Equation of a Plane

Introduction to the Plane and its Normal Form

A plane is a flat two-dimensional surface extending infinitely in all directions within itself. In three-dimensional geometry, a plane is uniquely determined if any one of the following is known:

1. The normal to the plane and its perpendicular distance from the origin.
2. A point through which it passes and a vector perpendicular to it.
3. Three non-collinear points lying on it.

Equation of a Plane in Normal Form

Let the perpendicular distance of the plane from the origin be $d$ with $d>0$, and let $\hat{n}$ be a unit normal vector to the plane directed from the origin toward the plane. If $P$ is any point on the plane with position vector $\vec{r}$, then the projection of $\vec{r}$ on $\hat{n}$ is constant and equal to $d$. Hence, the vector equation of the plane is $$\vec{r} \cdot \hat{n} = d$$

Cartesian Form

If $$\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}, \qquad \hat{n} = l\hat{i} + m\hat{j} + n\hat{k},$$ where $l,m,n$ are the direction cosines of the unit normal, then $$lx + my + nz = d$$ This is the Cartesian equation in normal form.

Here:

• $l^2 + m^2 + n^2 = 1$ because $\hat{n}$ is a unit vector.
• $d$ is the actual perpendicular distance from the origin to the plane.

Plane Perpendicular to a Given Vector and Passing Through a Given Point

Suppose a plane passes through the point $A$ with position vector $\vec{a}$ and is perpendicular to a given vector $\vec{N}$. Let $P$ be any point on the plane with position vector $\vec{r}$. Then the vector $$\vec{AP} = \vec{r} - \vec{a}$$ lies in the plane. Since $\vec{N}$ is normal to the plane, it is perpendicular to every vector lying in the plane. Therefore, $$(\vec{r} - \vec{a}) \cdot \vec{N} = 0$$ This is the vector equation of the plane.

Cartesian Form: If the given point is $A(x_1,y_1,z_1)$ and the direction ratios of the normal vector are $A,B,C$, then $$A(x-x_1) + B(y-y_1) + C(z-z_1) = 0$$ Expanding, $$Ax + By + Cz + D = 0$$ where $$D = -(Ax_1 + By_1 + Cz_1).$$ This is the most general Cartesian equation of a plane, where $(A,B,C)$ are the direction ratios of a normal to the plane.

Equation of a Plane Passing Through Three Non-Collinear Points

Let the plane pass through three non-collinear points $$A(x_1,y_1,z_1), \quad B(x_2,y_2,z_2), \quad C(x_3,y_3,z_3).$$ Then the vectors $$\vec{AB} = \vec{b}-\vec{a}, \qquad \vec{AC} = \vec{c}-\vec{a}$$ lie in the plane. Since the cross product of two non-parallel vectors in the plane gives a normal vector to the plane, a normal is $$(\vec{b}-\vec{a}) \times (\vec{c}-\vec{a}).$$ Hence the vector equation of the plane is $$(\vec{r}-\vec{a}) \cdot [(\vec{b}-\vec{a}) \times (\vec{c}-\vec{a})] = 0$$

Cartesian Form: If $P(x,y,z)$ is any general point on the plane, then the vectors $\vec{AP}, \vec{AB}, \vec{AC}$ are coplanar. Therefore, their scalar triple product is zero: $$\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0$$ This determinant form is valid only when the three given points are non-collinear. If the three points are collinear, no unique plane is determined.

Intercept Form of the Equation of a Plane

Suppose a plane cuts intercepts $a, b, c$ on the x, y, and z-axes respectively. Then the plane passes through the points $$(a,0,0), \quad (0,b,0), \quad (0,0,c).$$ Using these three intercept points in the equation of a plane, we obtain the intercept form: $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$ This form is valid only when none of the intercepts is zero.

Plane Passing Through the Intersection of Two Given Planes

Let two intersecting planes be $$\pi_1: A_1x + B_1y + C_1z + D_1 = 0$$ and $$\pi_2: A_2x + B_2y + C_2z + D_2 = 0.$$ Then every plane passing through their line of intersection is given by the linear combination $$(A_1x + B_1y + C_1z + D_1) + \lambda(A_2x + B_2y + C_2z + D_2) = 0$$ where $\lambda$ is a scalar parameter.

This works because every point on the common line satisfies both plane equations, so it also satisfies any linear combination of them.

An extra condition, such as passing through a particular point or being perpendicular to another plane, is then used to determine $\lambda$.

Example 1: Normal Form from Normal Vector and Distance

Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $3\hat{i} + 5\hat{j} - 6\hat{k}$.

