Three Dimensional Geometry: Planes

Equation of a Plane in Space (Vector Form - Normal Form, Point-Normal Form)

A plane is a fundamental geometric object in three-dimensional space. It can be thought of as a flat, two-dimensional surface that extends infinitely in all directions. To describe a plane algebraically using vectors, we need information that uniquely determines its position and orientation in space. Two common ways to define a plane lead to its vector equation:

• Knowing its distance from the origin and the direction perpendicular to it.
• Knowing a point on the plane and the direction perpendicular to the plane.

1. Vector Equation of a Plane in Normal Form

This form is based on the perpendicular distance of the plane from the origin and the unit vector pointing in the direction of this perpendicular.

Consider a plane whose shortest distance from the origin O is $p$, where $p$ is a non-negative scalar ($p \ge 0$). Let N be the foot of the perpendicular drawn from the origin O to the plane. The length of this perpendicular segment ON is equal to $p$, so $|\vec{ON}| = p$.

Let $\hat{n}$ be a unit vector in the direction of $\vec{ON}$. Since $\vec{ON}$ points from the origin towards the plane along the perpendicular, $\hat{n}$ is a unit vector normal (perpendicular) to the plane. Thus, $\vec{ON} = p\hat{n}$.

Let P be any arbitrary point on the plane. Let its position vector with respect to the origin O be $\vec{r} = \vec{OP}$.

Consider the vector $\vec{NP}$. Since both N (the foot of the perpendicular from O to the plane) and P (any point on the plane) lie on the plane, the vector $\vec{NP}$ lies entirely within the plane.

The vector $\vec{ON}$ (or the unit vector $\hat{n}$) is perpendicular to the plane. By definition of a normal vector, it is perpendicular to every vector that lies within the plane.

Therefore, the vector $\vec{NP}$ must be perpendicular to the vector $\vec{ON}$ (or to $\hat{n}$).

According to the property of the dot product, if two vectors are perpendicular, their dot product is zero:

Now, express the vector $\vec{NP}$ in terms of the position vectors of P and N:

Substitute this expression into the dot product equation:

Use the distributive property of the dot product:

Using the property $(k\vec{u}) \cdot \vec{v} = k(\vec{u} \cdot \vec{v})$:

Since $\hat{n}$ is a unit vector, the dot product of $\hat{n}$ with itself is equal to the square of its magnitude, which is 1:

Substitute this back into the equation:

This is the vector equation of a plane in normal form:

$$ \mathbf{\vec{r} \cdot \hat{n} = p} $$

Here:

• $\vec{r}$ is the position vector of any arbitrary point on the plane.
• $\hat{n}$ is a unit vector normal to the plane.
• $p$ is the perpendicular distance of the plane from the origin ($p \ge 0$).

2. Vector Equation of a Plane Perpendicular to a Given Vector and Passing Through a Given Point (Point-Normal Form)

This form defines a plane by specifying a point that lies on the plane and a vector that is perpendicular to the plane.

Suppose we want to find the equation of a plane that passes through a given point A and is perpendicular to a given non-zero vector $\vec{N}$.

Let O be the origin. Let the position vector of the given point A be $\vec{a} = \vec{OA}$.

Let $\vec{N}$ be the given vector normal (perpendicular) to the plane. $\vec{N}$ is the normal vector.

Let P be any arbitrary point on the plane. Let its position vector relative to the origin O be $\vec{r} = \vec{OP}$.

Consider the vector $\vec{AP}$. Since both A (a point on the plane) and P (any point on the plane) lie on the plane, the vector $\vec{AP}$ lies entirely within the plane.

The normal vector $\vec{N}$ is perpendicular to the plane. By definition, it is perpendicular to every vector that lies in the plane.

Therefore, the vector $\vec{AP}$ must be perpendicular to the normal vector $\vec{N}$.

This condition of perpendicularity implies their dot product is zero:

Now, express the vector $\vec{AP}$ in terms of the position vectors of P and A:

Substitute this expression for $\vec{AP}$ into the dot product equation:

$$ \mathbf{(\vec{r} - \vec{a}) \cdot \vec{N} = 0} $$

This is the vector equation of a plane in point-normal form.

This equation can be expanded using the distributive property of the dot product:

Rearranging the terms, we get an alternative form:

Since $\vec{a}$ and $\vec{N}$ are known, the dot product $\vec{a} \cdot \vec{N}$ is a fixed scalar constant. Let $d = \vec{a} \cdot \vec{N}$. Then the equation becomes:

$$ \mathbf{\vec{r} \cdot \vec{N} = d} $$

This form also represents a plane perpendicular to the vector $\vec{N}$. However, the scalar $d$ here is not necessarily the perpendicular distance from the origin. The perpendicular distance from the origin is $p = \frac{|d|}{|\vec{N}|} = \frac{|\vec{a} \cdot \vec{N}|}{|\vec{N}|} = |\vec{a} \cdot \hat{N}|$, which is the magnitude of the scalar projection of $\vec{a}$ onto the direction of $\vec{N}$.

