---
title: "Vectors, Lines and Planes in Space"
book: "High School Mathematics"
subject: math
language: en
chapter: 31
exercises: 9
source: https://one-course.com/books/math/2/en/chapter/31-vectors-lines-and-planes-in-space
---

# Chapter 31 — Vectors, Lines and Planes in Space

Three-dimensional geometry becomes computational once [vectors](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) and [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) are available: lines get parametric representations, planes get Cartesian [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation), and the [scalar product](#def-g12-space-dot) measures angles and distances. This chapter builds this toolbox and uses it to solve the classical [intersection](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) and distance problems.

## 31.1 Vectors in space

[Vectors](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) in space obey the same rules as in the plane: they add, they scale, and $\vect{AB} = \vect{CD}$ exactly when $ABDC$ is a parallelogram. In a [coordinate system](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) $(O; \vec\imath, \vec\jmath, \vec k)$, every [vector](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) has [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) $(x, y, z)$.

**Definition 31.1 (Collinearity, coplanarity).**

Two [vectors](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) are *collinear* if one is a multiple of the other. Three [vectors](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) $\vec u, \vec v, \vec w$ are *coplanar* if one of them can be written as a combination of the other two, say $\vec w = a\vec u + b\vec v$.

**Proposition 31.2.**

If $\vec u, \vec v, \vec w$ are not [coplanar](#def-g12-space-collinear), every [vector](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) of space decomposes uniquely as $x\vec u + y\vec v + z\vec w$.

**Idea of proof.** Through the tip of the [vector](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) to decompose, draw the line parallel to $\vec w$; it meets the plane of $\vec u, \vec v$ in a single point, which splits the [vector](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) into a component in that plane (uniquely $x\vec u + y\vec
v$, plane geometry) and a component along $\vec w$. Uniqueness: a difference of two decompositions would express one of the [vectors](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) in terms of the other two, contradicting non-coplanarity. ∎

## 31.2 Lines and planes

**Definition 31.3 (Parametric representation of a line).**

The line through $A(x_A, y_A, z_A)$ with [direction vector](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-direction) $\vec u (a, b, c) \neq \vec 0$ is the set of points

$$
M(x_A + ta,\; y_A + tb,\; z_A + tc), \qquad t \in \R .
$$

**Definition 31.4 (Plane).**

The plane through $A$ directed by two [non-collinear](#def-g12-space-collinear) [vectors](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) $\vec u, \vec v$ is the set of points $M$ with $\vect{AM} = s\vec u + t\vec v$, $(s, t) \in \R^2$.

![Left: the line through A with direction u, graduated by the parameter t. Right: the plane through A directed by u and v; every point M of the plane is reached as AM = s u + t v.](https://one-course.com/images/onecourse/chapters/math-2/g12-space/fig-fe0897e01060.svg)

![Left: the line through A with direction u, graduated by the parameter t. Right: the plane through A directed by u and v; every point M of the plane is reached as AM = s u + t v.](https://one-course.com/images/onecourse/chapters/math-2/g12-space/fig-7463ec207c91.svg)

*Left: the line through $A$ with direction $\vec u$, graduated by the parameter $t$. Right: the plane through $A$ directed by $\vec u$ and $\vec v$; every point $M$ of the plane is reached as $\vect{AM} = s\vec u + t\vec v$.*

**Proposition 31.5 (Relative positions).**

Two distinct planes are either parallel or intersect in a line. A line and a plane are either parallel (possibly contained) or intersect in a single point. Two lines in space may be intersecting, strictly parallel, identical, or *skew* ([non-coplanar](#def-g12-space-collinear)).

**Sketch.** These are incidence statements; each case analysis reduces, in [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system), to solving a [linear system](https://one-course.com/books/math/2/en/chapter/7-equations-of-lines-and-linear-systems#def-g10-lines-system) and counting its solutions — see [Method 31.10](#met-g12-space-intersections). ∎

## 31.3 Scalar product

**Definition 31.6 (Scalar product).**

The *scalar product* of $\vec u$ and $\vec v$ is

$$
\vec u \cdot \vec v = \tfrac12\left(\norm{\vec u + \vec v}^2
- \norm{\vec u}^2 - \norm{\vec v}^2\right),
$$

where $\norm{\vec u}$ is the length of $\vec u$. If both [vectors](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) are nonzero, $\vec u \cdot \vec v = \norm{\vec u}\,\norm{\vec v}\cos\theta$ where $\theta$ is the angle between them; and in an *[orthonormal](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system)* [coordinate system](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system),

$$
\vec u \cdot \vec v = xx' + yy' + zz' .
$$

Two [vectors](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) are *[orthogonal](https://one-course.com/books/math/2/en/chapter/16-the-scalar-product-in-the-plane#def-g11-scal-orthogonal)* if their scalar product is $0$.

