1.2 Lines, planes and volumes

Describe a set by its allowed directions

A line can be described by the points we can reach from one starting point while moving along one direction. A plane allows two independent directions. Equations offer a complementary description: they constrain which points are allowed. Moving between these two viewpoints is the main task of affine geometry. It also prepares the language of tangent lines and tangent planes.

The starting point describes location, while the direction vectors describe how the set extends. These are separate kinds of data. A line through the origin and a translated parallel line have the same directions, but are different sets of positions. Keeping the two roles separate makes it easier to find intersections, recognize equivalent representations and formulate distance problems.

Definition

Lines and hyperplanes

For $P_0\in\mathbb R^n$ and a nonzero vector $v$, the line through $P_0$ in direction $v$ is

$$L=\{P_0+tv:t\in\mathbb R\}.$$

For a nonzero vector $n\in\mathbb R^n$ and $c\in\mathbb R$, the hyperplane with normal $n$ is

$$H=\{x\in\mathbb R^n:\langle n,x\rangle=c\}.$$

If $P_0\in H$, its equation is also $\langle n,x-P_0\rangle=0$. A hyperplane has dimension $n-1$: it is a line in $\mathbb R^2$ and a plane in $\mathbb R^3$.

For two distinct points, choose the direction as their difference. The segment between them restricts the parameter to the interval from zero to one. A different starting point on the same line gives the same set after shifting the parameter; multiplying the direction by a nonzero scalar merely changes the rate or orientation of traversal. A parametrization is therefore a description of a line, rather than the line itself.

Worked example

Convert a line without dividing by zero

The line through $A=(2,-1,1)$ and $B=(4,0,1)$ has direction $(2,1,0)$. Thus

$$(x,y,z)=(2,-1,1)+t(2,1,0).$$

Eliminating $t$ gives $x-2y=4$ and $z=1$. Both equations are needed in three-dimensional space. A suitable symmetric description is $(x-2)/2=y+1$, together with $z=1$. The zero third direction component means that the third coordinate stays constant.

A perpendicular displacement gives the nearest point

The nearest-point problem asks which allowed position has the smallest distance from a given position. For a line through $P_0$, project $P-P_0$ onto the line direction to locate the nearest point; the remaining perpendicular component measures the distance. For a hyperplane, subtract the normal component from $P$ to locate its nearest point. In both cases an orthogonal decomposition supplies a proof of minimality, not merely a plausible candidate.

Theorem

Nearest point on a line

For $L=P_0+\mathbb Rv$ with $v\ne0$, put $w=P-P_0$. The unique closest point to $P$ is

$$Q=P_0+\frac{\langle w,v\rangle}{\|v\|^2}v, \qquad \operatorname{dist}(P,L)=\|P-Q\|.$$

Proof

Why no other line point is closer

The construction gives $P-Q\perp v$. Any other point on $L$ is $Q+sv$. Therefore

$$\|P-(Q+sv)\|^2=\|P-Q\|^2+s^2\|v\|^2.$$

The second term is nonnegative and vanishes only for $s=0$. Thus $Q$ is the unique minimum.

Worked example

A closest point on a spatial line

Let $P=(3,2,4)$ and $L=(1,0,1)+t(1,2,2)$. Then $w=(2,2,3)$, its dot product with the direction is $12$, and the squared direction length is $9$. Hence

$$Q=(1,0,1)+\tfrac43(1,2,2)=(7/3,8/3,11/3).$$

The residual is $(2/3,-2/3,1/3)$, which has dot product zero with $(1,2,2)$ and length $1$. Thus the distance is $1$. Checking membership of $Q$ and orthogonality of the residual verifies both the location and the minimizing property.

Theorem

Nearest point on a hyperplane

For $H=\{x:\langle n,x\rangle=c\}$ with $n\ne0$, the closest point to $P$ and the distance are

$$Q=P-\frac{\langle n,P\rangle-c}{\|n\|^2}n, \qquad \operatorname{dist}(P,H)=\frac{|\langle n,P\rangle-c|}{\|n\|}.$$

Proof

Separate normal and tangential directions

Substitution gives $\langle n,Q\rangle=c$, so $Q\in H$. The displacement $P-Q$ is parallel to $n$. For any $R\in H$, the difference $Q-R$ is orthogonal to $n$, since its inner product with $n$ is $c-c=0$. Consequently

$$\|P-R\|^2=\|P-Q\|^2+\|Q-R\|^2.$$

This is smallest only when $R=Q$. Taking the norm of the normal displacement gives the distance formula.

