting $\lambda=0$: $$Q = (2, -1, 2).$$

Step 4: Now find the distance between $$P(-1,-5,-10) \quad \text{and} \quad Q(2,-1,2).$$ Using the 3D distance formula, $$PQ = \sqrt{(2-(-1))^2 + (-1-(-5))^2 + (2-(-10))^2}.$$ So, $$PQ = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13.$$

Answer: 13 units.

Example 10: Ratio of Distances / Division of a Segment by a Plane

Determine the ratio in which the plane $2x - 3y + z = 4$ divides the line segment joining the points $A(2, 1, 3)$ and $B(3, -2, 5)$.

Solution: Step 1: First check whether either endpoint lies on the plane. For point $A(2,1,3)$, $$2(2) - 3(1) + 3 = 4 - 3 + 3 = 4.$$ So point $A$ lies on the plane.

For completeness, point $B(3,-2,5)$ gives $$2(3) - 3(-2) + 5 = 6 + 6 + 5 = 17 \ne 4.$$ So $B$ does not lie on the plane.

Step 2: Since the plane passes through $A$, the plane meets the segment $AB$ exactly at the endpoint $A$. Therefore the segment is divided at $A$ itself.

Step 3: In ratio language, this is a degenerate case and may be written as $$0:1,$$ meaning the dividing point coincides with $A$.

Answer: $0:1$ (The plane passes through point $A$).