1. Chapter at a Glance
To locate a bulb hanging in a room, two numbers are not enough — you need its distance from two walls and its height above the floor. Space is three-dimensional, so a point needs three coordinates (x, y, z). This short chapter extends coordinate geometry from the plane to space: the three axes and three coordinate planes, the eight octants, coordinates of a point, and the distance formula in three dimensions. It prepares you for vectors and 3-D geometry in Class 12.
- Three mutually perpendicular axes: x, y and z.
- Coordinate planes XY, YZ, ZX and the eight octants.
- Coordinates of a point; points on axes and planes.
- Distance between two points in space.
- Using distance to test collinearity and identify triangles.
2. Axes, Planes and Octants
Through a point O (origin) draw three mutually perpendicular lines: the x-axis, y-axis and z-axis. They form a right-handed system. Each pair of axes determines a coordinate plane.
| Location | Form of coordinates |
|---|---|
| Origin | (0, 0, 0) |
| On x-axis | (x, 0, 0) |
| On y-axis | (0, y, 0) |
| On z-axis | (0, 0, z) |
| In XY-plane | (x, y, 0) — z = 0 |
| In YZ-plane | (0, y, z) — x = 0 |
| In ZX-plane | (x, 0, z) — y = 0 |
The x-coordinate of P is its (signed) distance from the YZ-plane, the y-coordinate its distance from the ZX-plane, and the z-coordinate its distance from the XY-plane.
Octants
The three planes divide space into 8 octants:
| Octant | I | II | III | IV | V | VI | VII | VIII |
|---|---|---|---|---|---|---|---|---|
| x | + | − | − | + | + | − | − | + |
| y | + | + | − | − | + | + | − | − |
| z | + | + | + | + | − | − | − | − |
Name the octants of (−3, 1, 2), (−3, 1, −2), (2, −4, −7) and (−4, 2, −5).
Solution: (−, +, +) II; (−, +, −) VI; (+, −, −) VIII; (−, +, −) VI.
3. Distance Formula
For P(x₁, y₁, z₁) and Q(x₂, y₂, z₂), PQ is the diagonal of a cuboid with edges |x₂ − x₁|, |y₂ − y₁|, |z₂ − z₁|. Applying the Baudhāyana–Pythagoras theorem twice:
Distance from the origin: OP = √(x² + y² + z²).
Find the distance between P(1, −3, 4) and Q(−4, 1, 2).
Solution: √(25 + 16 + 4) = √45 = 3√5.
Show that P(−2, 3, 5), Q(1, 2, 3) and R(7, 0, −1) are collinear.
Solution: PQ = √(9 + 1 + 4) = √14; QR = √(36 + 4 + 16) = 2√14; PR = √(81 + 9 + 36) = 3√14. PQ + QR = PR → collinear.
Find the equation of the set of points P such that PA² + PB² = 2k², where A(3, 4, 5) and B(−1, 3, −7).
Solution: (x − 3)² + (y − 4)² + (z − 5)² + (x + 1)² + (y − 3)² + (z + 7)² = 2k² → 2x² + 2y² + 2z² − 4x − 14y + 4z + 109 = 2k².
4. Solved Problems (Board Pattern)
Show that (0, 7, −10), (1, 6, −6) and (4, 9, −6) are the vertices of an isosceles triangle.
Solution: AB = √(1 + 1 + 16) = √18; BC = √(9 + 9 + 0) = √18; AC = √(16 + 4 + 16) = 6. AB = BC → isosceles. (Also AB² + BC² = 36 = AC², so it is right-angled at B.)
Find the equation of the set of points equidistant from (1, 2, 3) and (3, 2, −1).
Solution: (x − 1)² + (y − 2)² + (z − 3)² = (x − 3)² + (y − 2)² + (z + 1)² → −2x − 6z + 10 = −6x + 2z + 10 → x − 2z = 0.
Verify that (−1, 2, 1), (1, −2, 5), (4, −7, 8) and (2, −3, 4) are vertices of a parallelogram.
Solution: AB = √(4 + 16 + 16) = 6, CD = √(4 + 16 + 16) = 6; BC = √(9 + 25 + 9) = √43, DA = √(9 + 25 + 9) = √43. Opposite sides equal; diagonals AC = √(25 + 81 + 49) = √155, BD = √(1 + 1 + 1) = √3 are unequal → a parallelogram (not a rectangle).
5. Master Formula Table
| Result | Formula |
|---|---|
| Distance | √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²] |
| From origin | √(x² + y² + z²) |
| Distance from XY-plane | |z| |
| Distance from x-axis | √(y² + z²) |
| Midpoint (enrichment) | ((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2) |
| Collinear | sum of two distances = third |
6. Master Summary Table
| Concept | Key fact | Exam focus |
|---|---|---|
| Axes, planes | 3 axes, 3 planes | Coordinates of points on them |
| Octants | 8, by signs | Name the octant |
| Distance | Pythagoras in 3-D | Lengths, triangle types |
| Collinearity | Additive distances | Prove points collinear |
| Locus | Equate distances | Equation of a set of points |
7. Mind Map
- x, y, z
- Right-handed
- XY, YZ, ZX
- 8 octants
- (x, y, z)
- On axes/planes
- √(Δx² + Δy² + Δz²)
- Collinear, locus
8. Tips and Tricks
- A coordinate that is 0 tells you the point lies in the plane of the other two axes.