Worked example

Foot of a perpendicular to a plane

For $H:x+2y-2z=4$, choose $n=(1,2,-2)$, whose norm is $3$. If $P=(1,2,-1)$, the signed residual in the plane equation is $7-4=3$. Therefore

$$Q=P-\tfrac13 n=(2/3,4/3,-1/3), \qquad \operatorname{dist}(P,H)=1.$$

Substituting $Q$ gives $2/3+8/3+2/3=4$. The sign of the residual selects which normal direction to subtract; the distance takes its absolute value.

Cross products, area and a normal direction

In three dimensions, two independent direction vectors determine a normal direction. The cross product constructs that normal while also measuring the area spanned by the two directions. It is a vector operation, whereas the inner product is scalar-valued. The order of the inputs matters because the normal has an orientation.

Definition

Cross product in three dimensions

For $u,v\in\mathbb R^3$,

$$u\times v=(u_2v_3-u_3v_2,\ u_3v_1-u_1v_3,\ u_1v_2-u_2v_1).$$

It is perpendicular to both inputs, is bilinear, and satisfies $v\times u=-(u\times v)$. Its orientation follows the right-hand rule.

Expanding the coordinate formula proves

$$\|u\times v\|^2=\|u\|^2\|v\|^2-\langle u,v\rangle^2.$$

For nonzero inputs this equals $\|u\|^2\|v\|^2\sin^2\theta$. Since the sine of an angle in $[0,\pi]$ is nonnegative, the cross-product length is base times perpendicular height. It is therefore the area of the parallelogram with adjacent sides $u,v$. A triangle with those two sides has half this area. If the inputs are dependent, the cross product is zero, correctly recording that the parallelogram has collapsed.

Given three noncollinear points, subtract one of them from the other two to form coinitial side vectors. Their cross product is nonzero and supplies a normal to the unique plane through the points. Subtracting the same base point is important: coordinate position vectors from the origin need not be directions lying in that plane.

Worked example

One computation supplies a plane, area and volume

Take $A=(0,1,0)$, $B=(2,1,1)$ and $C=(1,3,0)$. Their side vectors are $u=(2,0,1)$ and $v=(1,2,0)$, so

$$u\times v=(-2,1,4).$$

Its dot products with $u$ and $v$ are both zero. The triangle area is $\sqrt{21}/2$, and the plane equation is $-2x+y+4z=1$. If $D=(1,0,3)$, then $D-A=(1,-1,3)$ and

$$\langle u\times v,D-A\rangle=-2-1+12=9.$$

The parallelepiped volume is $9$ and the tetrahedron volume is $9/6=3/2$.

The scalar triple product is the determinant whose rows are the three edge vectors. Its magnitude is base area times perpendicular height: the cross product gives the oriented base normal, and the inner product extracts the normal component of the third edge. A tetrahedron has one third of the triangular base area times height, hence one sixth of the parallelepiped volume. Zero triple product means that the three edge vectors are dependent, so the four endpoint positions are coplanar. Reversing two edges changes the sign but preserves the geometric volume.

Intersections and relative position

To intersect a line with a hyperplane, substitute its parameter formula into the equation. This gives a single linear equation in the parameter. If the direction has a nonzero inner product with the normal, there is one intersection. If that inner product is zero, the equation has either no solution or every parameter as a solution. In the latter case the line lies in the hyperplane. A parallel direction alone therefore does not decide whether the sets meet; the starting position must also be checked.

The acute angle between a line direction and a plane is complementary to the acute angle between that direction and the normal. Thus its sine is the absolute dot product divided by the product of the lengths. Two nonparallel planes meet in a line whose direction is the cross product of their normals. To locate that line, solve their two equations for one point, then attach the direction vector. When normals are proportional, compare the constants in the same proportion to distinguish coincident planes from disjoint parallel planes.

Two lines in space require a further distinction. Nonparallel lines can fail to intersect because they are skew. Equating their parametrizations produces three coordinate equations in two parameters, and all three must hold at once. Solving just two equations can give a candidate that fails the third. Parallelism, intersection and coplanarity are related questions, but none should be substituted for another without checking the relevant equations.

Higher-dimensional affine sets

Definition

An affine plane of dimension k

If $v_1,\ldots,v_k$ are linearly independent in $\mathbb R^n$, then

$$P=P_0+\operatorname{span}\{v_1,\ldots,v_k\} =\left\{P_0+\sum_{i=1}^k t_i v_i:t_i\in\mathbb R\right\}$$

is an affine $k$-plane.

