Coordinate system in 3D — x, y, z axes, octants: Exercises

Before we start, one shared picture — keep it in your head for every problem:

Figure — Coordinate system in 3D — x, y, z axes, octants

Recall What the picture shows (click)

The three axes meet at the origin $O$ . The floor is the xy-plane ( $z = 0$ ), the back-left wall is the yz-plane ( $x = 0$ ), the back-right wall is the zx-plane ( $y = 0$ ). The first box up-front-right, where all three coordinates are positive, is Octant I. Signs of $(x, y, z)$ decide which of the 8 boxes you are in.

Level 1 — Recognition

(Can you read a point and say where it lives?)

Problem 1.1

For each point, say whether it lies on an axis, in a coordinate plane, or in general space, and name which one: $A (0, 0, - 4)$ , $B (5, 0, 3)$ , $C (0, 7, 0)$ , $D (2, - 1, 6)$ .

Recall Solution

The rule (from the parent note): count how many coordinates are zero.

$A (0, 0, - 4)$ : two zeros ( $x = 0, y = 0$ ) ⟹ on the $z$ -axis. What it looks like: sitting on the vertical ruler, $4$ units below $O$ .
$B (5, 0, 3)$ : exactly one zero ( $y = 0$ ) ⟹ in the zx-plane. It rests flat against the $y = 0$ wall.
$C (0, 7, 0)$ : two zeros ( $x = 0, z = 0$ ) ⟹ on the $y$ -axis, $7$ units along the $y$ ruler.
$D (2, - 1, 6)$ : no zeros ⟹ general space, signs $(+, -, +)$ ⟹ Octant IV.

Problem 1.2

Name the octant of each: $P (- 1, - 2, - 3)$ , $Q (4, - 5, 2)$ , $R (- 3, 6, - 1)$ .

Recall Solution

Match the sign-pattern to the octant table.

$P$ : $(-, -, -)$ ⟹ Octant VII.
$Q$ : $(+, -, +)$ ⟹ Octant IV.
$R$ : $(-, +, -)$ ⟹ Octant VI.

Why these: the top four octants (I–IV) have $z > 0$ ; here $P$ and $R$ have $z < 0$ so they live in the "basement" (V–VIII).

Level 2 — Application

(Plug into the distance rule and read signs under transformations.)

Problem 2.1

Find the distance of $P (1, 2, 2)$ from the origin.

Recall Solution

Use $O P = x^{2} + y^{2} + z^{2}$ (this is Pythagoras applied twice — see parent note §5). $O P = 1^{2} + 2^{2} + 2^{2} = 1 + 4 + 4 = 9 = 3.$ Why the : distance is a length, always the positive root.

Problem 2.2

A point $S$ has coordinates $(3, - 4, 12)$ . Find $O S$ , and state its octant.

Recall Solution

$O S = 3^{2} + (- 4)^{2} + 1 2^{2} = 9 + 16 + 144 = 169 = 13.$ Notice $(- 4)^{2} = + 16$ : squaring destroys the sign. Signs $(+, -, +)$ ⟹ Octant IV.

Problem 2.3

Point $T (2, - 5, 3)$ is reflected in the xy-plane. Write the new point and say which octant it lands in.

Recall Solution

Reflecting in the xy-plane ( $z = 0$ ) means "mirror across the floor": flip the sign of $z$ only. $T = (2, - 5, 3) ⟶ T^{'} = (2, - 5, - 3) .$ Original signs $(+, -, +)$ = Octant IV; after flipping $z$ we get $(+, -, -)$ = Octant VIII — the box directly below IV, exactly the "basement" idea.

Level 3 — Analysis

(Combine ideas — equidistance, locus, collinearity.)

Problem 3.1

Find the point on the $y$ -axis that is equidistant from $A (3, 1, 2)$ and $B (5, 5, 2)$ .

