Planes

Introduction

A plane is a surface that is completely flat. Mathematically, it is a surface such that if we take any two points on it, the line segment joining them also lies entirely on the surface.

We now seek to find an equation that describes a plane. That is, we aim to determine a condition that a point \((x, y, z)\) must satisfy to lie on a given plane. Alternatively, in vector form, if \(\mathbf{r}\) is the position vector of any point on the plane, we wish to derive the equation that \(\mathbf{r}\) must satisfy.

Equation of a Plane

There are several ways to express the equation of a plane. We begin by exploring a less common vector form.

Parametric Equation of a Plane

Consider a plane that passes through a fixed point \( A \) whose position vector is \( \mathbf{a} \). Suppose two non-collinear vectors \( \mathbf{b} \) and \( \mathbf{c} \) lie parallel to the plane. Since two non-collinear vectors define a plane, our plane must be parallel to the plane formed by \( \mathbf{b} \) and \( \mathbf{c} \). However, infinitely many such planes exist, all parallel to the plane of \( \mathbf{b} \) and \( \mathbf{c} \). To uniquely specify a single plane, we also require a fixed point \( A \) on it. This ensures a unique plane.

To determine the equation of this plane, let \( P \) be any point on the plane, with its position vector denoted by \( \mathbf{r} \). The displacement vector from \( A \) to \( P \) is

\[ \mathbf{AP} = \mathbf{OP} - \mathbf{OA} = \mathbf{r} - \mathbf{a}. \]

Since \( \mathbf{AP} \), \( \mathbf{b} \), and \( \mathbf{c} \) are coplanar, by the fundamental theorem of three-dimensional geometry, the vector \( \mathbf{r} - \mathbf{a} \) must be expressible as a linear combination of \( \mathbf{b} \) and \( \mathbf{c} \):

\[ \mathbf{r} - \mathbf{a} = \lambda \mathbf{b} + \mu \mathbf{c}, \quad \lambda, \mu \in \mathbb{R}. \]

Thus, the equation of the plane is given by

\[ \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c}, \quad \lambda, \mu \in \mathbb{R}. \]

By varying \( \lambda \) and \( \mu \), we obtain different points \( \mathbf{r} \) on the plane. These parameters provide a parametric representation of the plane, ensuring that each point on the plane corresponds to a unique pair of values \( (\lambda, \mu) \).

Two points and a vector parallel to the plane are known

If two points on the plane are known, along with a vector parallel to the plane, we can determine its equation as follows.

Let \( A \) and \( B \) be two points on the plane with position vectors \( \mathbf{a} \) and \( \mathbf{b} \), respectively. Suppose a vector \( \mathbf{c} \) is also given, which is parallel to the plane.

Now, let \( P \) be any point on the plane with position vector \( \mathbf{r} \). The displacement vectors are:

\[ \mathbf{AP} = \mathbf{r} - \mathbf{a}, \quad \mathbf{AB} = \mathbf{b} - \mathbf{a}. \]

Since \( \mathbf{AP} \), \( \mathbf{AB} \), and \( \mathbf{c} \) are coplanar, we apply the fundamental theorem of three-dimensional geometry, which states that a vector lying in the plane of two other vectors can be expressed as their linear combination. Thus,

\[ \mathbf{r} - \mathbf{a} = \lambda (\mathbf{b} - \mathbf{a}) + \mu \mathbf{c}, \quad \lambda, \mu \in \mathbb{R}. \]

Rearranging, we obtain the equation of the plane:

\[ \mathbf{r} = \mathbf{a} + \lambda (\mathbf{b} - \mathbf{a}) + \mu \mathbf{c}, \quad \lambda, \mu \in \mathbb{R}. \]

Here, \( \lambda \) and \( \mu \) are independent parameters, allowing us to generate all points \( \mathbf{r} \) on the plane. This provides a parametric representation of the plane when two specific points and a parallel vector are known.

Parametric forms of planes are not very commonly used, but whenever they appear, one should be able to interpret them correctly.

The equation

\[ \mathbf{r} = \mathbf{j} + \mathbf{k} + s (\mathbf{i} - \mathbf{j}) + t (2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) \]

represents a plane that is parallel to the vectors \( \mathbf{i} - \mathbf{j} \) and \( 2\mathbf{i} - \mathbf{j} + 3\mathbf{k} \) and passes through the point whose position vector is \( \mathbf{j} + \mathbf{k} \).

Here, \( s \) and \( t \) are independent parameters that generate all points on the plane.

Three points on the plane given

If we know the position vectors of three non-collinear points on a plane, then the plane is uniquely determined.

Let \( A \), \( B \), and \( C \) be three points on the plane with position vectors \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \), respectively. Let \( P \) be any point on the plane with position vector \( \mathbf{r} \).

Since the vectors \( \mathbf{PA} \), \( \mathbf{AB} \), and \( \mathbf{AC} \) lie in the plane, they are coplanar. The displacement vectors are:

\[ \mathbf{PA} = \mathbf{r} - \mathbf{a}, \quad \mathbf{AB} = \mathbf{b} - \mathbf{a}, \quad \mathbf{AC} = \mathbf{c} - \mathbf{a}. \]

By the fundamental theorem of three-dimensional geometry, since these vectors are coplanar, \( \mathbf{PA} \) must be a linear combination of \( \mathbf{AB} \) and \( \mathbf{AC} \), meaning there exist scalars \( \lambda, \mu \in \mathbb{R} \) such that:

\[ \mathbf{r} - \mathbf{a} = \lambda (\mathbf{b} - \mathbf{a}) + \mu (\mathbf{c} - \mathbf{a}). \]

Rearranging,

\[ \mathbf{r} = \mathbf{a} + \lambda (\mathbf{b} - \mathbf{a}) + \mu (\mathbf{c} - \mathbf{a}), \quad \lambda, \mu \in \mathbb{R}. \]

This is the parametric equation of the plane passing through the three given points. Different values of \( \lambda \) and \( \mu \) generate different points \( \mathbf{r} \) on the plane.

Scalar Triple Product Form of the Plane Equation

Given a point \( A \) with position vector \( \mathbf{a} \) on the plane and two non-collinear vectors \( \mathbf{b} \) and \( \mathbf{c} \) parallel to the plane, let \( \mathbf{r} \) be the position vector of an arbitrary point on the plane.