The number of independent directions is the dimension, regardless of how many coordinates describe the surrounding space. A plane in four-dimensional space still needs only two independent parameters. In an equation description, a consistent linear system with rank equal to the ambient dimension minus two describes such a plane. Redundant equations do not remove further directions, so counting written equations without checking independence can give the wrong dimension.

Gaussian elimination converts equations into a parametrization by identifying free variables. Conversely, eliminating the parameters gives equations satisfied by every point in the affine set. An inconsistent system describes the empty set, not an affine plane. For a consistent system, the solution set is one particular solution plus the null space of the coefficient matrix. The particular solution locates the set; a basis of the null space supplies its independent directions. This is the algebraic form of the location-and-directions viewpoint with which we began.

A geometry checklist for calculations

Distance formulas must also be independent of harmless changes in representation. Multiplying a plane equation by a nonzero real number changes both its normal and its constant. The numerator in the distance formula acquires the absolute value of that factor, as does the denominator, so the distance stays unchanged. The foot-point formula is unchanged as well. This matters when two calculations use normals of different lengths or opposite signs: their signed residuals can differ, but the actual foot point and distance must agree.

For a line in three dimensions, there is an alternative distance calculation. The magnitude of the cross product of its direction with the displacement from a point on the line is the area of a parallelogram. Dividing that area by the direction's length gives its perpendicular height, which is exactly the point-to-line distance. The projection formula works in every dimension; the cross-product version is convenient in three dimensions when the area calculation is short. Both express the same geometric height and should give the same result.

When a distance question involves two sets, look for a displacement perpendicular to the allowed directions of both. For two skew lines, the cross product of their independent directions supplies a possible common perpendicular direction. Finding points on the two lines whose difference has that direction leads to a linear system for their parameters. The logic is the same as in the point-to-line problem, but now both endpoints can move. A final orthogonality check against both line directions is what confirms that the segment is the common perpendicular, rather than merely a segment joining the lines.

These representation checks help distinguish arithmetic from geometry. A computation may be written with large coefficients because of an inconvenient choice of normal or direction. Rescaling can simplify the arithmetic without changing the set. Choose smaller coefficients when possible, then verify the resulting point against the original equation rather than relying only on the transformed form.

Start by identifying whether the data describe positions, directions or normals. A normal is perpendicular to a plane's directions, whereas a direction vector lies along the object. Using one where the other is needed reverses the intended geometry. Next choose parameters when you want to generate points, and equations when you want to impose constraints or test membership. For distance questions, first construct a candidate and then prove the residual is perpendicular to every allowed displacement.

Finally, match the output to the question. A plane equation is not a point; a cross product is not an area until its norm is taken; a determinant is a signed volume until its absolute value is taken. Equivalent parametrizations may look different but generate the same set. To compare them, check one common point and the direction space. These checks also reveal why a correct numerical distance is only part of a nearest-point problem when the point itself is requested.

Common mistake

Count constraints in the ambient space

One nontrivial linear equation in three variables describes a plane. A spatial line generally needs two independent equations. A parameter formula describes a line only when its direction is nonzero; two plane parameters describe a plane only when their directions are independent.

Exercises and solutions

Checkpoint

1. Construct a plane and a triangle area.

Find the plane through $(1,0,0)$, $(1,2,0)$ and $(1,0,3)$, and the area of the triangle they form.

Solution · 1. Form sides from one vertex

The side vectors are $(0,2,0)$ and $(0,0,3)$; their cross product is $(6,0,0)$. The plane is $x=1$ and the area is $6/2=3$.

Checkpoint

2. Find a distance and a foot point.

Find the closest point to the origin on $2x-y+2z=6$ and its distance from the origin.

Solution · 2. Move along the normal

The normal is $(2,-1,2)$ and has squared length $9$. The closest point is $(6/9)(2,-1,2)=(4/3,-2/3,4/3)$, whose length is $2$. Substitution into the plane equation gives $6$.

Checkpoint

3. Decide whether two nonparallel lines intersect.

Compare $L=(t,0,0)$ and $M=(0,1+s,1)$ for real parameters $s,t$.

Solution · 3. The third coordinate prevents intersection

Their directions are $(1,0,0)$ and $(0,1,0)$, so they are not parallel. Every point of $L$ has third coordinate $0$, whereas every point of $M$ has third coordinate $1$. They do not intersect and are skew.