The normal form $\vec{r} \cdot \hat{n} = p$ is a special case of $\vec{r} \cdot \vec{N} = d$ where $\vec{N} = \hat{n}$ (a unit vector) and $d=p$ (the perpendicular distance). In general form $\vec{r} \cdot \vec{N} = d$, $\vec{N}$ is any normal vector and $d$ is related to the distance from origin and the magnitude of $\vec{N}$.

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Equation of a Plane in Space (Cartesian Form - General Form, Normal Form, Intercept Form)

Just as straight lines can be represented in Cartesian coordinates, so too can planes in three-dimensional space. The vector equations of a plane can be converted into algebraic equations involving the $(x, y, z)$ coordinates of any point lying on the plane. There are several common forms of the Cartesian equation of a plane, each derived from a particular way of defining the plane.

1. General Form of the Equation of a Plane

The general form of the Cartesian equation of a plane is a linear equation in variables $x, y, z$. This form can be derived from the point-normal vector form of the plane's equation, $(\vec{r} - \vec{a}) \cdot \vec{N} = 0$.

Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ be the position vector of any arbitrary point P$(x, y, z)$ on the plane.

Let $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ be the position vector of a known point A$(x_1, y_1, z_1)$ that lies on the plane.

Let $\vec{N} = A\hat{i} + B\hat{j} + C\hat{k}$ be a non-zero vector normal (perpendicular) to the plane. The components $A, B, C$ are the direction ratios of the normal vector.

The vector $\vec{r} - \vec{a}$ is the vector $\vec{AP} = (x - x_1)\hat{i} + (y - y_1)\hat{j} + (z - z_1)\hat{k}$, which lies in the plane.

The point-normal equation is $(\vec{r} - \vec{a}) \cdot \vec{N} = 0$. Substitute the component forms:

Using the component formula for the dot product:

Expand this equation:

Rearrange the terms:

Let $D = -Ax_1 - By_1 - Cz_1$. Since $A, B, C, x_1, y_1, z_1$ are constants, $D$ is also a constant. The equation simplifies to:

$$ \mathbf{Ax + By + Cz + D = 0} $$

This is the general linear equation of a plane in Cartesian coordinates. Any equation of the first degree in $x, y, z$ always represents a plane.

In the general form $Ax + By + Cz + D = 0$:

• The coefficients $A, B, C$ are the direction ratios of a vector normal to the plane.
• The constant $D$ is related to the distance of the plane from the origin and the specific normal vector chosen.

2. Cartesian Equation in Normal Form

This form is derived from the vector equation $\vec{r} \cdot \hat{n} = p$, where $\hat{n}$ is a unit normal vector and $p$ is the perpendicular distance from the origin.

Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ be the position vector of any point P$(x, y, z)$ on the plane.

Let $\hat{n} = l\hat{i} + m\hat{j} + n\hat{k}$ be the unit vector normal to the plane from the origin. Here, $l, m, n$ are the direction cosines of the normal vector, and they satisfy the condition $l^2 + m^2 + n^2 = 1$.

Substitute these component forms into the vector equation $\vec{r} \cdot \hat{n} = p$:

Using the component formula for the dot product:

This is the Cartesian equation of a plane in normal form:

$$ \mathbf{lx + my + nz = p} $$

Here:

• $l, m, n$ are the direction cosines of the normal to the plane.
• $p$ is the perpendicular distance of the plane from the origin ($p \ge 0$).

Converting General Form to Normal Form

Given the general equation of a plane $Ax + By + Cz + D = 0$. We want to convert it to the normal form $lx + my + nz = p$.

We know that $A, B, C$ are direction ratios of the normal vector $\vec{N} = A\hat{i} + B\hat{j} + C\hat{k}$. The direction cosines of the normal are $l = \pm \frac{A}{\sqrt{A^2+B^2+C^2}}$, $m = \pm \frac{B}{\sqrt{A^2+B^2+C^2}}$, $n = \pm \frac{C}{\sqrt{A^2+B^2+C^2}}$.

Rewrite the general equation by moving the constant term to the right side: $Ax + By + Cz = -D$.

To make the coefficients the direction cosines, we divide the entire equation by the magnitude of the normal vector, $|\vec{N}| = \sqrt{A^2+B^2+C^2}$. However, we must be careful to ensure the constant term on the right side (which will be $p$) is non-negative.