Since the vectors \( \mathbf{r} - \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \) are coplanar, instead of expressing \( \mathbf{r} - \mathbf{a} \) as a linear combination of \( \mathbf{b} \) and \( \mathbf{c} \), we use the fact that three vectors are coplanar if and only if their scalar triple product is zero.

\[ (\mathbf{r} - \mathbf{a}) \cdot (\mathbf{b} \times \mathbf{c}) = 0 \]

which in determinant form is written as

\[ [\mathbf{r} - \mathbf{a}, \mathbf{b}, \mathbf{c}] = 0. \]

Expanding using determinant properties,

\[ [\mathbf{r}, \mathbf{b}, \mathbf{c}] - [\mathbf{a}, \mathbf{b}, \mathbf{c}] = 0 \]

which simplifies to

\[ [\mathbf{r}, \mathbf{b}, \mathbf{c}] = [\mathbf{a}, \mathbf{b}, \mathbf{c}]. \]

This provides an alternative implicit form of the plane equation using the scalar triple product.

Normal Form of the Plane Equation

The normal line of a plane is a line that is perpendicular to every line lying on the plane. This line is said to be perpendicular to the plane itself. A vector parallel to this normal line is called a normal vector to the plane.

We can express the equation of a plane using its normal line or normal vector. The normal vector determines the orientation of the plane, but to uniquely determine the plane, we also require a specific point on it. Given a point on the plane and a normal vector, we can easily derive the equation of the plane.

Let \( \mathbf{n} \) be a normal vector to the plane, and let \( \mathbf{a} \) be the position vector of a known point \( A \) on the plane. Suppose \( P \) is any other general point on the plane, with position vector \( \mathbf{r} \).

Since \( \mathbf{n} \) is perpendicular to every vector lying on the plane, the displacement vector \( \mathbf{AP} \), given by

\[ \mathbf{AP} = \mathbf{r} - \mathbf{a}, \]

must also be perpendicular to \( \mathbf{n} \). Thus,

\[ \mathbf{AP} \cdot \mathbf{n} = 0. \]

This gives the equation of the plane in normal form:

\[ (\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0. \]

This is the vector equation of a plane using a normal vector.

For example:

Suppose a plane is passing through a point whose position vector is

\[ \mathbf{a} = \mathbf{i} - \mathbf{j} - 2\mathbf{k}, \]

and the plane has a normal vector

\[ \mathbf{n} = 2\mathbf{i} + \mathbf{j} + \mathbf{k}. \]

Then the equation of the plane is given by

\[ (\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0 \implies \mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}. \]

Substituting the given vectors,

\[ \mathbf{r} \cdot (2\mathbf{i} + \mathbf{j} + \mathbf{k}) = (\mathbf{i} - \mathbf{j} - 2\mathbf{k}) \cdot (2\mathbf{i} + \mathbf{j} + \mathbf{k}) \implies \mathbf{r} \cdot (2\mathbf{i} + \mathbf{j} + \mathbf{k}) = 2 - 1 - 2 = -1. \]

Thus,

\[ \mathbf{r} \cdot (2\mathbf{i} + \mathbf{j} + \mathbf{k}) = -1. \]

Now, if \( \mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \), that is, if the terminal point is \( (x, y, z) \), then

\[ (x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) \cdot (2\mathbf{i} + \mathbf{j} + \mathbf{k}) = -1 \implies 2x + y + z = -1. \]

This is the Cartesian form of the equation of the plane. We will study this form in detail soon.

General Vector Form of Plane

Now, observe that the equation

\[ (\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0 \]

can always be rewritten as

\[ \mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}. \]

The quantity \( \mathbf{a} \cdot \mathbf{n} \) is a scalar constant that depends only on the given point \( \mathbf{a} \) on the plane and the normal vector \( \mathbf{n} \). Let this constant be denoted by \( d \). Then the equation becomes

\[ \mathbf{r} \cdot \mathbf{n} = d. \]

Thus, any plane whose equation is written using a normal vector reduces to the general form

\[ \mathbf{r} \cdot \mathbf{n} = d, \]

where \( \mathbf{n} \) is a fixed non-zero vector (the normal vector), and \( d \in \mathbb{R} \) is a constant. This compact expression captures all planes in space through a linear condition on the dot product of the position vector \( \mathbf{r} \) with the normal vector.

Example

Write the equation of a plane passing through a point \( A \) with position vector \( \mathbf{a} \), and parallel to two non-collinear vectors \( \mathbf{b} \) and \( \mathbf{c} \).

Since the plane is parallel to both \( \mathbf{b} \) and \( \mathbf{c} \), any normal vector \( \mathbf{n} \) to the plane must be perpendicular to both. A natural choice is

\[ \mathbf{n} = \mathbf{b} \times \mathbf{c}, \]

since the cross product \( \mathbf{b} \times \mathbf{c} \) is orthogonal to both \( \mathbf{b} \) and \( \mathbf{c} \), and hence perpendicular to the plane they span.

The equation of the plane is then

\[ (\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0 \implies (\mathbf{r} - \mathbf{a}) \cdot (\mathbf{b} \times \mathbf{c}) = 0 \implies \mathbf{r} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}). \]

Thus, the required equation is

\[ \mathbf{r} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}), \]

which is the standard scalar triple product form of the plane equation.

Equation of a Plane Passing Through Three Non-Collinear Points

Let three non-collinear points \( A, B, C \) lie on the plane, with respective position vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \). When we refer to points \( A, B, C \), we mean the points whose position vectors are \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) respectively.

To write the equation of the plane, we require a point on the plane (which we already have as \( \mathbf{a} \)) and a normal vector. To find the normal vector, observe that the vectors

\[ \mathbf{b} - \mathbf{a} \quad \text{and} \quad \mathbf{c} - \mathbf{a} \]

lie on the plane and are non-collinear since \( A, B, C \) are non-collinear. Hence, the normal vector can be taken as

\[ \mathbf{n} = (\mathbf{b} - \mathbf{a}) \times (\mathbf{c} - \mathbf{a}). \]

Thus, the equation of the plane passing through \( \mathbf{a} \) is

\[ (\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0 \implies (\mathbf{r} - \mathbf{a}) \cdot \left[(\mathbf{b} - \mathbf{a}) \times (\mathbf{c} - \mathbf{a})\right] = 0. \]

Expanding the cross product,

\[ (\mathbf{b} - \mathbf{a}) \times (\mathbf{c} - \mathbf{a}) = \mathbf{b} \times \mathbf{c} - \mathbf{b} \times \mathbf{a} - \mathbf{a} \times \mathbf{c} + \mathbf{a} \times \mathbf{a}. \]

Since \( \mathbf{a} \times \mathbf{a} = \mathbf{0} \), this reduces to

\[ \mathbf{b} \times \mathbf{c} + \mathbf{a} \times \mathbf{b} + \mathbf{c} \times \mathbf{a}. \]

So the equation becomes

\[ (\mathbf{r} - \mathbf{a}) \cdot \left(\mathbf{b} \times \mathbf{c} + \mathbf{a} \times \mathbf{b} + \mathbf{c} \times \mathbf{a}\right) = 0. \]