- Compare squared distances to check right angles quickly.
- Octants I–IV have z > 0 (above XY-plane); V–VIII have z < 0 below them, in the same x–y order.
- Saying the distance of (x, y, z) from the x-axis is |x| — it is √(y² + z²).
- Forgetting the z-term in the distance formula.
- Mixing up which plane has which coordinate zero.
9. Important One-liners
- A point in space needs three coordinates.
- The three coordinate planes divide space into eight octants.
- Any point on the z-axis has the form (0, 0, z).
- The distance of P(x, y, z) from the YZ-plane is |x|.
- 3-D distance formula extends the 2-D one with a (z₂ − z₁)² term.
10. Exam-pattern Questions with Detailed Answers
A point on the y-axis has the form:
- (x, 0, 0)
- (0, y, 0)
- (0, 0, z)
- (x, y, 0)
Show answer
x = z = 0 on the y-axis.
Answer: (B)
Find the distance of (2, 3, 6) from the origin.
Show answer
√(4 + 9 + 36) = 7.
Answer: 7
In which octant is (4, −2, −5)?
Show answer
(+, −, −) → octant VIII.
Answer: VIII
Find the distance between (2, −1, 3) and (−2, 1, 3).
Show answer
√(16 + 4 + 0) = 2√5.
Answer: 2√5
Find the distance of (3, −4, 5) from the z-axis and from the XY-plane.
Show answer
From z-axis √(9 + 16) = 5; from XY-plane |5| = 5.
Answer: 5 and 5
Find k if the distance between (2, k, 1) and (−1, 2, 7) is 7.
Show answer
9 + (k − 2)² + 36 = 49 → (k − 2)² = 4 → k = 4 or 0.
Answer: k = 0 or 4
Find the point on the y-axis equidistant from (3, 1, 2) and (5, 5, 2).
Show answer
(0, y, 0): 9 + (y − 1)² + 4 = 25 + (y − 5)² + 4 → −2y + 14 = −10y + 54 → y = 5. (0, 5, 0).
Answer: (0, 5, 0)
Are (0, 7, 10), (−1, 6, 6) and (−4, 9, 6) the vertices of a right triangle?
Show answer
AB² = 1 + 1 + 16 = 18, BC² = 9 + 9 + 0 = 18, AC² = 16 + 4 + 16 = 36 = AB² + BC². Yes, right-angled at B (and isosceles).
Answer: Yes, right-angled isosceles
Find the equation of the set of points P such that the sum of distances from A(4, 0, 0) and B(−4, 0, 0) is 10.
Show answer
√[(x − 4)² + y² + z²] = 10 − √[(x + 4)² + y² + z²]. Squaring and simplifying: 20√[(x + 4)² + y² + z²] = 100 + 16x; squaring again gives 9x² + 25y² + 25z² = 225.
Answer: 9x² + 25y² + 25z² − 225 = 0
Show that A(1, 2, 3), B(−1, −2, −1), C(2, 3, 2) and D(4, 7, 6) are vertices of a parallelogram but not a rectangle.
Show answer
AB = √(4 + 16 + 16) = 6 = CD; BC = √(9 + 25 + 9) = √43 = DA. Diagonals AC = √(1 + 1 + 1) = √3, BD = √(25 + 81 + 49) = √155 — unequal. Parallelogram, not a rectangle.
Answer: Proved
The centroid of a triangle is (1, 1, 1) and two vertices are (3, −5, 7) and (−1, 7, −6). Find the third vertex (centroid = average of vertices).
Show answer
x: (3 − 1 + x)/3 = 1 → x = 1; y: (−5 + 7 + y)/3 = 1 → y = 1; z: (7 − 6 + z)/3 = 1 → z = 2. (1, 1, 2).
Answer: (1, 1, 2)
A cuboid has one vertex at the origin and edges of lengths 3, 4, 12 along the axes. Find the coordinates of the opposite vertex and the length of its diagonal.
Show answer
Opposite vertex (3, 4, 12). Diagonal √(9 + 16 + 144) = 13.
Answer: (3, 4, 12); 13
11. Quick Revision Summary
- Three perpendicular axes; three coordinate planes; eight octants.
- Point (x, y, z): distances from YZ-, ZX-, XY-planes.
- On axes two coordinates are 0; in a plane one coordinate is 0.
- Distance = √[(Δx)² + (Δy)² + (Δz)²].
- Use distances to test collinearity and classify triangles/quadrilaterals.