Alternative University

Analytic Geometry


Euclidean Space
“A surface in the space is a plane only when it is the first-order algebraic surface… any surface of the first order is a plane. Plane is infinite. Another definition of plane: a plane is a two-dimensional, doubly ruled surface spanned by two linearly independent vectors.”
Krivoshapko & Ivanov, Encyclopedia of Analytical Surfaces

In the Euclidean space of introductory mathematics, a plane is a flat surface extending infinitely in all directions. We say it is doubly ruled because it can be rolled in different directions (infinite directions in fact).

Ruled surfaces, of differential geometry o, can be rolled, because they can be generated by straight lines moving in a direction that is not collinear or parallel with the straight line (hence can be rolled in that direction).

For a plane, any straight line that is coplanar with the plane can be used to generate the plane as a ruled surface.

Equations of a Plane

The equation of a plane is of “order 1” because the equation variables have an exponent equal to one (no equation variable exponents greater than 1). The following is an algebraic equation of a plane in three-dimensional (x, y, z) space:

( x ╱ a ) + ( y ╱ b ) + ( z ╱ c )  =  1

[equation 1]

where a is the x-intercept, b is the y-intercept, and c is the z-intercept; in other words, the plane contains the points (a, 0, 0), (0, b, 0) and (0, 0, c).

Here is another equation of the plane:

Ax + By + Cz + D  =  0

[equation 2]

where −D/A is the x-intercept, −D/B is the y-intercept, and −D/C is the z-intercept.

A useful feature of this formula is that using those coefficients provides the following vector that is perpendicular to the plane:

N  =  ( A, B, C )

where N is not generally a unit vector. Before normalizing N, let us first look at vector algebra products that are useful for working with planes.

Vector Algebra Products

Vector algebra products are products generated by processing two vectors together to form a product. Since there are two vectors as input for this processing, the processing is referred to as a binary operation.

There are two products of binary operations of vectors: one generates a scalar result; the other generates another (new) vector. Hence, the first of these products is called a scalar product since it generates a scalar, and the second is called a vector product since it generates a vector.

Scalar Product
(Dot Product, Inner Product)

The scalar product is also referred to as a dot product or inner product, and like multiplication is denoted with a dot (·) or no symbol.

In this operation, corresponding components of the two input vectors are multiplied together, and those products are summed.

For example, given two vectors u = (a₁, a₂) and v = (b₁, b₂), the dot product of those two vectors is:

u·v = a₁b₁ + a₂b₂

[equation 3]

Likewise, for vectors u = (a₁, a₂, a₃) and v = (b₁, b₂, b₃), the dot product is:

uv = a₁b₁ + a₂b₂ + a₃b₃

[equation 4]

And so on, for any number of components (not just two or three components as shown here).

An important property of the dot product is that it is zero if the vectors are perpendicular.

Another useful feature of the dot product is that the dot product of a vector with itself gives the square of the vector length, which is the radicand of the Pythagorean theorem.

For example, following is the dot product of u = (a₁, a₂, a₃) with itself, and the length (norm) of that vector:

Vector Product
(Cross Product)

The vector product (also called cross product) uses the large times symbol (×).

Here is the formula for the cross product of two vectors u = (a₁, a₂, a₃) and v = (b₁, b₂, b₃), pronounced “u cross v”:

u×v = ( a₂b₃−a₃b₂, a₃b₁−a₁b₃, a₁b₂−a₂b₁ )

[equation 5]

The cross product produces a new vector that is perpendicular to the two input vectors. Since the two input vectors span a plane (any two noncollinear vectors span a plane), the new vector is perpendicular to that plane.

Optional Technical Note: Notice that there is symmetry of input vector component subscripts in each cross product component of the cross product formula (equation 5 above).


Problem: Show that the vector N = (A, B, C) is perpendicular to the plane Ax + By + Cz + D = 0.

Solution: We use the cross product formula (eq. 5), since the cross product of two vectors is perpendicular to the two vectors and hence perpendicular to the plane that is spanned by the two vectors. We create two vectors on the plane to produce a cross product that is collinear/parallel with N, showing that N is perpendicular to the plane since it is collinear/parallel with a cross product that is perpendicular to the plane.

To create two vectors on a plane requires at least three noncollinear points on the plane (three points that are not on the same line). Here we can use the x-intercept, y-intercept and z-intercept as three noncollinear points on the plane.

The formula Ax + By + Cz + D = 0 is a standard equation of the plane (eq. 2 above), which has −D/A as the x-intercept, −D/B as the y-intercept, and −D/C as the z-intercept.