Expanding the dot product:

\[ \mathbf{r} \cdot (\mathbf{b} \times \mathbf{c}) + \mathbf{r} \cdot (\mathbf{a} \times \mathbf{b}) + \mathbf{r} \cdot (\mathbf{c} \times \mathbf{a}) - \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) - \mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) - \mathbf{a} \cdot (\mathbf{c} \times \mathbf{a}) = 0. \]

Now recall that \( \mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 0 \) and \( \mathbf{a} \cdot (\mathbf{c} \times \mathbf{a}) = 0 \), since a vector is always perpendicular to its own cross product with any other vector. Hence, the equation simplifies to

\[ \mathbf{r} \cdot (\mathbf{b} \times \mathbf{c}) + \mathbf{r} \cdot (\mathbf{a} \times \mathbf{b}) + \mathbf{r} \cdot (\mathbf{c} \times \mathbf{a}) = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}). \]

Using scalar triple product notation:

\[ [\mathbf{r} \ \mathbf{b} \ \mathbf{c}] + [\mathbf{r} \ \mathbf{a} \ \mathbf{b}] + [\mathbf{r} \ \mathbf{c} \ \mathbf{a}] = [\mathbf{a} \ \mathbf{b} \ \mathbf{c}]. \]

This is the equation of the plane passing through three non-collinear points \( A, B, C \) written in terms of scalar triple products.

Cartesian Form of the Equation of a Plane

As discussed earlier, to write the equation of a plane, we require two pieces of data:

– a point on the plane, and
– a vector normal to the plane.

Let a line normal to the plane have direction ratios \( (a, b, c) \), and let it pass through a point \( A(x_1, y_1, z_1) \) on the plane. Let \( P(x, y, z) \) be any other point on the plane. Then the direction ratios of the vector \( \vec{AP} \) are

\[ (x - x_1,\ y - y_1,\ z - z_1). \]

Since vector \( \vec{AP} \) lies on the plane and the normal vector is perpendicular to every vector on the plane, it follows that

\[ a(x - x_1) + b(y - y_1) + c(z - z_1) = 0. \]

This is the Cartesian equation of the plane. It is a first-degree equation in \( x, y, z \), and all such equations represent planes in three-dimensional space.

Example:

Suppose a plane has a normal vector with direction ratios \( (1, -1, 2) \), and passes through the point \( (0, 1, -2) \). Then the equation of the plane is:

\[ 1(x - 0) + (-1)(y - 1) + 2(z + 2) = 0 \implies x - y + 1 + 2z + 4 = 0 \implies x - y + 2z + 5 = 0. \]

Thus, the required Cartesian equation of the plane is

\[ \boxed{x - y + 2z + 5 = 0}. \]

General Cartesian Form

If we expand the general equation of a plane written using a point and a normal vector:

\[ a(x - x_1) + b(y - y_1) + c(z - z_1) = 0, \]

we obtain:

\[ ax + by + cz - (ax_1 + by_1 + cz_1) = 0. \]

The last term is constant, so denote it by \( d = - (ax_1 + by_1 + cz_1) \). Hence, the equation becomes

\[ ax + by + cz + d = 0. \]

This shows that any plane can be represented by an equation of the form

\[ ax + by + cz + d = 0. \]

But is the converse also true? That is, does any equation of the form \( ax + by + cz + d = 0 \), where \( a, b, c \) are not all zero, always represent a plane?
Yes, it does. Let us prove this.

Assume that

\[ ax + by + cz + d = 0 \]

defines some surface in \( \mathbb{R}^3 \). Let \( P_1 = (x_1, y_1, z_1) \) and \( P_2 = (x_2, y_2, z_2) \) be any two points satisfying this equation. Then:

\[ a x_1 + b y_1 + c z_1 + d = 0, \quad a x_2 + b y_2 + c z_2 + d = 0. \]

Subtracting the two equations,

\[ a(x_1 - x_2) + b(y_1 - y_2) + c(z_1 - z_2) = 0. \]

Now, let

\[ \vec{P_1P_2} = (x_2 - x_1)\mathbf{i} + (y_2 - y_1)\mathbf{j} + (z_2 - z_1)\mathbf{k}, \]

and let

\[ \vec{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}. \]

The above scalar equation

\[ a(x_1 - x_2) + b(y_1 - y_2) + c(z_1 - z_2) = 0 \]

is precisely

\[ \vec{P_1P_2} \cdot \vec{n} = 0, \]

i.e., the vector joining any two points on the surface is orthogonal to the fixed vector \( \vec{n} \). Hence, \( \vec{n} \) is perpendicular to every line segment joining two points on the surface. This implies that the entire surface lies in a plane orthogonal to \( \vec{n} \).

Thus,

\[ ax + by + cz + d = 0 \]

always represents a plane (provided \( a, b, c \) are not all zero).

Interpreting \( ax + by + cz + d = 0 \):

One should be able to interpret the equation

\[ ax + by + cz + d = 0 \]

geometrically. This is the Cartesian form of the equation of a plane. It represents a plane in three-dimensional space, provided that \( a, b, c \) are not all zero.

The key point is that the vector

\[ \vec{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \]

is normal (perpendicular) to the plane. That is, the plane has normal direction ratios \( (a, b, c) \), or equivalently, it has a normal vector \( \vec{n} \) pointing in the direction \( a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \).

So whenever one sees an equation of the form

\[ ax + by + cz + d = 0, \]

they should immediately interpret it as describing a plane whose normal vector is

\[ \vec{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}. \]

This vector determines the orientation of the plane.

Other Important Interpretations of the Plane Equation \( ax + by + cz + d = 0 \)

1. If \( d = 0 \):

   The equation becomes

   \[ ax + by + cz = 0, \]

   which clearly passes through the origin \( (0, 0, 0) \). So whenever \( d = 0 \), the plane passes through the origin.

2. If \( c = 0 \):

   The equation becomes

   \[ ax + by + d = 0. \]

   There is no \( z \)-term, which means \( z \) can take any real value independently. This looks like a straight line equation in the \( xy \)-plane, but in three dimensions, it describes a plane parallel to the \( z \)-axis.

   To justify this rigorously: if a plane is parallel to the \( z \)-axis, then its normal vector must be perpendicular to the \( z \)-axis. Here, the normal vector is

   \[ \vec{n} = a\mathbf{i} + b\mathbf{j}, \quad \text{and the direction of the \( z \)-axis is } \mathbf{k}. \]

   Since

   \[ \vec{n} \cdot \mathbf{k} = 0, \]

   the plane is indeed perpendicular to \( \mathbf{k} \), meaning it is parallel to the \( z \)-axis.

3. If \( b = 0 \):

   The equation becomes

   \[ ax + cz + d = 0. \]

   There is no \( y \)-term, so \( y \) is unrestricted. The plane is therefore parallel to the \( y \)-axis.

   Similarly, if \( a = 0 \), the plane is parallel to the \( x \)-axis.