Divide the equation $Ax + By + Cz = -D$ by $\pm \sqrt{A^2+B^2+C^2}$:

The coefficients are now the direction cosines $l, m, n$. The right side is the perpendicular distance $p$. We choose the sign of $\sqrt{A^2+B^2+C^2}$ such that the right side is non-negative.

• If $-D$ is positive (i.e., $D$ is negative), we use the positive sign: $+\sqrt{A^2+B^2+C^2}$.
• If $-D$ is negative (i.e., $D$ is positive), we use the negative sign: $-\sqrt{A^2+B^2+C^2}$.
• If $D=0$, the plane passes through the origin, $p=0$. Either sign can be used for the denominator; the DCs $(l, m, n)$ and $(-l, -m, -n)$ simply represent the two opposite directions of the normal.

Thus, the direction cosines of the normal are $l = \frac{A}{\pm \sqrt{A^2+B^2+C^2}}$, $m = \frac{B}{\pm \sqrt{A^2+B^2+C^2}}$, $n = \frac{C}{\pm \sqrt{A^2+B^2+C^2}}$, and the perpendicular distance from the origin is $p = \frac{|-D|}{\sqrt{A^2+B^2+C^2}} = \frac{|D|}{\sqrt{A^2+B^2+C^2}}$. The sign in the denominator is chosen to make $p \ge 0$. Specifically, the denominator should have the opposite sign to $D$ (or be positive if $D=0$).

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3. Intercept Form of the Equation of a Plane

If a plane intersects the coordinate axes at distinct points (other than the origin), its equation can be expressed in terms of these intercepts. Let the plane make non-zero intercepts $a, b, c$ on the X, Y, and Z axes respectively.

• The intercept on the X-axis is $a$. This means the plane passes through the point $(a, 0, 0)$.
• The intercept on the Y-axis is $b$. This means the plane passes through the point $(0, b, 0)$.
• The intercept on the Z-axis is $c$. This means the plane passes through the point $(0, 0, c)$.

Assuming the plane does not pass through the origin and is not parallel to any axis, its equation can be written in the intercept form:

$$ \mathbf{\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1} $$

This form is useful when the intercepts are known or need to be found. Note that this form is valid only if $a, b, c$ are non-zero. If any intercept is zero (plane passes through the origin) or infinite (plane is parallel to an axis), this form cannot be used.

Converting General Form to Intercept Form

Given the general equation of a plane $Ax + By + Cz + D = 0$. Assume that $A, B, C,$ and $D$ are all non-zero (this ensures the plane doesn't pass through the origin and isn't parallel to any axis).

Move the constant term $D$ to the right side of the equation:

To make the right side equal to 1 (like in the intercept form), divide the entire equation by $-D$:

Now, rewrite each term on the left side to match the structure $\frac{x}{\text{intercept}}$:

Comparing this with the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, we can identify the intercepts:

• Intercept on X-axis, $a = -D/A$.
• Intercept on Y-axis, $b = -D/B$.
• Intercept on Z-axis, $c = -D/C$.

This conversion is only possible if $A, B, C, D$ are all non-zero. If $D=0$, the plane passes through the origin and doesn't have defined non-zero intercepts. If $A=0$, the plane is parallel to the X-axis and doesn't intercept it at a finite non-zero value, etc.

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Equation of a Plane Passing Through Three Non-Collinear Points

A plane in three-dimensional space is uniquely determined if we are given three points that do not lie on the same straight line (i.e., are non-collinear). We can derive the equation of such a plane using the concept of coplanarity of vectors.

Let the three given non-collinear points be A, B, and C. Let their position vectors with respect to a fixed origin O be $\vec{a}$, $\vec{b}$, and $\vec{c}$ respectively.

Let P be any arbitrary point on the plane containing the points A, B, and C. Let the position vector of P with respect to the origin O be $\vec{r}$.

Vector Equation

Consider the three vectors formed by taking A as the initial point and B, C, and P as the terminal points:

• Vector from A to B: $\vec{AB} = \vec{OB} - \vec{OA} = \vec{b} - \vec{a}$
• Vector from A to C: $\vec{AC} = \vec{OC} - \vec{OA} = \vec{c} - \vec{a}$
• Vector from A to P: $\vec{AP} = \vec{OP} - \vec{OA} = \vec{r} - \vec{a}$

Since the points A, B, C, and P all lie on the same plane, the vectors $\vec{AP}$, $\vec{AB}$, and $\vec{AC}$, which originate from a common point A and lie within the plane, must be coplanar.

The condition for three vectors to be coplanar is that their scalar triple product is zero.