Recall Solution

WHAT we want: a point on the $y$ -axis. Any such point has $x = 0$ and $z = 0$ , so call it $P (0, y, 0)$ — this is why only one unknown $y$ survives. WHY equidistant: set $P A = P B$ , or better $P A^{2} = P B^{2}$ (squaring avoids ugly roots and is legal since both distances are positive). $P A^{2} = (0 - 3)^{2} + (y - 1)^{2} + (0 - 2)^{2} = 9 + (y - 1)^{2} + 4.$ $P B^{2} = (0 - 5)^{2} + (y - 5)^{2} + (0 - 2)^{2} = 25 + (y - 5)^{2} + 4.$ Set equal: $9 + (y - 1)^{2} = 25 + (y - 5)^{2} .$ Expand: $9 + y^{2} - 2 y + 1 = 25 + y^{2} - 10 y + 25$ . The $y^{2}$ cancels (that's why this is solvable linearly): $10 - 2 y = 50 - 10 y \Rightarrow 8 y = 40 \Rightarrow y = 5.$ Answer: $P (0, 5, 0)$ .

Problem 3.2

Show that $A (1, 2, 3)$ , $B (2, 4, 5)$ , $C (4, 8, 9)$ are collinear — i.e. lie on one straight line.

Recall Solution

Idea: three points are collinear when the longest distance equals the sum of the two shorter ones (the triangle "flattens"). $A B = 1^{2} + 2^{2} + 2^{2} = 9 = 3.$ $B C = 2^{2} + 4^{2} + 4^{2} = 36 = 6.$ $A C = 3^{2} + 6^{2} + 6^{2} = 81 = 9.$ Since $A B + B C = 3 + 6 = 9 = A C$ , the points are collinear, with $B$ between $A$ and $C$ .

Problem 3.3

Find all points on the $z$ -axis whose distance from $A (3, 4, 0)$ is $13$ .

Recall Solution

A point on the $z$ -axis is $P (0, 0, z)$ . $P A^{2} = (0 - 3)^{2} + (0 - 4)^{2} + (z - 0)^{2} = 9 + 16 + z^{2} = 25 + z^{2} .$ Set $P A = 13 \Rightarrow P A^{2} = 169$ : $25 + z^{2} = 169 \Rightarrow z^{2} = 144 \Rightarrow z = \pm 12.$ Both signs are valid — two points: $(0, 0, 12)$ and $(0, 0, - 12)$ , one above and one below the floor.

Level 4 — Synthesis

(Build a figure and reason with several tools at once.)

Problem 4.1

The point $A (2, 3, 6)$ and the origin $O$ are two opposite corners of a rectangular box whose edges are parallel to the axes. (a) Name the other six corners. (b) Verify the space-diagonal length equals $O A$ .

Recall Solution

Picture the box: one corner at $O (0, 0, 0)$ , the opposite corner at $A (2, 3, 6)$ . Every corner uses each coordinate as either its min ( $0$ ) or its max ( $2, 3, 6$ ).

(a) The eight corners are all combinations $(0 or 2, 0 or 3, 0 or 6)$ : $(0, 0, 0), (2, 0, 0), (0, 3, 0), (0, 0, 6), (2, 3, 0), (2, 0, 6), (0, 3, 6), (2, 3, 6) .$ Removing $O$ and $A$ leaves the other six: $(2, 0, 0), (0, 3, 0), (0, 0, 6), (2, 3, 0), (2, 0, 6), (0, 3, 6)$ .

(b) Space diagonal $=$ distance $O$ to $A$ : $O A = 2^{2} + 3^{2} + 6^{2} = 4 + 9 + 36 = 49 = 7.$ The box has edge lengths $2, 3, 6$ , and the box-diagonal formula $a^{2} + b^{2} + c^{2} = 4 + 9 + 36 = 7$ agrees. ✔

Problem 4.2

Find the point equidistant from all four points $O (0, 0, 0)$ , $A (4, 0, 0)$ , $B (0, 4, 0)$ , $C (0, 0, 4)$ .