4. If \( b = 0 \) and \( c = 0 \):

   The equation becomes

   \[ ax + d = 0, \]

   so \( x \) is fixed, and \( y, z \) are free. The plane is parallel to both \( y \)-axis and \( z \)-axis, and hence perpendicular to the \( x \)-axis.

   Similarly:

   • If \( a = 0 \) and \( c = 0 \), the plane is perpendicular to the \( y \)-axis.
   • If \( a = 0 \) and \( b = 0 \), the plane is perpendicular to the \( z \)-axis.

5. All of \( a, b, c \) cannot be zero simultaneously.

   If \( a = b = c = 0 \), the equation becomes

   \[ 0x + 0y + 0z + d = 0 \implies d = 0, \]

   which is either always true or always false depending on the value of \( d \), and hence does not define a plane. Therefore, at least one of \( a, b, c \) must be non-zero for the equation to represent a plane.

Example

Find the equation of the plane passing through the three points

\[ A(1, 1, 1), \quad B(2, 3, 2), \quad C(-1, -2, 0). \]

Solution:

To write the equation of a plane, we need two things:

1. A point on the plane (don’t worry—we’ve got three! So we’re rich).
2. A normal vector (this we don’t have yet, but we will dig it out).

Let us compute two vectors that lie on the plane:

\[ \vec{AB} = \vec{OB} - \vec{OA} = (2 - 1)\mathbf{i} + (3 - 1)\mathbf{j} + (2 - 1)\mathbf{k} = \mathbf{i} + 2\mathbf{j} + \mathbf{k}, \]

\[ \vec{AC} = \vec{OC} - \vec{OA} = (-1 - 1)\mathbf{i} + (-2 - 1)\mathbf{j} + (0 - 1)\mathbf{k} = -2\mathbf{i} -3\mathbf{j} - \mathbf{k}. \]

Now, the normal vector to the plane is given by the cross product

\[ \vec{n} = \vec{AB} \times \vec{AC}. \]

Compute this using the determinant:

\[ \vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 1 \\ -2 & -3 & -1 \end{vmatrix} = \mathbf{i}(2 \cdot (-1) - 1 \cdot (-3)) - \mathbf{j}(1 \cdot (-1) - 1 \cdot (-2)) + \mathbf{k}(1 \cdot (-3) - 2 \cdot (-2)). \]

Simplifying:

\[ \vec{n} = \mathbf{i}(-2 + 3) - \mathbf{j}(-1 + 2) + \mathbf{k}(-3 + 4) = \mathbf{i}(1) - \mathbf{j}(1) + \mathbf{k}(1) = \mathbf{i} - \mathbf{j} + \mathbf{k}. \]

So the normal vector has direction ratios \( a = 1, b = -1, c = 1 \).
Now we use point \( A(1, 1, 1) \) (any of the three would work) and write the equation of the plane using the Cartesian form:

\[ a(x - x_1) + b(y - y_1) + c(z - z_1) = 0. \]

Substituting:

\[ 1(x - 1) - 1(y - 1) + 1(z - 1) = 0 \implies x - 1 - y + 1 + z - 1 = 0 \implies x - y + z -1 = 0. \]

Thus, the required equation of the plane is

\[ \boxed{x - y + z = 1}. \]

Intercept Form of the Equation of a Plane

If a plane cuts the x-axis at point \( P(p, 0, 0) \), we say its x-intercept is \( p \).
If it cuts the y-axis at point \( Q(0, q, 0) \), its y-intercept is \( q \).
If it cuts the z-axis at point \( R(0, 0, r) \), its z-intercept is \( r \).

Given these three intercepts \( p, q, r \ne 0 \), the points \( P, Q, R \) are all on the plane. Since we are given three non-collinear points, the plane is uniquely determined.

To write the equation of the plane, we need:

1. A point on the plane (we have three),
2. A normal vector (we’ll compute this now).

Let us construct two vectors in the plane:

\[ \vec{PQ} = \vec{OQ} - \vec{OP} = (0, q, 0) - (p, 0, 0) = -p\,\mathbf{i} + q\,\mathbf{j}, \]

\[ \vec{PR} = \vec{OR} - \vec{OP} = (0, 0, r) - (p, 0, 0) = -p\,\mathbf{i} + r\,\mathbf{k}. \]

The normal vector to the plane is then given by the cross product:

\[ \vec{n} = \vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -p & q & 0 \\ -p & 0 & r \end{vmatrix} = \mathbf{i}(q \cdot r - 0) - \mathbf{j}( -p \cdot r - 0) + \mathbf{k}( -p \cdot 0 + p \cdot q) \]

\[ = q r\,\mathbf{i} + p r\,\mathbf{j} + p q\,\mathbf{k}. \]

So the normal vector is \( \vec{n} = qr\,\mathbf{i} + pr\,\mathbf{j} + pq\,\mathbf{k} \), and we can now write the equation of the plane using point \( P(p, 0, 0) \) and normal vector \( \vec{n} \):

\[ a(x - p) + b(y - 0) + c(z - 0) = 0, \]

where \( a = qr, b = pr, c = pq \). Substituting:

\[ qr(x - p) + pr(y) + pq(z) = 0. \]

Divide the entire equation by \( pqr \ne 0 \):

\[ \frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 1. \]

This is the intercept form of the equation of a plane. It represents a plane that cuts the x-, y-, and z-axes at \( p, q, r \) respectively (none of which are zero).

Normalized Form of the Equation of a Plane

The normalized form of a plane reveals a lot about the orientation and position of the plane in space.

Suppose the equation of a plane is given in vector form as

\[ \vec{r} \cdot \vec{n} = d, \]

or equivalently in Cartesian form as

\[ ax + by + cz + d = 0. \]

Can we understand the orientation of the plane just by looking at this equation? One way is to rewrite the equation in intercept form, which gives immediate geometric insight. But a more refined and analytically rich way is to write the equation in normalized form.

Assume the plane is given in the form

\[ \vec{r} \cdot \vec{n} = d, \]

where \( \vec{n} \) is the normal vector and \( d > 0 \). (If \( d < 0 \), we simply multiply both sides by \( -1 \) to make the right-hand side positive. For instance,

\[ \vec{r} \cdot (\mathbf{i} + \mathbf{j} - \mathbf{k}) = -3 \quad \text{can be written as} \quad \vec{r} \cdot (-\mathbf{i} - \mathbf{j} + \mathbf{k}) = 3.) \]

Now, divide both sides by the magnitude of the normal vector \( |\vec{n}| \):

\[ \vec{r} \cdot \frac{\vec{n}}{|\vec{n}|} = \frac{d}{|\vec{n}|}. \]

Let

\[ \hat{n} = \frac{\vec{n}}{|\vec{n}|}, \quad d_0 = \frac{d}{|\vec{n}|}. \]

Then the equation becomes

\[ \vec{r} \cdot \hat{n} = d_0. \]

This is the normalized form of the plane. It satisfies two properties:

– \( \hat{n} \) is a unit vector normal to the plane,
– \( d_0 > 0 \) is a scalar.