Solution: Step 1: The given normal vector is $$\vec{n} = 3\hat{i} + 5\hat{j} - 6\hat{k}.$$ Its magnitude is $$|\vec{n}| = \sqrt{3^2 + 5^2 + (-6)^2} = \sqrt{9+25+36} = \sqrt{70}.$$

Step 2: The corresponding unit normal vector is $$\hat{n} = \frac{3\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{70}}.$$

Step 3: Since the perpendicular distance from the origin is 7, use the normal form $$\vec{r} \cdot \hat{n} = d.$$ Thus, $$\vec{r} \cdot \left(\frac{3}{\sqrt{70}}\hat{i} + \frac{5}{\sqrt{70}}\hat{j} - \frac{6}{\sqrt{70}}\hat{k}\right) = 7.$$

Answer: $\vec{r} \cdot \left( \frac{3}{\sqrt{70}}\hat{i} + \frac{5}{\sqrt{70}}\hat{j} - \frac{6}{\sqrt{70}}\hat{k} \right) = 7$

Example 2: Cartesian to Normal Form

Find the direction cosines of the unit normal to the plane $2x + 3y - z = 5$ and find its distance from the origin.

Solution: Step 1: Compare the plane with the standard form $Ax+By+Cz+D=0$. Here, the normal vector is $$\vec{n} = 2\hat{i} + 3\hat{j} - \hat{k}.$$

Step 2: Compute the magnitude of the normal vector: $$|\vec{n}| = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4+9+1} = \sqrt{14}.$$

Step 3: The unit normal vector is $$\hat{n} = \frac{1}{\sqrt{14}}(2\hat{i} + 3\hat{j} - \hat{k}).$$ So the direction cosines are $$l = \frac{2}{\sqrt{14}}, \quad m = \frac{3}{\sqrt{14}}, \quad n = -\frac{1}{\sqrt{14}}.$$

Step 4: Divide the plane equation by $\sqrt{14}$: $$\frac{2}{\sqrt{14}}x + \frac{3}{\sqrt{14}}y - \frac{1}{\sqrt{14}}z = \frac{5}{\sqrt{14}}.$$ This is the normal form, so the distance from the origin is $$d = \frac{5}{\sqrt{14}}.$$

Answer: Direction cosines: $\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, -\frac{1}{\sqrt{14}}$. Distance: $\frac{5}{\sqrt{14}}$.

Example 3: Plane Passing Through a Point

Find the Cartesian equation of the plane passing through the point $(1, 4, 6)$ and whose normal vector has direction ratios $2, -1, 3$.

Solution: Step 1: The point on the plane is $$(x_1,y_1,z_1) = (1,4,6),$$ and the normal vector has direction ratios $$(A,B,C) = (2,-1,3).$$

Step 2: Use the point-normal form of the plane: $$A(x-x_1) + B(y-y_1) + C(z-z_1) = 0.$$ Substituting the values, $$2(x-1) - (y-4) + 3(z-6) = 0.$$

Step 3: Expand and simplify: $$2x - 2 - y + 4 + 3z - 18 = 0$$ $$2x - y + 3z - 16 = 0.$$

Answer: $2x - y + 3z - 16 = 0$

Example 4: Plane Passing Through Three Points

Find the vector and Cartesian equations of the plane passing through the points $A(1, 1, -1), B(6, 4, -5),$ and $C(-4, -2, 3)$.

Solution: Step 1: Compute the vectors $$\vec{AB} = (6-1)\hat{i} + (4-1)\hat{j} + (-5+1)\hat{k} = 5\hat{i} + 3\hat{j} - 4\hat{k},$$ $$\vec{AC} = (-4-1)\hat{i} + (-2-1)\hat{j} + (3+1)\hat{k} = -5\hat{i} - 3\hat{j} + 4\hat{k}.$$

Step 2: Observe that $$\vec{AC} = -\vec{AB}.$$ Hence the three points are collinear.

Step 3: A unique plane through three points exists only when the three points are non-collinear. Since the three given points are collinear, infinitely many planes pass through the common line containing them.

Answer: No unique plane exists because the given three points are collinear.

Example 5: Finding Intercepts

Find the intercepts cut off by the plane $2x + y - z = 5$.

Solution: Step 1: To find the x-intercept, set $y=0$ and $z=0$: $$2x = 5 \implies x = \frac{5}{2}.$$ So the x-intercept is $\frac{5}{2}$.

Step 2: To find the y-intercept, set $x=0$ and $z=0$: $$y = 5.$$ So the y-intercept is $5$.

Step 3: To find the z-intercept, set $x=0$ and $y=0$: $$-z = 5 \implies z = -5.$$ So the z-intercept is $-5$.

Step 4: In intercept form, the plane can be written as $$\frac{x}{5/2} + \frac{y}{5} + \frac{z}{-5} = 1.$$

Answer: $5/2, 5, -5$

Example 6: Plane through the Intersection of Two Planes

Find the equation of the plane passing through the line of intersection of the planes $x + y + z = 1$ and $2x + 3y + 4z = 5$ which passes through the point $(1, 1, 1)$.