Say we have one vector u that extends from the x-intercept to the y-intercept, and another vector v from the x-intercept to the z-intercept:

 =  ( −D/A, 0, 0 ) − ( 0, −D/B, 0 )
 =  ( −D/A, D/B, 0 )
 =  ( −D/A, 0, 0 ) − ( 0, 0, −D/C )
 =  ( −D/A, 0, D/C )

since subtracting points creates a vector between the points (because subtracting coordinate axis values of the points provides the coordinate axis offsets between those points, each such coordinate axis offset a component of the vector between the two points).

Using the cross product formula (eq. 5 above), “u cross v”:

 =  (a₂b₃−a₃b₂, a₃b₁−a₁b₃, a₁b₂−a₂b₁)
 =  ( D/B·D/C, D/A·D/C, D/B·D/A )

is perpendicular to the plane since it is perpendicular to the vectors on the plane that produced the cross product.

Scaling a vector does not change the line it is on, since scalar multiplication produces a vector that is on the same line (a one-dimensional vector space). Hence, we can scale this vector with any nonzero scalar, as many times as we want, and the scaled vector will still be on the same line perpendicular to the plane.

First scale by C/D:

(C/D)( D/B·D/C, D/A·D/C, D/B·D/A )
 =  ( D/B, D/A, C/D·D/B·D/A )
 =  ( D/B, D/A, C/B·D/A )

then scale by A/D:

(A/D)( D/B, D/A, C/B·D/A )
 =  ( A/D·D/B, 1, C/B )
 =  ( A/B, 1, C/B )

and scale by B:

B( A/B, 1, C/B )
 =  ( A, B, C )

showing N = (A, B, C) is collinear with a cross product (vector) that is perpendicular to the plane Ax + By + Cz + D = 0, therefore is itself perpendicular to the plane.

Normal of a Plane

As explained in the previous lesson, a unit vector o is a vector with length (norm) equal to 1, and converting a vector to a unit vector (with the same direction but length 1) is called normalization, which is accomplished by dividing the vector components by the norm of the vector.

Normalizing N creates a unit vector, usually denoted n, that is the same direction as N but with vector length (norm) equal to 1:

where the norm of N, denoted |N|, is the principal square root of the dot product of N with itself:

The unit vector n defines a plane that is perpendicular to the vector. We refer to this unit vector as the normal of the plane. All lines on the plane are perpendicular to this vector.

The normal can be used to generate two linearly independent (noncollinear) vectors on the plane that span the plane. Generating the two vectors can be done by first creating a single vector on the plane that is perpendicular to the normal, then taking the cross product of that vector with the normal to generate a second vector on the plane.

By definition of the cross product, the second vector on the plane will be perpendicular to both the normal and the first vector on the plane, thus can span the plane with that first vector.

This allows the plane to be defined only by a normal (unit vector perpendicular to the plane), from which two linearly independent vectors can be generated that are perpendicular to the normal and thus span the plane as just described.

Any vector that is perpendicular to the normal (unit vector n) will be on the plane. To create the first vector on the plane, from the normal, we can manipulate the vector components of a copy of the normal, knowing that the dot product of two perpendicular vectors equals zero.

For example, one of the components could be zeroed out, and the other two components swapped with one of them negated. Then dotting the normal with such a vector would show they are perpendicular, generating the first vector on the plane, from which the cross product of the norm and that first vector would create a second vector on the plane.

This method of generating a plane is useful in ray tracing o.

Parallel Planes

Say we have a plane in 3D space:

Figure 1: Rectangle representing a plane in 3D space. Coordinate axes are in background.

The plane could be denoted P and defined by the formula:

P = r + hn

where r is a position (radius) vector with base (tail) at the origin of a coordinate system, n is the normal of the plane (unit vector perpendicular to the plane), and hn is a scaled vector (scalar multiplication of the vector n with a scalar h) with the tail of hn at the head of r, and the head of hn at a point on the plane, the scalar h specifying how far the point on the plane is along the line of n  from the head of r  (making that point the n-intercept of the plane for the position r):

Figure 2: Vector formula r + hn defining the plane P.

For example, the radius vector could be a vector from the center of a sphere or ellipsoid to a position on the surface of the sphere or ellipsoid, and the scalar h could be the distance from that position to a point along the line of n above or below the surface of the sphere or ellipsoid.

Setting different values for h defines different planes that are parallel to each other since they are all perpendicular to the same normal.

If two planes are defined this way, one plane with h equal to h1 and the other plane with h equal to h2, those planes are parallel (because they have the same normal), and the distance between those two planes would be the absolute value of the difference of the two h values:

| h2 − h1 |

The order of subtraction does not matter because this is the distance without direction (an absolute value — the Pytharorean theorem in one dimension). The following gives the same distance:

| h1 − h2 |

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2021–Sep–22  20:32  UTC