But this form tells much more.

Recall that \( \vec{r} \) is the position vector of any point on the plane. Then \( \vec{r} \cdot \hat{n} \) is the projection of \( \vec{r} \) along the direction of \( \hat{n} \). Since

\[ \vec{r} \cdot \hat{n} = d_0 > 0, \]

this projection is positive, meaning that the angle between \( \vec{r} \) and \( \hat{n} \) is acute. Wherever \( \vec{r} \) lies on the plane, the angle between it and the unit normal vector \( \hat{n} \) is always acute.

Now observe this geometrically: the projection of \( \vec{r} \) onto \( \hat{n} \) is always a segment of length \( d_0 \), and since \( \vec{r} \cdot \hat{n} = d_0 \) for all \( \vec{r} \) on the plane, this implies the entire plane lies at a fixed perpendicular distance \( d_0 \) from the origin.

Thus:

• \( \hat{n} \) points from the origin toward the plane,
• \( d_0 \) is the perpendicular distance from the origin to the plane.

Hence, the normalized form

\[ \vec{r} \cdot \hat{n} = d_0, \quad \text{with } d_0 > 0, \]

encodes direction, orientation, and distance all at once.

For example, consider the plane given by the equation

\[ \vec{r} \cdot (\mathbf{i} - \mathbf{j} + \mathbf{k}) + 3 = 0. \]

This can be rewritten as

\[ \vec{r} \cdot (-\mathbf{i} + \mathbf{j} - \mathbf{k}) = 3. \]

Now compute the magnitude of the normal vector:

\[ |-\mathbf{i} + \mathbf{j} - \mathbf{k}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}. \]

Divide both sides by \( \sqrt{3} \) to obtain the normalized form:

\[ \vec{r} \cdot \left( \frac{-\mathbf{i} + \mathbf{j} - \mathbf{k}}{\sqrt{3}} \right) = \sqrt{3}. \]

Thus, the plane lies at a distance \( \sqrt{3} \) from the origin, and the unit normal vector

\[ \hat{n} = \frac{-\mathbf{i} + \mathbf{j} - \mathbf{k}}{\sqrt{3}} \]

points from the origin toward the plane.

The direction of this vector shows that the plane is oriented to cut:

– the negative x-axis (since the \( i \)-component is negative),
– the positive y-axis (since the \( j \)-component is positive),
– the negative z-axis (since the \( k \)-component is negative).

This geometric picture is captured in the figure above: the red vector \( \vec{n} \) is the unit normal from the origin to the plane, and the perpendicular length from the origin to the plane is \( \sqrt{3} \).

So, in normalized form, the equation

\[ \vec{r} \cdot \hat{n} = d_0 \]

gives both the orientation (via \( \hat{n} \)) and the position (via \( d_0 \)) of the plane in a geometrically meaningful way.

Normalized Cartesian Form

If the equation of the plane is given in Cartesian form as

\[ ax + by + cz + d = 0, \]

then we first rearrange it as

\[ ax + by + cz = -d. \]

To express this in normalized form, we must ensure that the right-hand side is positive. If \( -d > 0 \), proceed directly. If \( -d < 0 \), multiply both sides by \( -1 \) to make the right-hand side positive.

Now divide both sides by

\[ \sqrt{a^2 + b^2 + c^2}, \]

which is the magnitude \( |\vec{n}| \) of the normal vector \( \vec{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \).

We get:

\[ \frac{ax + by + cz}{\sqrt{a^2 + b^2 + c^2}} = \frac{-d}{\sqrt{a^2 + b^2 + c^2}}. \]

Let

\[ \hat{n} = \frac{a\mathbf{i} + b\mathbf{j} + c\mathbf{k}}{\sqrt{a^2 + b^2 + c^2}}, \quad d_0 = \frac{-d}{\sqrt{a^2 + b^2 + c^2}}. \]

Then the equation becomes

\[ \vec{r} \cdot \hat{n} = d_0, \]

which is the normalized form of the plane. It gives a unit normal vector \( \hat{n} \) pointing from the origin toward the plane, and \( d_0 > 0 \) is the perpendicular distance of the plane from the origin.

What Is the Meaning of Normalization?

To normalize something means to adjust it to a common or standard scale so that fair comparisons or consistent interpretations can be made. Normalization removes the effect of differing units, magnitudes, or contexts, allowing you to compare or analyze things meaningfully.

Let us take a simple and relatable example.

Suppose:

– Student A scores 300 out of 500 in one board.
– Student B scores 500 out of 650 in another board.

Now, if someone asks: Who performed better?, we cannot directly compare these numbers, because the total marks are different.

So we normalize the scores to a common scale—say, out of 100—by converting them into percentages:

\[ \text{Student A: } \frac{300}{500} \times 100 = 60\%, \quad \text{Student B: } \frac{500}{650} \times 100 \approx 76.92\%. \]

Now, on the same scale, Student B clearly performed better. This is the essence of normalization:

→ Remove scale bias,
→ Bring all data to a common basis,
→ Enable meaningful comparison.

When we normalize a vector \( \vec{n} \), we divide it by its magnitude \( |\vec{n}| \), giving a unit vector \( \hat{n} = \vec{n}/|\vec{n}| \).

This gives a direction vector of length 1, which is like saying:

→ Forget the length; tell me only the direction.

Similarly, when we write the normalized form of a plane, we divide both sides of \( \vec{r} \cdot \vec{n} = d \) by \( |\vec{n}| \), so that the normal vector becomes unit, and the right-hand side becomes the actual distance from origin, measured in a standardized way.

So, just as you cannot compare marks across different boards without converting to percentages, you cannot compare or interpret planes and vectors geometrically unless you normalize them—bringing all to the same scale and same footing.

Interaction of a Point with a Plane

When a point \( P \) lies near a plane \( \Pi \), we are often interested in certain geometric quantities that describe the relationship between the point and the plane. These include:

– the perpendicular distance from point \( P \) to the plane \( \Pi \),
– the foot of the perpendicular from \( P \) onto the plane \( \Pi \),
– the image of point \( P \) with respect to the plane \( \Pi \) (i.e., its reflection across the plane).

We shall now learn how to compute all of these using both the vector form and the Cartesian form of the equation of a plane.

Plane in Vector Form

Consider a point \( P \) having position vector \( \vec{p} \), and a plane given by the equation

\[ (\vec{r} - \vec{a}) \cdot \vec{n} = 0, \]

representing a plane that passes through point \( A \), whose position vector is \( \vec{a} \), and has a normal vector \( \vec{n} \).