**Proposition 31.7 (Properties).**

The [scalar product](#def-g12-space-dot) is symmetric ($\vec u\cdot\vec v = \vec v\cdot\vec u$), bilinear ($(a\vec u + b\vec v)\cdot \vec w = a\,\vec u\cdot\vec w +
b\,\vec v\cdot \vec w$), and $\vec u \cdot \vec u = \norm{\vec u}^2 \geq 0$.

**Proof.** In an [orthonormal system](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system), all three properties are immediate on the formula $xx' + yy' + zz'$, which itself follows from the defining formula and the Pythagorean computation of lengths: $\norm{\vec u}^2 = x^2 + y^2 + z^2$. ∎

**Definition 31.8 (Normal vector).**

A *normal vector* of a plane $\mathcal P$ is a nonzero [vector](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) [orthogonal](https://one-course.com/books/math/2/en/chapter/16-the-scalar-product-in-the-plane#def-g11-scal-orthogonal) to every [vector](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) lying in $\mathcal P$ (it suffices that it be [orthogonal](https://one-course.com/books/math/2/en/chapter/16-the-scalar-product-in-the-plane#def-g11-scal-orthogonal) to two [non-collinear](#def-g12-space-collinear) directions of $\mathcal P$).

![A normal vector n of the plane P is orthogonal to every direction of P. The point H, foot of the perpendicular from M_0, realizes the distance from M_0 to the plane.](https://one-course.com/images/onecourse/chapters/math-2/g12-space/fig-c59e2488424e.svg)

*A [normal vector](#def-g12-space-normal) $\vec n$ of the plane $\mathcal P$ is [orthogonal](https://one-course.com/books/math/2/en/chapter/16-the-scalar-product-in-the-plane#def-g11-scal-orthogonal) to every direction of $\mathcal P$. The point $H$, foot of the perpendicular from $M_0$, realizes the distance from $M_0$ to the plane.*

**Theorem 31.9 (Cartesian equation of a plane).**

In an [orthonormal](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) [coordinate system](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system), the plane through $A$ with [normal vector](#def-g12-space-normal) $\vec n(a, b, c)$ has [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation)

$$
a x + b y + c z + d = 0,
$$

where $d = -(ax_A + by_A + cz_A)$. Conversely, every such [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) with $(a,b,c) \neq (0,0,0)$ defines a plane with [normal vector](#def-g12-space-normal) $(a, b, c)$.

**Proof.** $M \in \mathcal P \iff \vect{AM} \perp \vec n \iff
a(x - x_A) + b(y - y_A) + c(z - z_A) = 0$, which expands to the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation). Conversely, pick any solution point $A$ of the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation); the same computation backwards shows the solution set is $\{M : \vect{AM}\cdot\vec n = 0\}$, the plane through $A$ normal to $\vec n$. ∎

**Method 31.10 (Intersections in practice).**

- *Line $\cap$ plane* : substitute the parametric [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) of the line into the plane [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) ; solve for $t$ (one solution: one point; $0 =$ nonzero: parallel; $0 = 0$ : line contained).
- *Plane $\cap$ plane* : solve the system of the two [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) , parametrizing by one free coordinate; the solution is a line (or the planes are parallel when the [normal vectors](#def-g12-space-normal) are [collinear](#def-g12-space-collinear) ).
- *Orthogonality checks* : line $\perp$ plane iff its direction is [collinear](#def-g12-space-collinear) with the normal; two planes are perpendicular iff their normals are [orthogonal](https://one-course.com/books/math/2/en/chapter/16-the-scalar-product-in-the-plane#def-g11-scal-orthogonal) .