Substituting the expressions for these vectors in terms of position vectors:

$$ \mathbf{(\vec{r} - \vec{a}) \cdot ((\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})) = 0} $$

This is the vector equation of the plane passing through three non-collinear points A, B, and C with position vectors $\vec{a}, \vec{b}, \vec{c}$. It states that the vector from A to any point P on the plane is coplanar with the vectors $\vec{AB}$ and $\vec{AC}$.

Cartesian Equation

To find the Cartesian equation of the plane passing through three non-collinear points, we use the coordinates of the points. Let the coordinates of the points be:

• A: $(x_1, y_1, z_1)$
• B: $(x_2, y_2, z_2)$
• C: $(x_3, y_3, z_3)$
• P: $(x, y, z)$ (any point on the plane)

The component forms of the vectors $\vec{AP}, \vec{AB}, \vec{AC}$ are:

• $\vec{AP} = (x-x_1)\hat{i} + (y-y_1)\hat{j} + (z-z_1)\hat{k}$
• $\vec{AB} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$
• $\vec{AC} = (x_3-x_1)\hat{i} + (y_3-y_1)\hat{j} + (z_3-z_1)\hat{k}$

The condition that these three vectors are coplanar is that their scalar triple product is zero. Using the determinant formula for the scalar triple product in terms of components:

Substitute the components into the determinant:

$$ \mathbf{\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0} $$

This is the Cartesian equation of the plane passing through three non-collinear points $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$. Expanding this $3 \times 3$ determinant will result in a linear equation in $x, y, z$, which is the general form of the plane equation $Ax + By + Cz + D = 0$. The coefficients $A, B, C$ are the components of the vector $(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})$, which is a normal vector to the plane.

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Equation of a Plane Passing Through the Intersection of Two Planes

When two distinct, non-parallel planes intersect in three-dimensional space, their intersection forms a straight line. It is often necessary to find the equation of a third plane that contains this line of intersection and satisfies some additional condition (like passing through a specific point or being perpendicular to another plane).

Concept of the Family of Planes

Let the equations of the two intersecting planes be given in general Cartesian form:

Any point $(x, y, z)$ that lies on the line of intersection of $P_1$ and $P_2$ must satisfy both equations simultaneously, i.e., $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x + B_2y + C_2z + D_2 = 0$.

Consider the equation formed by taking a linear combination of the two plane equations:

where $\lambda$ is a scalar parameter.

If a point $(x, y, z)$ lies on the line of intersection, then $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x + B_2y + C_2z + D_2 = 0$. Substituting these values into the linear combination equation, we get $0 + \lambda(0) = 0$, which is true for any value of $\lambda$. This means that any point on the line of intersection satisfies the combined equation for any value of $\lambda$.

Furthermore, the combined equation is linear in $x, y, z$: $(A_1 + \lambda A_2)x + (B_1 + \lambda B_2)y + (C_1 + \lambda C_2)z + (D_1 + \lambda D_2) = 0$. This is the general form of a plane equation (provided $A_1+\lambda A_2$, $B_1+\lambda B_2$, $C_1+\lambda C_2$ are not all zero for a given $\lambda$).

Thus, the equation $(A_1x + B_1y + C_1z + D_1) + \lambda (A_2x + B_2y + C_2z + D_2) = 0$ represents a family of planes that all pass through the line of intersection of $P_1$ and $P_2$. Different values of the parameter $\lambda$ correspond to different planes in this family.

Vector Form

If the equations of the two intersecting planes are given in vector form as $\vec{r} \cdot \vec{n_1} = d_1$ and $\vec{r} \cdot \vec{n_2} = d_2$, the equation of any plane passing through their line of intersection can be written by taking a linear combination of these equations:

Rearranging the terms using the linearity of the dot product:

This is the vector equation of a plane passing through the intersection of two planes $\vec{r} \cdot \vec{n_1} = d_1$ and $\vec{r} \cdot \vec{n_2} = d_2$, where $\lambda$ is a scalar parameter. The vector $\vec{n_1} + \lambda \vec{n_2}$ is a normal vector to the resulting plane.

Cartesian Form

If the equations of the two intersecting planes are given in Cartesian form as $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x + B_2y + C_2z + D_2 = 0$, the equation of any plane passing through their line of intersection is given by the linear combination:

$$ \mathbf{(A_1x + B_1y + C_1z + D_1) + \lambda (A_2x + B_2y + C_2z + D_2) = 0} $$

where $\lambda$ is a scalar parameter. This is the Cartesian form of the family of planes. To find the equation of a specific plane from this family, an additional condition is required. For example, if the plane is known to pass through a specific point $(x_0, y_0, z_0)$, substitute these coordinates into the equation, and solve for $\lambda$. Once $\lambda$ is found, substitute it back into the equation to get the equation of the particular plane.

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