Drop a perpendicular from point \( P \) to the plane, and let it meet the plane at point \( N \). Then triangle \( \triangle APN \) is a right-angled triangle at \( N \), and the vector \( \vec{n} \) lies along the direction of \( \vec{PN} \), the perpendicular from \( P \) to the plane.

The vector from \( A \) to \( P \) is

\[ \vec{AP} = \vec{p} - \vec{a}. \]

The length of the perpendicular segment \( \vec{PN} \) is the magnitude of the projection of vector \( \vec{AP} \) along \( \vec{n} \). That is,

\[ |\vec{PN}| = \left| \frac{ (\vec{p} - \vec{a}) \cdot \vec{n} }{ |\vec{n}| } \right| = \frac{ |(\vec{p} - \vec{a}) \cdot \vec{n}| }{|\vec{n}| }. \]

This expression gives the perpendicular distance from point \( P \) to the plane.

Calculation of Foot:

Let us now compute the foot of the perpendicular from the point \( P \).

Let \( N \) be the foot of the perpendicular from \( P \) onto the plane. Then the vector \( \vec{NP} \) is the projection of \( \vec{AP} \) along the normal direction, i.e.,

\[ \vec{NP} = \left( \frac{(\vec{p} - \vec{a}) \cdot \vec{n}}{|\vec{n}|^2} \right) \vec{n}. \]

Since

\[ \vec{OP} - \vec{ON} = \vec{NP}, \]

we get:

\[ \vec{ON} = \vec{OP} - \vec{NP} = \vec{p} - \left( \frac{(\vec{p} - \vec{a}) \cdot \vec{n}}{|\vec{n}|^2} \right) \vec{n}. \]

Hence, the position vector of the foot \( N \) of point \( P \) on the plane is:

\[ \boxed{ \vec{r}_N = \vec{p} - \frac{(\vec{p} - \vec{a}) \cdot \vec{n}}{|\vec{n}|^2} \vec{n} }. \]

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Calculation of Perpendicular Distance and Foot when the Plane is in General Vector Form

But in most problems, the plane is not given in the form \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \). Instead, it is typically given in the form

\[ \vec{r} \cdot \vec{n} = d, \]

which does not explicitly mention any point \( \vec{a} \) on the plane.

Let us compute the perpendicular distance from a point \( P \), having position vector \( \vec{p} \), to the plane \( \Pi \) with equation \( \vec{r} \cdot \vec{n} = d \).

From point \( P \), drop a perpendicular to the plane. Let this perpendicular meet the plane at point \( N \), the foot of the perpendicular. Since the line segment \( PN \) is perpendicular to the plane, it must be parallel to the normal vector \( \vec{n} \).

Hence, the line passing through point \( P \) and directed along \( \vec{n} \) is given by

\[ \vec{r} = \vec{p} + \lambda \vec{n}, \quad \lambda \in \mathbb{R}. \]

To find the foot \( N \), we need the value of \( \lambda \) such that this point lies on the plane. Substituting into the plane equation:

\[ (\vec{p} + \lambda \vec{n}) \cdot \vec{n} = d \implies \vec{p} \cdot \vec{n} + \lambda |\vec{n}|^2 = d \implies \lambda = \frac{d - \vec{p} \cdot \vec{n}}{|\vec{n}|^2}. \]

Therefore, the position vector of point \( N \) is

\[ \vec{r}_N = \vec{p} + \frac{d - \vec{p} \cdot \vec{n}}{|\vec{n}|^2} \vec{n}. \]

Now, the vector from \( P \) to \( N \) is

\[ \vec{PN} = \vec{r}_N - \vec{p} = \frac{d - \vec{p} \cdot \vec{n}}{|\vec{n}|^2} \vec{n}. \]

Hence, the distance from point \( P \) to the plane is

\[ |\vec{PN}| = \left| \frac{d - \vec{p} \cdot \vec{n}}{|\vec{n}|^2} \vec{n} \right| = \left| \frac{d - \vec{p} \cdot \vec{n}}{|\vec{n}|^2} \right| \cdot |\vec{n}| = \frac{ |d - \vec{p} \cdot \vec{n}| }{|\vec{n}| }. \]

This is the general formula for the perpendicular distance from a point to a plane given in the form \( \vec{r} \cdot \vec{n} = d \).

While computing the perpendicular distance from point \( P \) to the plane \( \vec{r} \cdot \vec{n} = d \), we also obtained the foot of the perpendicular, that is, the point \( N \) on the plane where the perpendicular from \( P \) meets the plane.

This point \( N \) has position vector

\[ \boxed {\vec{r}_N = \vec{p} + \frac{d - \vec{p} \cdot \vec{n}}{|\vec{n}|^2} \vec{n}}. \]

Cartesian Form of Plane: Foot of the Perpendicular and Distance from a Point to a Plane

Consider a plane \( \Pi \) given by

\[ ax + by + cz + d = 0, \]

and a point \( P(x_0, y_0, z_0) \) in space. Drop a perpendicular \( PN \) from the point \( P \) to the plane. Let the foot of this perpendicular be \( N \). Since \( PN \) is normal to the plane, its direction ratios are \( (a, b, c) \).

Let us now write the equation of the line passing through \( P \) and perpendicular to the plane. The direction ratios are \( (a, b, c) \), and the line passes through \( (x_0, y_0, z_0) \), so the line is given by:

\[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} = \lambda. \]

This gives the coordinates of any point on the line as:

\[ x = a\lambda + x_0,\quad y = b\lambda + y_0,\quad z = c\lambda + z_0. \]

Since point \( N \) lies on this line and also lies on the plane, it must satisfy the plane equation. Substituting into the plane equation:

\[ a(a\lambda + x_0) + b(b\lambda + y_0) + c(c\lambda + z_0) + d = 0, \]

\[ \lambda(a^2 + b^2 + c^2) + ax_0 + by_0 + cz_0 + d = 0, \]

\[ \lambda = \frac{-(ax_0 + by_0 + cz_0 + d)}{a^2 + b^2 + c^2}. \]

This gives us the value of \( \lambda \) corresponding to the foot of the perpendicular.

The coordinates of the foot \( N \) are then:

\[ x = a\lambda + x_0,\quad y = b\lambda + y_0,\quad z = c\lambda + z_0. \]

Now, the distance from point \( P \) to the plane is simply the length of vector \( \vec{PN} \), which is:

\[ \text{Distance} = \sqrt{(a\lambda)^2 + (b\lambda)^2 + (c\lambda)^2} = |\lambda| \sqrt{a^2 + b^2 + c^2}. \]

Substituting the expression for \( \lambda \), we get:

\[ \boxed{\text{Distance} = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}}. \]

This is the well-known formula for the perpendicular distance from a point to a plane in Cartesian form.

