Basic Definitions

Every linear equation $Ax+By+Cz+D=0$, with $A$, $B$, and $C$ not all zero, represents a plane; conversely, every plane has an equation of this form. Here $A,\,B,\,C$ are the coordinates of the normal vector:

A point $M_0(x_0,y_0,z_0)$ lies on the plane $A_1x+B_1y+C_1z+D_1=0$ precisely when its coordinates satisfy the equation:

The normal equation of a plane has the form

where $\cos\alpha,\,\cos\beta,\,\cos\gamma$ are the direction cosines of the unit normal and $p\ge 0$ is the distance from the origin to the plane. It is obtained from the general equation $Ax+By+Cz+D=0$ by multiplying through by the normalizing factor

i.e. the normalizing factor is taken with the sign opposite to the free term $D$. Then

The intercept form of the equation of a plane is

where $a=-\dfrac{D}{A}$, $b=-\dfrac{D}{B}$, $c=-\dfrac{D}{C}$. The numbers $a,\,b,\,c$ are respectively the abscissa, ordinate, and applicate (the $x$-, $y$- and $z$-intercepts) of the points where the plane meets the coordinate axes. This form requires $A,\,B,\,C,\,D$ all nonzero; a plane through the origin or parallel to a coordinate axis has no such form.

The equation of the plane passing through the point $M_0(x_0,y_0,z_0)$ and perpendicular to the vector $\mathbf{n}=A\mathbf{i}+B\mathbf{j}+C\mathbf{k}$ has the form

or, in vector form, $\mathbf{n}\cdot\overrightarrow{M_0M}=0$.

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The equation of the plane passing through three given non-collinear points $M_1(x_1,y_1,z_1)$, $M_2(x_2,y_2,z_2)$, $M_3(x_3,y_3,z_3)$ has the form

or, in vector form, as the vanishing of the mixed product (scalar triple product) $\overrightarrow{M_1M_2}\cdot\bigl(\overrightarrow{M_1M_3}\times\overrightarrow{M_1M}\bigr)=0$. If the three points are collinear, $\overrightarrow{M_1M_2}$ and $\overrightarrow{M_1M_3}$ are proportional, the determinant vanishes for every $(x,y,z)$, and the points do not determine a unique plane.

For an arbitrary value of the parameter $\lambda$, the equation

defines a plane passing through the line of intersection of the planes

As $\lambda$ runs over all real numbers this gives every plane of the pencil except the second plane $A_2x+B_2y+C_2z+D_2=0$ itself (its $\lambda\to\infty$ limit). To include it, use the homogeneous form

with $(\lambda_1,\lambda_2)\neq(0,0)$.

Write $L_i(M_0)=A_ix_0+B_iy_0+C_iz_0+D_i$. The condition for a point $M_0(x_0,y_0,z_0)$ to lie on a plane of the pencil is

When $L_2(M_0)\neq0$ this gives $\lambda_{M_0}=-\dfrac{L_1(M_0)}{L_2(M_0)}$. To avoid the division — and to cover $L_2(M_0)=0$, where the pencil plane through $M_0$ is the second plane $L_2=0$ itself — use the equivalent homogeneous equation of that plane:

If $L_1(M_0)=L_2(M_0)=0$, the point $M_0$ lies on the base line and every plane of the pencil passes through it.

The plane passing through the point $M_0(x_0,y_0,z_0)$ and the line defined as the intersection of the planes

is the pencil member through $M_0$ (assuming $M_0$ is not on the line). Using $L_i(M_0)=A_ix_0+B_iy_0+C_iz_0+D_i$, it is

which for $L_2(M_0)\neq0$ is the same as

with $\lambda_{M_0}=-L_1(M_0)/L_2(M_0)$.

The angle $\varphi$ between the planes $A_1x+B_1y+C_1z+D_1=0$ and $A_2x+B_2y+C_2z+D_2=0$ is determined by the formula

taking $\varphi$ as the angle between the planes, $0^\circ\le\varphi\le90^\circ$, where $\mathbf{n}_1=(A_1,B_1,C_1)$ and $\mathbf{n}_2=(A_2,B_2,C_2)$ are the plane normals. They are parallel (or coincident) when the normals are proportional,

perpendicular when

and intersecting in a line when the normals are not proportional, $\mathbf{n}_1\times\mathbf{n}_2\neq\mathbf{0}$.

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The signed distance (deviation) $\delta$ of a point $M_0(x_0,y_0,z_0)$ from the plane $Ax+By+Cz+D=0$ is found by the formula

where the sign before the radical is taken opposite to the sign of the free term $D$. When $D=0$ either sign is admissible; choose an orientation consistently. Ordinary distance is unchanged: the (ordinary, nonnegative) distance from the point $M_0$ to the plane equals $d=|\delta|$.

