High School Mathematics · Grades 10–12

31Vectors, Lines and Planes in Space

Three-dimensional geometry becomes computational once vectors and coordinates are available: lines get parametric representations, planes get Cartesian equations, and the scalar product measures angles and distances. This chapter builds this toolbox and uses it to solve the classical intersection and distance problems.

31.1 Vectors in space

Vectors 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 $(O; \vec\imath, \vec\jmath, \vec k)$ , every vector has coordinates $(x, y, z)$ .

Definition 31.1 (Collinearity, coplanarity)

Two vectors are collinear if one is a multiple of the other. Three vectors $\vec u, \vec v, \vec w$ 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, every vector of space decomposes uniquely as $x\vec u + y\vec v + z\vec w$ .

Idea of proof. Through the tip of the 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 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 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 $\vec u (a, b, c) \neq \vec 0$ 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 vectors $\vec u, \vec v$ 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. — 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).

Sketch. These are incidence statements; each case analysis reduces, in coordinates, to solving a linear system and counting its solutions — see Method 31.10. ∎

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 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 coordinate system,

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

Two vectors are orthogonal if their scalar product is $0$ .

Proposition 31.7 (Properties)

The scalar product 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, 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 orthogonal to every vector lying in $\mathcal P$ (it suffices that it be orthogonal to two non-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. — A normal vector $\vec n$ of the plane $\mathcal P$ is 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 coordinate system, the plane through $A$ with normal vector $\vec n(a, b, c)$ has equation

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

where $d = -(ax_A + by_A + cz_A)$ . Conversely, every such equation with $(a,b,c) \neq (0,0,0)$ defines a plane with normal vector $(a, b, c)$ .

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. Conversely, pick any solution point $A$ of the 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 of the line into the plane 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, parametrizing by one free coordinate; the solution is a line (or the planes are parallel when the normal vectors are collinear).
Orthogonality checks: line $\perp$ plane iff its direction is collinear with the normal; two planes are perpendicular iff their normals are orthogonal.

Proposition 31.11 (Distance from a point to a plane)

In an orthonormal 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 projection of $M_0$ on $\mathcal P$ : the foot of the perpendicular, so $\vect{HM_0}$ is 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 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 is orthonormal.

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

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 of the plane through $A(1, 1, 0)$ with normal vector $\vec n(2, -1, 3)$ . Does the point $B(0, 2, 1)$ belong to it?

Solution

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

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

Solution of Exercise 31.4.

Points of $\mathcal D$ : $(1 + t,\, 2 - t,\, 2t)$ . Substituting into the equation of $\mathcal P$ :

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

Unique intersection 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

Solution of Exercise 31.5.

The normal vectors $(1, -1, 2)$ and $(2, 1, 1)$ are not collinear, so the planes intersect in a line. Adding the two equations: $3x + 3z = 5$ , so $x = \frac53 - z$ ; then the first 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 $(-1, 1, 1)$ . (Both equations 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).

Solution

Solution of Exercise 31.6.

Directions $\vec u_1(1, 2, -1)$ and $\vec u_2(1, 1, 2)$ are not 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 $\mathcal P : 2x - y + 2z - 3 = 0$ .

Compute the distance from $M_0(3, 1, 1)$ to $\mathcal P$ .
Determine the coordinates of the orthogonal projection $H$ of $M_0$ on $\mathcal P$ , and check the distance $M_0H$ .

Solution

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 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 from $A$ ).

Show that the diagonal $(AG)$ is orthogonal to the plane $(BDE)$ .
Compute the distance from $A$ to the plane $(BDE)$ , and the point where $(AG)$ pierces it.

Solution

Solution of Exercise 31.8.

Coordinates: $G(1,1,1)$ , and the plane $(BDE)$ passes through $B(1,0,0)$ , $D(0,1,0)$ , $E(0,0,1)$ .

1. The plane $(BDE)$ has equation $x + y + z = 1$ (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 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 $D(2,2,2)$ .

Verify that $\vect{AB}$ , $\vect{AC}$ , $\vect{AD}$ are not coplanar (the points form a genuine tetrahedron).
Find a vector $\vec n$ orthogonal to both $\vect{BC}$ and $\vect{BD}$ , and deduce a Cartesian equation of the plane $(BCD)$ .
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 $\vec u, \vec v$ .
Deduce the volume of the tetrahedron ( $V = \frac13 \times \text{base area} \times \text{height}$ ).

Solution

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 give $a - b = 1$ , $b = 1$ , $-a = 1$ : the first two force $a = 2$ , contradicting $a = -1$ . Not coplanar.

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}.

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 and the scalar product (Theorem 31.9, Proposition 31.11). Use throughout the cube of Exercise 31.8: 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.

Write the equation of the plane through $B(1,0,0)$ with normal vector $(1,1,1)$ , a parametric representation of the line through $A$ directed by $(1,1,1)$ , and their intersection point.
Compute the distance from $A$ to that plane (Proposition 31.11).
Are the vectors $(1,-1,0)$ and $(1,1,-2)$ orthogonal?
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?
List the midpoints of the six edges $[BC], [CD], [DH], [HE], [EF], [FB]$ of the cube, and verify that all six satisfy the equation $x + y + z = \frac32$ .

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

Show that $P$ is perpendicular to the great diagonal $(AG)$ and passes through the cube’s center.
Compute the distances between consecutive midpoints of question 5 around the section: what do you find?
Compute the angle of the section’s polygon at one vertex (two consecutive edge-vectors and a dot product). Conclude: the section is a regular hexagon.
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.)
Where does the diagonal $(AG)$ pierce $P$ ? Verify that the piercing point is the midpoint of $[AG]$ .
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 $c = \frac32$ .
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.

Compute the volume of the corner tetrahedron $ABDE$ (base, height, and the one-third formula the Middle School volume admitted — Exercise 31.9 computes such volumes in general).
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.
Compute the distance from the cube’s center to the plane $(BDE)$ .
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.

Find where the line through $B(1,0,0)$ directed by $(0,1,1)$ pierces the hexagon’s plane $P$ .
Show that the lines $(AB)$ and $(EG)$ neither meet nor are parallel: skew lines, the phenomenon that cannot happen in a plane.
Compute the orthogonal projection of $B$ onto $P$ (walk from $B$ along the normal until the plane), and check your point satisfies $P$ ’s equation.
Finale — the space surveyor’s kit: parametric lines for piercing, normal vectors for angles and distances, plane equations 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

Solution of Problem 31.1.

1. Plane: $x + y + z = 1$ . Line: $(t, t, t)$ . Intersection: $3t = 1$ : $\left(\frac13, \frac13, \frac13\right)$ .

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

3. $1 - 1 + 0 = 0$ : orthogonal.

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: $\left(1, \frac12, 0\right)$ , $\left(\frac12, 1, 0\right)$ , $\left(0, 1, \frac12\right)$ , $\left(0, \frac12, 1\right)$ , $\left(\frac12, 0, 1\right)$ , $\left(1, 0, \frac12\right)$ : 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 differ by vectors like $\left(-\frac12, \frac12, 0\right)$ : length $\frac{\sqrt2}{2}$ each time — six equal sides.

8. At $\left(\frac12, 1, 0\right)$ : edge-vectors $\left(\frac12, -\frac12, 0\right)$ and $\left(-\frac12, 0, \frac12\right)$ : dot product $-\frac14$ , norms $\frac{\sqrt2}{2}$ : $\cos = -\frac12$ : 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 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 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 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: 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 measure angles and distances (questions 2, 8, 15, 16); plane equations 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.