Since the value of \( \lambda \) corresponding to the foot of the perpendicular \( N \) is

\[ \lambda = \frac{ - (a x_0 + b y_0 + c z_0 + d) }{ a^2 + b^2 + c^2 }, \]

we can now directly substitute this value into the symmetric form of the line:

\[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} = \lambda. \]

Thus, the coordinates of the foot \( N \) are given by

\[ \boxed{\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} = -\frac{a x_0 + b y_0 + c z_0 + d}{a^2 + b^2 + c^2}}. \]

An example is definitely a must to understand how to use this foot of the perpendicular formula.

Consider the plane

\[ 2x - y - 2z - 4 = 0, \]

and let us compute the foot of the perpendicular from the point \( P(-1, 1, 1) \) to this plane.

According to the formula, the coordinates of the foot are given by

\[ \frac{x + 1}{2} = \frac{y - 1}{-1} = \frac{z - 1}{-2} = -\frac{2(-1) - (1) - 2(1) - 4}{2^2 + (-1)^2 + (-2)^2}. \]

Compute the right-hand side directly:

\[ = -\frac{-2 - 1 - 2 - 4}{4 + 1 + 4} = -\frac{-9}{9} = 1. \]

Now equating each expression to 1:

\[ \frac{x + 1}{2} = 1 \implies x = 1, \quad \frac{y - 1}{-1} = 1 \implies y = 0, \quad \frac{z - 1}{-2} = 1 \implies z = -1. \]

Hence, the foot of the perpendicular from point \( (-1, 1, 1) \) to the plane \( 2x - y - 2z - 4 = 0 \) is

\[ \boxed{(1, 0, -1)}. \]

Observe that the rightmost expression in the foot formula,

\[ -\frac{2(-1) - (1) - 2(1) - 4}{4 + 1 + 4} = -\frac{-9}{9} = 1, \]

comes from substituting the point \( (-1, 1, 1) \) into the left-hand side of the plane equation

\[ 2x - y - 2z - 4. \]

Substituting:

\[ 2(-1) - (1) - 2(1) - 4 = -2 - 1 - 2 - 4 = -9. \]

So the numerator of the expression is simply the negative of the value obtained when the point is substituted into the left-hand side of the plane. The denominator is the sum of squares of the coefficients of \( x, y, z \), i.e., \( 2^2 + (-1)^2 + (-2)^2 = 9 \).

This pattern always holds:

\[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} = -\frac{ax_0 + by_0 + cz_0 + d}{a^2 + b^2 + c^2}, \]

where the numerator is the negative of the left-hand side of the plane equation evaluated at the point, and the denominator is the square of the length of the normal vector.

Image of a Point in a Plane

The calculation of the image of a point with respect to a plane is quite straightforward.

Let \( Q \) be the image of a point \( P \) in a plane \( \Pi \). Then, by definition, the midpoint of the segment \( PQ \) lies on the plane and is the foot of the perpendicular from \( P \) to the plane. Let this foot be denoted by \( N \).

Since \( N \) is the midpoint of \( P \) and \( Q \), we have:

\[ Q = 2N - P. \]

So to compute the image point \( Q \), we first compute the foot \( N \) of point \( P \) on the plane, and then use the midpoint formula to compute \( Q \).

There is also a direct formula to calculate the image of point \( P(x_0, y_0, z_0) \) in the plane

\[ ax + by + cz + d = 0, \]

without computing the foot explicitly.

The image point \( Q(x, y, z) \) satisfies the symmetric equation:

\[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} = -\frac{2(ax_0 + by_0 + cz_0 + d)}{a^2 + b^2 + c^2}. \]

This formula is similar to the one used for the foot of the perpendicular, but now includes a factor of 2 in the numerator. This factor appears because the image point lies on the opposite side of the plane, at the same distance as \( P \), along the line perpendicular to the plane.

You are encouraged to derive this formula yourself in the same we derived foot formula above.

Ratio in Which a Plane Divides a Line Segment

We wish to calculate the ratio in which a plane

\[ \Pi : ax + by + cz + d = 0 \]

divides a line segment joining two points

\[ A(x_1, y_1, z_1) \quad \text{and} \quad B(x_2, y_2, z_2). \]

The division may be internal or external, depending on the sign of the ratio. Let the plane \( \Pi \) intersect the line segment \( AB \) at a point \( P \). Assume that \( P \) divides \( AB \) in the ratio \( \lambda : 1 \), where \( \lambda \in \mathbb{R} \).

Then, by the section formula, the coordinates of point \( P \) are:

\[ \left( \frac{\lambda x_2 + x_1}{\lambda + 1},\ \frac{\lambda y_2 + y_1}{\lambda + 1},\ \frac{\lambda z_2 + z_1}{\lambda + 1} \right). \]

Since \( P \) lies on the plane \( \Pi \), it must satisfy the plane equation. Substituting the coordinates of \( P \) into the plane equation:

\[ a \left( \frac{\lambda x_2 + x_1}{\lambda + 1} \right) + b \left( \frac{\lambda y_2 + y_1}{\lambda + 1} \right) + c \left( \frac{\lambda z_2 + z_1}{\lambda + 1} \right) + d = 0. \]

Combine all terms over the common denominator \( \lambda + 1 \):

\[ \frac{ \lambda (a x_2 + b y_2 + c z_2) + (a x_1 + b y_1 + c z_1) }{\lambda + 1} + d = 0. \]

Multiply both sides by \( \lambda + 1 \):

\[ \lambda (a x_2 + b y_2 + c z_2) + (a x_1 + b y_1 + c z_1) + d(\lambda + 1) = 0. \]

Group and simplify:

\[ \lambda (a x_2 + b y_2 + c z_2 + d) + (a x_1 + b y_1 + c z_1 + d) = 0. \]

Isolating \( \lambda \):

\[ \lambda = -\frac{a x_1 + b y_1 + c z_1 + d}{a x_2 + b y_2 + c z_2 + d}. \]

This can be abbreviated as

\[ \lambda = -\frac{\Pi(x_1, y_1, z_1)}{\Pi(x_2, y_2, z_2)}, \]

where \( \Pi(x, y, z) \) denotes the left-hand side of the plane equation evaluated at that point.

Thus, to find the ratio in which the plane divides the segment \( AB \), we evaluate the plane equation at both endpoints, and take the negative ratio.