The equation of the straight line passing through two distinct points $M_1(x_1,y_1,z_1)$ and $M_2(x_2,y_2,z_2)$ has the form

where a zero denominator is read as the matching numerator being zero (e.g. $x_2=x_1$ means the line lies in the plane $x=x_1$). Equivalently, in vector form — with no such restriction —

where

The equation of the straight line passing through the point $M_1(x_1,y_1,z_1)$ parallel to the vector $\mathbf{S}=l\mathbf{i}+m\mathbf{j}+n\mathbf{k}$ has the form

This is the canonical equation of the line — also called the symmetric form of the line; the nonzero vector $\mathbf{S}$ is called the direction vector of the line $L$. A zero denominator is read as the matching numerator being zero (that coordinate is then constant); the parametric form VI (c) expresses the same line with no assumption on the components.

The parametric equation of a straight line is

It is obtained from the canonical equation by introducing the parameter $t$: setting $\dfrac{x-x_1}{l}=\dfrac{y-y_1}{m}=\dfrac{z-z_1}{n}=t$ and solving each ratio for the corresponding coordinate. Unlike the canonical form, this holds for any direction vector $\mathbf{S}=(l,m,n)\neq\mathbf{0}$, including one with zero components.

A straight line in space can be defined by the equations of two planes:

This is the general equation of a straight line, provided the planes actually meet in a line, i.e. $\mathbf{n}_1\times\mathbf{n}_2\neq\mathbf{0}$ (parallel distinct planes have no common line, and coincident planes do not determine a unique one). To bring it to canonical form:

1) Find a point $M$ on the line by fixing the coordinate whose complementary $2\times2$ minor is nonzero — at least one such coordinate exists because $\mathbf{n}_1\times\mathbf{n}_2\neq\mathbf{0}$. Fixing $z=z_0$ works when $\begin{vmatrix} A_1 & B_1 \\ A_2 & B_2 \end{vmatrix}\neq0$; then solve the system for the remaining two coordinates:

2) Find the direction vector $\mathbf{S}$, parallel to the line ($\mathbf{S}\perp\mathbf{n}_1$ and $\mathbf{S}\perp\mathbf{n}_2$), as the vector product (cross product)

Hence $\dfrac{x-x_0}{l}=\dfrac{y-y_0}{m}=\dfrac{z-z_0}{n}$ — the canonical form of the line.

If a line is given by the canonical equation $\dfrac{x-x_0}{l}=\dfrac{y-y_0}{m}=\dfrac{z-z_0}{n}$, then the pair of equations

defines the same line as the intersection of two planes. This split assumes $l,m,n$ are all nonzero; if one vanishes, keep its constant-coordinate equation instead — e.g. $l=0$ gives $x=x_0$ together with $\dfrac{y-y_0}{m}=\dfrac{z-z_0}{n}$.

The angle between two lines in space, given by their canonical equations $\dfrac{x-x_1}{l_1}=\dfrac{y-y_1}{m_1}=\dfrac{z-z_1}{n_1}$ and $\dfrac{x-x_2}{l_2}=\dfrac{y-y_2}{m_2}=\dfrac{z-z_2}{n_2}$, is determined by the formula

with $\varphi$ the angle between the lines, $0^\circ\le\varphi\le90^\circ$, where $\mathbf{S}_1=(l_1,m_1,n_1)$ and $\mathbf{S}_2=(l_2,m_2,n_2)$ are the (nonzero) direction vectors. The lines are parallel when the direction vectors are proportional,

and perpendicular when

Two lines given by their canonical equations $\dfrac{x-x_1}{l_1}=\dfrac{y-y_1}{m_1}=\dfrac{z-z_1}{n_1}$ and $\dfrac{x-x_2}{l_2}=\dfrac{y-y_2}{m_2}=\dfrac{z-z_2}{n_2}$ are coplanar if and only if

If the numbers $l_1,m_1,n_1$ are not proportional to $l_2,m_2,n_2$, this coplanarity condition is the necessary and sufficient condition for the two lines to intersect.

The angle between a line $\dfrac{x-x_1}{l}=\dfrac{y-y_1}{m}=\dfrac{z-z_1}{n}$ and a plane $Ax+By+Cz+D=0$ is determined by the formula

with $\varphi$ the angle between the line and the plane, $0^\circ\le\varphi\le90^\circ$, where $\mathbf{S}=(l,m,n)$ and $\mathbf{n}=(A,B,C)$ are both nonzero. The direction of the line is parallel to the plane when $\mathbf{n}\cdot\mathbf{S}=0$, that is,

The line then either lies in the plane, if $Ax_1+By_1+Cz_1+D=0$, or is parallel to the plane and does not meet it, if $Ax_1+By_1+Cz_1+D\neq0$ (see VIII). The line is perpendicular to the plane when $\mathbf{S}$ and $\mathbf{n}$ are proportional,

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To find the point of intersection of a line $\dfrac{x-x_0}{l}=\dfrac{y-y_0}{m}=\dfrac{z-z_0}{n}$ with a plane $Ax+By+Cz+D=0$, write the line in parametric form

and substitute into the equation of the plane. From it we determine the parameter $t$:

Substituting this value back into the parametric equations gives the coordinates $x,y,z$ of the intersection point.

• If $Al+Bm+Cn\neq0$, the line intersects the plane.
• If $Al+Bm+Cn=0$ and $Ax_0+By_0+Cz_0+D\neq0$, the line is parallel to the plane.
• If $Al+Bm+Cn=0$ and $Ax_0+By_0+Cz_0+D=0$, the line lies in the plane.