**Proposition 31.11 (Distance from a point to a plane).**

In an [orthonormal system](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system), the distance from $M_0(x_0, y_0, z_0)$ to the plane $\mathcal P : ax + by + cz + d = 0$ is

$$
\operatorname{dist}(M_0, \mathcal P)
= \frac{\abs{ax_0 + by_0 + cz_0 + d}}{\sqrt{a^2 + b^2 + c^2}}.
$$

**Proof.** Let $H$ be the [orthogonal](https://one-course.com/books/math/2/en/chapter/16-the-scalar-product-in-the-plane#def-g11-scal-orthogonal) projection of $M_0$ on $\mathcal P$: the foot of the perpendicular, so $\vect{HM_0}$ is [collinear](#def-g12-space-collinear) with the unit normal $\frac{\vec n}{\norm{\vec n}}$, and the distance is $\abs{\vect{HM_0}\cdot \frac{\vec n}{\norm{\vec n}}}$. For any $H \in \mathcal P$, $\vect{HM_0}\cdot\vec n = ax_0 + by_0 + cz_0 - (ax_H + by_H + cz_H)
= ax_0 + by_0 + cz_0 + d$ (using the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) at $H$), whence the formula after dividing by $\norm{\vec n} = \sqrt{a^2+b^2+c^2}$. ∎

**Example 31.12.**

Distance from the origin to the plane $x + 2y + 2z - 6 = 0$: $\frac{\abs{-6}}{\sqrt{1 + 4 + 4}} = \frac63 = 2$.

## 31.4 Exercises

In all exercises the [coordinate system](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) is [orthonormal](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system).

**Exercise 31.1 ★.**

Let $A(1, 0, 2)$, $B(3, 1, 1)$ and $C(2, -1, 3)$. Are the points $A$, $B$, $C$ aligned? Give a parametric representation of the line $(AB)$.

**Solution of Exercise 31.1.**

$\vect{AB}(2, 1, -1)$ and $\vect{AC}(1, -1, 1)$ are not proportional ($\frac21 \neq \frac{1}{-1}$), so the points are not aligned. Line $(AB)$:

$$
(x, y, z) = (1 + 2t,\; t,\; 2 - t), \qquad t \in \R .
$$

**Exercise 31.2 ★.**

Determine a Cartesian [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of the plane through $A(1, 1, 0)$ with [normal vector](#def-g12-space-normal) $\vec n(2, -1, 3)$. Does the point $B(0, 2, 1)$ belong to it?

**Solution of Exercise 31.2.**

$2(x - 1) - (y - 1) + 3(z - 0) = 0$, *i.e.*

$$
2x - y + 3z - 1 = 0 .
$$

For $B(0,2,1)$: $0 - 2 + 3 - 1 = 0$: yes, $B$ lies on the plane.

**Exercise 31.3 ★.**

Compute the angle at $A$ (to the nearest degree) of the triangle with vertices $A(0,0,0)$, $B(1, 1, 0)$ and $C(1, 0, 1)$.

**Solution of Exercise 31.3.**

$\vect{AB}(1,1,0)$, $\vect{AC}(1,0,1)$:

$$
\cos\widehat{A}
= \frac{\vect{AB}\cdot\vect{AC}}{\norm{\vect{AB}}\,\norm{\vect{AC}}}
= \frac{1}{\sqrt2\,\sqrt2} = \frac12,
$$

so $\widehat A = 60^\circ$.

**Exercise 31.4 ★★.**

Consider the line $\mathcal D$ through $A(1, 2, 0)$ with direction $\vec u(1, -1, 2)$, and the plane $\mathcal P : 2x + y - z + 1 = 0$. Determine $\mathcal D \cap \mathcal P$.

**Solution of Exercise 31.4.**

Points of $\mathcal D$: $(1 + t,\, 2 - t,\, 2t)$. Substituting into the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of $\mathcal P$:

$$
2(1+t) + (2 - t) - 2t + 1 = 5 - t = 0 \iff t = 5 .
$$

Unique [intersection](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) point: $(6, -3, 10)$.

**Exercise 31.5 ★★.**

Let $\mathcal P : x - y + 2z = 1$ and $\mathcal Q : 2x + y + z = 4$. Show that $\mathcal P$ and $\mathcal Q$ intersect in a line and give a parametric representation of it.

**Solution of Exercise 31.5.**

The [normal vectors](#def-g12-space-normal) $(1, -1, 2)$ and $(2, 1, 1)$ are not [collinear](#def-g12-space-collinear), so the planes intersect in a line. Adding the two [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation): $3x + 3z = 5$, so $x = \frac53 - z$; then the first [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) gives $y = x + 2z - 1 = \frac23 + z$. With $z = t$:

$$
(x, y, z) = \left(\frac53 - t,\; \frac23 + t,\; t\right), \qquad t \in \R,
$$

a line with [direction vector](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-direction) $(-1, 1, 1)$. (Both [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) check identically.)