Solution: Step 1: Write the two planes in standard form: $$\pi_1: x+y+z-1=0,$$ $$\pi_2: 2x+3y+4z-5=0.$$

Step 2: The general plane through their line of intersection is $$(x+y+z-1) + \lambda(2x+3y+4z-5) = 0.$$

Step 3: Since the required plane passes through $(1,1,1)$, substitute these coordinates: $$(1+1+1-1) + \lambda(2+3+4-5) = 0$$ $$2 + 4\lambda = 0$$ $$\lambda = -\frac{1}{2}.$$

Step 4: Substitute this value into the family: $$(x+y+z-1) - \frac{1}{2}(2x+3y+4z-5) = 0.$$ Multiply by 2: $$2x+2y+2z-2 - 2x - 3y - 4z + 5 = 0$$ $$-y - 2z + 3 = 0.$$ So, $$y + 2z = 3.$$

Answer: $y + 2z = 3$

Example 7: Plane Parallel to Another Plane

Find the equation of the plane passing through the point $(1, 4, -2)$ and parallel to the plane $2x - y + 3z = 7$.

Solution: Step 1: Parallel planes have the same normal vector. Therefore the required plane must be of the form $$2x - y + 3z = d.$$

Step 2: Since the plane passes through $(1,4,-2)$, substitute these values: $$2(1) - 4 + 3(-2) = d$$ $$2 - 4 - 6 = d$$ $$d = -8.$$

Step 3: Therefore the equation is $$2x - y + 3z = -8$$ or equivalently $$2x - y + 3z + 8 = 0.$$

Answer: $2x - y + 3z + 8 = 0$

Example 8: Identifying Coordinate Planes

Find the equation of the yz-plane.

Solution: Step 1: Any point lying on the yz-plane has x-coordinate equal to 0.

Step 2: Therefore the equation of the yz-plane is simply $$x = 0.$$ This plane contains the y-axis and z-axis and is perpendicular to the x-axis.

Answer: $x = 0$

Example 9: Normal Vector from Plane Equation

Find the unit normal vector to the plane $\vec{r} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) + 5 = 0$.

Solution: Step 1: Rewrite the plane equation as $$\vec{r} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = -5.$$ A normal vector is therefore $$\vec{N} = 2\hat{i} + \hat{j} - 2\hat{k}.$$

Step 2: Its magnitude is $$|\vec{N}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4+1+4} = 3.$$

Step 3: Hence a unit normal vector is $$\hat{n} = \frac{1}{3}(2\hat{i} + \hat{j} - 2\hat{k}).$$ Since a plane has two opposite normals, another unit normal is $$-\frac{1}{3}(2\hat{i} + \hat{j} - 2\hat{k}).$$

Answer: $\pm \frac{1}{3}(2\hat{i} + \hat{j} - 2\hat{k})$

Example 10: Plane Passing Through Intersection and Perpendicular to a Plane

Find the equation of the plane passing through the line of intersection of the planes $\vec{r}\cdot(\hat{i} + \hat{j} + \hat{k}) = 1$ and $\vec{r}\cdot(2\hat{i} + 3\hat{j} - \hat{k}) + 4 = 0$ and perpendicular to the plane $\vec{r}\cdot(\hat{i} - \hat{j} + \hat{k}) = 0$.

Solution: Step 1: Write the three planes in Cartesian form: $$\pi_1: x+y+z-1=0,$$ $$\pi_2: 2x+3y-z+4=0,$$ $$\pi_3: x-y+z=0.$$

Step 2: The family of planes through the intersection of $\pi_1$ and $\pi_2$ is $$(x+y+z-1) + \lambda(2x+3y-z+4) = 0.$$ Expanding, $$(1+2\lambda)x + (1+3\lambda)y + (1-\lambda)z + (-1+4\lambda)=0.$$ Its normal vector is $$\vec{N} = (1+2\lambda)\hat{i} + (1+3\lambda)\hat{j} + (1-\lambda)\hat{k}.$$

Step 3: The normal vector of $\pi_3$ is $$\vec{n}_3 = \hat{i} - \hat{j} + \hat{k}.$$ For the required plane to be perpendicular to $\pi_3$, their normal vectors must be perpendicular: $$\vec{N} \cdot \vec{n}_3 = 0.$$ So, $$(1+2\lambda)(1) + (1+3\lambda)(-1) + (1-\lambda)(1) = 0$$ $$1+2\lambda -1 -3\lambda +1 -\lambda = 0$$ $$1 - 2\lambda = 0$$ $$\lambda = \frac{1}{2}.$$

Step 4: Substitute this into the family equation: $$(x+y+z-1) + \frac{1}{2}(2x+3y-z+4)=0.$$ Multiply by 2: $$2x+2y+2z-2 + 2x+3y-z+4 = 0$$ $$4x + 5y + z + 2 = 0.$$

Answer: $4x + 5y + z + 2 = 0$