Recall Solution

WHY this is solvable: "equidistant from four points" gives us up to three equations for the unknown $P (x, y, z)$ — enough to pin one point. Let $P (x, y, z)$ with $P O^{2} = P A^{2} = P B^{2} = P C^{2}$ .

$P O^{2} = x^{2} + y^{2} + z^{2} .$ $P A^{2} = (x - 4)^{2} + y^{2} + z^{2} .$ Set $P O^{2} = P A^{2}$ : the $y^{2}, z^{2}$ cancel: $x^{2} = (x - 4)^{2} = x^{2} - 8 x + 16 \Rightarrow 8 x = 16 \Rightarrow x = 2.$ By identical symmetry $P O^{2} = P B^{2} \Rightarrow y = 2$ and $P O^{2} = P C^{2} \Rightarrow z = 2$ . Answer: $P (2, 2, 2)$ — the centre of the "corner tetrahedron". (Check: $P O = 12 = 23$ , and $P A = 4 + 4 + 4 = 23$ ✔.)

Level 5 — Mastery

(Prove a general statement / handle every case.)

Problem 5.1

Prove that the point $P (x, y, z)$ lies on the zx-plane if and only if it is equidistant from the two points $Q (0, d, 0)$ and $Q^{'} (0, - d, 0)$ (for any fixed $d \neq = 0$ ).

Recall Solution

Set up. Compute both squared distances: $P Q^{2} = x^{2} + (y - d)^{2} + z^{2}, P Q^{'2} = x^{2} + (y + d)^{2} + z^{2} .$ ( $\Rightarrow$ ) Equidistant forces the plane. Suppose $P Q = P Q^{'}$ , so $P Q^{2} = P Q^{'2}$ : $(y - d)^{2} = (y + d)^{2} \Rightarrow y^{2} - 2 d y + d^{2} = y^{2} + 2 d y + d^{2} \Rightarrow - 4 d y = 0.$ Since $d \neq = 0$ , we get $y = 0$ — i.e. $P$ lies in the zx-plane. ( $\Leftarrow$ ) The plane forces equidistance. If $y = 0$ then $(y - d)^{2} = d^{2} = (y + d)^{2}$ , so $P Q^{2} = P Q^{'2}$ and hence $P Q = P Q^{'}$ . Both directions hold, so the statement is proved. ∎ What it looks like: $Q$ and $Q^{'}$ are mirror images across the zx-plane; the set of points equally far from a pair of mirror points is exactly the mirror plane between them.

Problem 5.2 (all-cases)

Classify, by number and sign of zero/non-zero coordinates, every possible "home" of a point $(x, y, z)$ in 3D. How many distinct location-types are there?

Recall Solution

Sort by how many coordinates are zero — this covers all cases exhaustively:

zeros	non-zeros	lives in	how many such regions
$3$	$0$	the origin	$1$
$2$	$1$	on an axis (a half-axis, one per sign)	$3 axes \times 2 signs = 6$ half-axes
$1$	$2$	in a coordinate plane (a quadrant of it)	$3 planes \times 4 quadrants = 12$
$0$	$3$	in an octant	$2^{3} = 8$

Grand total of distinct location-types: $1 + 6 + 12 + 8 = 27$ . Why $27$ is satisfying: it's $3^{3}$ — each coordinate independently is $-$ , $0$ , or $+$ , three choices each, three coordinates ⟹ $3 \times 3 \times 3 = 27$ . Every point of space belongs to exactly one of these $27$ types.

Connections

Distance Formula in 3D — every L2–L4 computation is this formula between two general points.
Section Formula in 3D — the collinearity of §3.2 sets up dividing a segment.
Direction Cosines and Direction Ratios — describe the direction of the diagonal in §4.1.
Quadrants in 2D Coordinate Geometry — the $12$ plane-quadrants in §5.2 are these, one set per coordinate plane.
Vectors in 3D — position vectors turn "equidistant" conditions into vector equations.
Equation of a Plane — the locus in §5.1 (a mirror plane) is a first taste of plane equations.