**Exercise 31.6 ★★.**

Show that the lines

$$
\mathcal D_1 : (x, y, z) = (1 + t,\; 2t,\; 3 - t), \qquad
\mathcal D_2 : (x, y, z) = (2 + s,\; 1 + s,\; 1 + 2s)
$$

are skew (not [coplanar](#def-g12-space-collinear)).

**Solution of Exercise 31.6.**

Directions $\vec u_1(1, 2, -1)$ and $\vec u_2(1, 1, 2)$ are not [collinear](#def-g12-space-collinear), so the lines are not parallel. Intersecting would require

$$
1 + t = 2 + s, \qquad 2t = 1 + s, \qquad 3 - t = 1 + 2s .
$$

The first two give $t = 1 + s$ and $2(1+s) = 1 + s$, so $s = -1$, $t = 0$; but then the third reads $3 = -1$, absurd. No common point and not parallel: the lines are skew.

**Exercise 31.7 ★★.**

Let $\mathcal P : 2x - y + 2z - 3 = 0$.

1. Compute the distance from $M_0(3, 1, 1)$ to $\mathcal P$ .
2. Determine the [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) of the [orthogonal](https://one-course.com/books/math/2/en/chapter/16-the-scalar-product-in-the-plane#def-g11-scal-orthogonal) projection $H$ of $M_0$ on $\mathcal P$ , and check the distance $M_0H$ .

**Solution of Exercise 31.7.**

*1.* $\operatorname{dist} = \dfrac{\abs{2\cdot3 - 1 + 2\cdot1 - 3}}
{\sqrt{4 + 1 + 4}} = \dfrac{4}{3}$.

*2.* $H = M_0 + t\,\vec n$ with $\vec n(2, -1, 2)$, choosing $t$ so that $H \in \mathcal P$:

$$
2(3 + 2t) - (1 - t) + 2(1 + 2t) - 3 = 4 + 9t = 0 \iff t = -\frac49 .
$$

Hence $H\left(\frac{19}{9}, \frac{13}{9}, \frac19\right)$, and

$$
M_0H = \norm{t\,\vec n} = \frac49 \times 3 = \frac43,
$$

matching the distance formula.

**Exercise 31.8 ★★★.**

The cube $ABCDEFGH$ has side $1$; place [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) so that $A(0,0,0)$, $B(1,0,0)$, $D(0,1,0)$, $E(0,0,1)$ (with $C = B + D$, $F = B + E$, $G = B + D + E$, $H = D + E$ as [vectors](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) from $A$).

1. Show that the diagonal $(AG)$ is [orthogonal](https://one-course.com/books/math/2/en/chapter/16-the-scalar-product-in-the-plane#def-g11-scal-orthogonal) to the plane $(BDE)$ .
2. Compute the distance from $A$ to the plane $(BDE)$ , and the point where $(AG)$ pierces it.

**Solution of Exercise 31.8.**

[Coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system): $G(1,1,1)$, and the plane $(BDE)$ passes through $B(1,0,0)$, $D(0,1,0)$, $E(0,0,1)$.

*1.* The plane $(BDE)$ has [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $x + y + z = 1$ (satisfied by the three points, and they are not aligned), so $\vec n(1,1,1)$ is normal to it. Since $\vect{AG} = (1,1,1) = \vec n$, the diagonal $(AG)$ is [orthogonal](https://one-course.com/books/math/2/en/chapter/16-the-scalar-product-in-the-plane#def-g11-scal-orthogonal) to the plane $(BDE)$.

*2.* Distance from $A(0,0,0)$: $\frac{\abs{0 + 0 + 0 - 1}}{\sqrt3} = \frac{1}{\sqrt3} = \frac{\sqrt3}{3}$. The line $(AG)$ is $(t, t, t)$; it meets the plane when $3t = 1$: at the point $\left(\frac13, \frac13, \frac13\right)$, the centroid of the triangle $BDE$.

**Exercise 31.9 ★★★.**

*(Volume of a tetrahedron.)* Let $A(1,1,1)$, $B(2,1,0)$, $C(0,2,1)$ and $D(2,2,2)$.