Example

Let there be a plane having equation:

\[ \Pi: x - 2y + z - 3 = 0, \]

and consider the points

\[ A(1, 2, 3), \quad B(4, 1, -1). \]

We want to find the ratio \( \lambda:1 \) in which the plane divides the line segment \( AB \). As discussed, the ratio is given by

\[ \lambda = -\frac{\Pi(x_1, y_1, z_1)}{\Pi(x_2, y_2, z_2)}. \]

First evaluate the plane expression at point \( A(1, 2, 3) \):

\[ \Pi(1, 2, 3) = 1 - 2(2) + 3 - 3 = 1 - 4 + 3 - 3 = -3. \]

Next, at point \( B(4, 1, -1) \):

\[ \Pi(4, 1, -1) = 4 - 2(1) + (-1) - 3 = 4 - 2 - 1 - 3 = -2. \]

Now compute the ratio:

\[ \lambda = -\frac{-3}{-2} = -\frac{3}{2}. \]

So the plane divides the line \( AB \) in the ratio

\[ \boxed{-3:2}. \]

Since the ratio is negative, the point of intersection lies on the extension of segment \( AB \) beyond point \( A \), and the division is external.

Relative Location of Points with Respect to a Plane

Consider two points

\[ A(x_1, y_1, z_1) \quad \text{and} \quad B(x_2, y_2, z_2) \]

in space, and let there be a plane

\[ \Pi : ax + by + cz + d = 0. \]

We wish to determine whether the points \( A \) and \( B \) lie on the same side of the plane, on opposite sides, or if either of them lies on the plane.

This concept has wide relevance—not just in geometry, but also in machine learning, where it is used to classify objects. For instance, an algorithm might learn to distinguish a cat from a dog by representing them as points in a high-dimensional space, and then using a plane (called a decision boundary) to separate the two classes. This geometric idea is precisely what's happening here.

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Let us now analyze the situation.

Evaluate the left-hand side of the plane equation at both points:

\[ \Pi(A) = ax_1 + by_1 + cz_1 + d, \quad \Pi(B) = ax_2 + by_2 + cz_2 + d. \]

Now suppose \( A \) and \( B \) lie on opposite sides of the plane. Then the plane divides the line segment \( AB \) internally, and the ratio in which it divides is

\[ \lambda = -\frac{ax_1 + by_1 + cz_1 + d}{ax_2 + by_2 + cz_2 + d}. \]

Since this is an internal division, the ratio \( \lambda \) must be positive, so:

\[ -\frac{\Pi(A)}{\Pi(B)} > 0 \implies \frac{\Pi(A)}{\Pi(B)} < 0 \implies \Pi(A) \cdot \Pi(B) < 0. \]

That is, the two evaluated expressions must have opposite signs, meaning the two points are on opposite sides of the plane.

Similarly, if both points lie on the same side of the plane, then the plane divides the line joining them externally, and the ratio \( \lambda \) is negative, so:

\[ -\frac{\Pi(A)}{\Pi(B)} < 0 \implies \Pi(A) \cdot \Pi(B) > 0. \]

Thus:

• If \( \Pi(A) \cdot \Pi(B) < 0 \), the points lie on opposite sides of the plane.
• If \( \Pi(A) \cdot \Pi(B) > 0 \), the points lie on the same side.

In simple terms, if two points are on the same side of a plane, then when we substitute their coordinates into the left-hand side of the plane equation

\[ ax + by + cz + d, \]

both will give values with the same sign—either both positive or both negative.

On the other hand, if the two points are on opposite sides of the plane, then substituting their coordinates into the left-hand side of the plane equation will give opposite signs—one positive and one negative.

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This leads to an important geometric conclusion:

A plane divides the entire space into two distinct regions—
– In one region, every point gives a positive value when substituted into the plane equation.
– In the other region, every point gives a negative value.
– The points on the plane itself are precisely those for which the value is zero.

This is the fundamental idea behind how a plane partitions space.

In vector form, a plane is written as

\[ \vec{r} \cdot \vec{n} = d, \]

where \( \vec{n} \) is the normal vector to the plane and \( d \) is a constant.

Just like in Cartesian form, this equation divides space into three parts:

– For points on the plane:

\[ \vec{r} \cdot \vec{n} = d. \]

– For points on one side of the plane:

\[ \vec{r} \cdot \vec{n} > d. \]

– For points on the other side of the plane:

\[ \vec{r} \cdot \vec{n} < d. \]

Thus, the plane divides space into two regions:

– One region where every point satisfies \( \vec{r} \cdot \vec{n} > d \),
– Another region where every point satisfies \( \vec{r} \cdot \vec{n} < d \).

Distance of a Point from a Plane Measured Along a Given Line

Consider a plane given by

\[ ax + by + cz + d = 0, \]

and a point \( P(x_0, y_0, z_0) \) in space.

We already have the standard formula to measure the perpendicular distance from \( P \) to the plane—that is, the shortest distance measured along the normal to the plane.

But suppose we want to measure the distance from \( P \) to the plane along some other line, not necessarily perpendicular to the plane. Let this line have direction ratios \( l, m, n \).

That is, we draw a line passing through point \( P \) with direction ratios \( (l, m, n) \), and let this line intersect the plane at a point \( Q \). The length of the segment \( PQ \) is then what we call the distance of point \( P \) from the plane measured along the line \( l, m, n \).

We begin by writing the equation of the line \( PQ \) in symmetric form:

\[ \frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n} = \lambda. \]

Hence, the coordinates of point \( Q \) can be expressed as:

\[ x = l\lambda + x_0, \quad y = m\lambda + y_0, \quad z = n\lambda + z_0, \]

for some \( \lambda \in \mathbb{R} \).

Now, since point \( Q \) lies on the plane, it must satisfy the plane equation. Substituting:

\[ a(l\lambda + x_0) + b(m\lambda + y_0) + c(n\lambda + z_0) + d = 0. \]

Expanding and isolating \( \lambda \):

\[ \lambda(al + bm + cn) + (ax_0 + by_0 + cz_0 + d) = 0, \]

\[ \lambda = -\frac{ax_0 + by_0 + cz_0 + d}{al + bm + cn}. \]

Now we compute the exact length of \( PQ \). The coordinates of \( Q \) are

\[ x = l\lambda + x_0,\quad y = m\lambda + y_0,\quad z = n\lambda + z_0, \]

so the segment \( PQ \) has length

\[ PQ = \sqrt{(l\lambda + x_0 - x_0)^2 + (m\lambda + y_0 - y_0)^2 + (n\lambda + z_0 - z_0)^2} = \sqrt{(l\lambda)^2 + (m\lambda)^2 + (n\lambda)^2} = |\lambda| \sqrt{l^2 + m^2 + n^2}. \]

Substituting the value of \( \lambda \):

\[ PQ = \left| \frac{ax_0 + by_0 + cz_0 + d}{al + bm + cn} \right| \cdot \sqrt{l^2 + m^2 + n^2}. \]

If the direction ratios \( l, m, n \) are direction cosines, i.e.,

\[ l^2 + m^2 + n^2 = 1, \]

then the formula simplifies to:

\[ PQ = \left| \frac{ax_0 + by_0 + cz_0 + d}{al + bm + cn} \right|. \]

This gives the distance from a point to a plane measured not perpendicularly, but along a given line.