1. Verify that $\vect{AB}$ , $\vect{AC}$ , $\vect{AD}$ are not [coplanar](#def-g12-space-collinear) (the points form a genuine tetrahedron).
2. Find a [vector](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) $\vec n$ [orthogonal](https://one-course.com/books/math/2/en/chapter/16-the-scalar-product-in-the-plane#def-g11-scal-orthogonal) to both $\vect{BC}$ and $\vect{BD}$ , and deduce a Cartesian [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of the plane $(BCD)$ .
3. Compute the distance from $A$ to the plane $(BCD)$ and the area of the triangle $BCD$ , using the formula $\text{Area} = \frac12\sqrt{\norm{\vec u}^2\norm{\vec v}^2 -  (\vec u \cdot \vec v)^2}$ for the triangle spanned by $\vec u, \vec v$ .
4. Deduce the volume of the tetrahedron ( $V = \frac13 \times \text{base area} \times \text{height}$ ).

**Solution of Exercise 31.9.**

*1.* $\vect{AB}(1, 0, -1)$, $\vect{AC}(-1, 1, 0)$, $\vect{AD}(1, 1, 1)$. If $\vect{AD} = a\vect{AB} + b\vect{AC}$, the [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) give $a - b = 1$, $b = 1$, $-a = 1$: the first two force $a = 2$, contradicting $a = -1$. Not [coplanar](#def-g12-space-collinear).

*2.* $\vect{BC}(-2, 1, 1)$, $\vect{BD}(0, 1, 2)$. Solving $\vec n \cdot \vect{BC} = \vec n \cdot \vect{BD} = 0$: from $b + 2c = 0$, $b = -2c$; then $-2a - 2c + c = 0$ gives $c = -2a$. With $a = 1$: $\vec n(1, 4, -2)$. Plane through $B(2,1,0)$:

$$
x + 4y - 2z - 6 = 0 .
$$

*3.* Distance from $A(1,1,1)$: $\dfrac{\abs{1 + 4 - 2 - 6}}{\sqrt{21}} = \dfrac{3}{\sqrt{21}}$. Area of $BCD$: $\norm{\vect{BC}}^2 = 6$, $\norm{\vect{BD}}^2 = 5$, $\vect{BC}\cdot\vect{BD} = 3$, so

$$
\text{Area} = \frac12\sqrt{6 \times 5 - 9} = \frac{\sqrt{21}}{2}.
$$

*4.*

$$
V = \frac13 \times \frac{\sqrt{21}}{2} \times \frac{3}{\sqrt{21}}
= \frac12 .
$$

## 31.5 Problem: Slicing the cube

**Problem 31.1.**

Weekend problem — a plane cuts a cube into a perfect hexagon, a regular tetrahedron hides in the corners, and the diagonal’s angle shapes every diamond

Slice a cube and you expect squares and rectangles — yet one famous cut produces a *regular hexagon*, and four clever corner cuts leave behind a *regular tetrahedron*. Both surprises, and the $109.5^\circ$ angle that carbon atoms worship, fall to this chapter’s toolkit: parametric lines, plane [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) and the [scalar product](#def-g12-space-dot) ([Theorem 31.9](#thm-g12-space-planeeq), [Proposition 31.11](#prop-g12-space-distance)). Use throughout the cube of [Exercise 31.8](#exo-g12-space-8): side $1$, $A(0,0,0)$, $B(1,0,0)$, $D(0,1,0)$, $E(0,0,1)$, $C(1,1,0)$, $F(1,0,1)$, $H(0,1,1)$, $G(1,1,1)$.

**Part I — Fluency.**

1. Write the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of the plane through $B(1,0,0)$ with [normal vector](#def-g12-space-normal) $(1,1,1)$ , a parametric representation of the line through $A$ directed by $(1,1,1)$ , and their [intersection](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) point.
2. Compute the distance from $A$ to that plane ( [Proposition 31.11](#prop-g12-space-distance) ).
3. Are the [vectors](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) $(1,-1,0)$ and $(1,1,-2)$ [orthogonal](https://one-course.com/books/math/2/en/chapter/16-the-scalar-product-in-the-plane#def-g11-scal-orthogonal) ?
4. Position of the planes $x + y + z = 1$ and $2x + 2y + 2z = 5$ ? Of the line through $A$ directed by $(1,1,0)$ relative to the first plane?
5. List the [midpoints](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of the six edges $[BC], [CD], [DH], [HE], [EF], [FB]$ of the cube, and verify that all six satisfy the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $x + y + z = \frac32$ .

**Part II — The hexagonal section.** Let $P$ be the plane $x + y + z = \frac32$.

6. Show that $P$ is perpendicular to the great diagonal $(AG)$ and passes through the cube’s center.
7. Compute the distances between consecutive [midpoints](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of question 5 around the section: what do you find?
8. Compute the angle of the section’s polygon at one vertex (two consecutive [edge-vectors](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) and a dot product). Conclude: the section is a *regular hexagon* .
9. Compute the hexagon’s perimeter and area, and compare the area with that of a face of the cube. (For the record: the largest planar section of the unit cube is the $1 \times \sqrt2$ diagonal rectangle, of area $\sqrt2$ — the regular hexagon takes silver.)
10. Where does the diagonal $(AG)$ pierce $P$ ? Verify that the piercing point is the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $[AG]$ .
11. Slide the cut: describe the sections $x + y + z = c$ as $c$ grows from $0$ to $3$ — compute the section for $c = \frac12$ (which polygon, which size?), and narrate the morphing that passes through your hexagon at $c = \frac32$ .
12. Why can no plane cut a cube in a polygon with *seven* sides? (Where must each side of the section lie?)

**Part III — The hidden tetrahedron.**

13. Compute the volume of the corner tetrahedron $ABDE$ (base, height, and the one-third formula the Middle School volume admitted — [Exercise 31.9](#exo-g12-space-9) computes such volumes in general).
14. The four vertices $B$ , $D$ , $E$ , $G$ : compute all six pairwise distances and conclude that $BDEG$ is a *regular* tetrahedron. Then obtain its volume by subtracting corner tetrahedra from the cube.
15. Compute the distance from the cube’s center to the plane $(BDE)$ .
16. The diamond angle: compute the angle between the diagonals $\vect{AG}$ and $\vect{BH}$ of the cube. Its supplement, about $109.5^\circ$ , is the angle between bonds in diamond and methane — why do four electron pairs around a carbon atom choose the corners of question 14’s tetrahedron?

**Part IV — Genuinely three-dimensional.**

17. Find where the line through $B(1,0,0)$ directed by $(0,1,1)$ pierces the hexagon’s plane $P$ .
18. Show that the lines $(AB)$ and $(EG)$ neither meet nor are parallel: *skew* lines, the phenomenon that cannot happen in a plane.
19. Compute the [orthogonal](https://one-course.com/books/math/2/en/chapter/16-the-scalar-product-in-the-plane#def-g11-scal-orthogonal) projection of $B$ onto $P$ (walk from $B$ along the normal until the plane), and check your point satisfies $P$ ’s [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) .
20. Finale — the space surveyor’s kit: parametric lines for piercing, [normal vectors](#def-g12-space-normal) for angles and distances, plane [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) for sections; and the cube as the playground where they produced a hexagon, a regular tetrahedron, the diamond angle and a pair of skew lines. One sentence each.

**Solution of Problem 31.1.**

**1.** Plane: $x + y + z = 1$. Line: $(t, t, t)$. [Intersection](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion): $3t = 1$: $\left(\frac13, \frac13, \frac13\right)$.

**2.** $\dfrac{\abs{0 + 0 + 0 - 1}}{\sqrt3} =
\dfrac{1}{\sqrt3} = \dfrac{\sqrt3}{3}$.

**3.** $1 - 1 + 0 = 0$: [orthogonal](https://one-course.com/books/math/2/en/chapter/16-the-scalar-product-in-the-plane#def-g11-scal-orthogonal).

**4.** Same normal $(1,1,1)$, but $\frac52 \neq 1$: strictly parallel planes. The line: direction $(1,1,0) \cdot (1,1,1) = 2 \neq 0$: not parallel to the plane, so it pierces it at one point.

**5.** [Midpoints](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint): $\left(1, \frac12, 0\right)$, $\left(\frac12, 1, 0\right)$, $\left(0, 1, \frac12\right)$, $\left(0, \frac12, 1\right)$, $\left(\frac12, 0, 1\right)$, $\left(1, 0, \frac12\right)$: each has coordinate sum $\frac32$.

**6.** $P$’s normal is $(1,1,1)$, the direction of $(AG)$: perpendicular. The center $\left(\frac12,\frac12,\frac12\right)$ has coordinate sum $\frac32$: on $P$.

**7.** Consecutive [midpoints](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) differ by [vectors](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) like $\left(-\frac12, \frac12, 0\right)$: length $\frac{\sqrt2}{2}$ each time — six equal sides.

**8.** At $\left(\frac12, 1, 0\right)$: [edge-vectors](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) $\left(\frac12, -\frac12, 0\right)$ and $\left(-\frac12, 0, \frac12\right)$: dot product $-\frac14$, [norms](https://one-course.com/books/math/2/en/chapter/16-the-scalar-product-in-the-plane#def-g11-scal-dot) $\frac{\sqrt2}{2}$: $\cos = -\frac12$: angle $120^\circ$. Equal sides, equal $120^\circ$ angles: a regular hexagon.

**9.** Perimeter $6 \times \frac{\sqrt2}{2} = 3\sqrt2
\approx 4.24$. Area $6 \times \frac{\sqrt3}{4} \times
\frac12 = \frac{3\sqrt3}{4} \approx 1.30$: larger than a face (area $1$) — a slice bigger than any side of the box; only the diagonal rectangle ($\sqrt2 \approx 1.41$) beats it.

**10.** $(t,t,t)$ meets $P$ at $t = \frac12$: the center — which is the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $[AG]$ ($A$ at $t = 0$, $G$ at $t = 1$).

**11.** For $c = \frac12$: the plane cuts the three edges at $A$’s corner, at $\left(\frac12,0,0\right)$, $\left(0,\frac12,0\right)$, $\left(0,0,\frac12\right)$: an equilateral triangle of side $\frac{\sqrt2}{2}$. As $c$ grows, the triangle grows, its corners get truncated into a hexagon (regular exactly at $c = \frac32$), then the picture shrinks symmetrically back to a triangle at $G$’s corner: the crystal-cut morphing.

**12.** Each side of the section is the [intersection](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) of the cutting plane with one *face* of the cube, and a plane meets a face (a flat polygon) in at most one segment: at most $6$ sides. Seven is impossible; three through six all occur (questions 8 and 11 show two of them).

**13.** Base $ABD$: right triangle of area $\frac12$; height $AE = 1$: volume $\frac13 \times \frac12 \times 1 = \frac16$.

**14.** Every pair among $B, D, E, G$ differs in exactly two [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) by $1$: all six distances equal $\sqrt2$ — regular. The cube splits into $BDEG$ plus four corner tetrahedra congruent to $ABDE$: volume $1 - 4 \times \frac16 = \frac13$.

**15.** Plane $(BDE)$: $x + y + z = 1$; center $\left(\frac12,\frac12,\frac12\right)$: distance $\frac{\abs{\frac32 - 1}}{\sqrt3} = \frac{1}{2\sqrt3} =
\frac{\sqrt3}{6}$.

**16.** $\vect{AG} = (1,1,1)$, $\vect{BH} = (-1,1,1)$: $\cos\theta = \frac{-1 + 1 + 1}{3} =
\frac13$: $\theta \approx 70.5^\circ$, supplement $\approx 109.5^\circ$. Four electron pairs repel one another and spread as far apart as possible around the carbon: the optimum is the regular tetrahedron’s corners, whose center-to-vertex directions meet at exactly this angle — diamond is question 14 crystallized.

**17.** $(1, t, t)$ on $P$: $1 + 2t = \frac32$: $t = \frac14$: the point $\left(1, \frac14, \frac14\right)$.

**18.** $(AB)$: points $(t, 0, 0)$; $(EG)$: points $(s, s, 1)$. Meeting would need $0 = 1$ in the third coordinate: never. Parallel would need $(1,0,0)$ and $(1,1,0)$ [collinear](#def-g12-space-collinear): no. Neither meeting nor parallel: skew — two corridors on different floors.

**19.** $B + t(1,1,1) = (1 + t, t, t)$ on $P$: $1 + 3t = \frac32$: $t = \frac16$: projection $\left(\frac76, \frac16, \frac16\right)$, whose coordinate sum is $\frac32$: on $P$, as required.

**20.** Parametric lines pierce (questions 1, 10, 17); [normal vectors](#def-g12-space-normal) measure angles and distances (questions 2, 8, 15, 16); plane [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) carve sections (the triangle-to- hexagon morphing). And the playground repaid the visit: a regular hexagon in a box of squares, a regular tetrahedron in the corners, the angle of diamond, and two lines that ignore each other — geometry only space can offer.