Extend coordinate geometry into 3-dimensional space — understand axes, coordinate planes, octants, and the distance formula for points in space.
3
Subtopics
9
MCQs
NCERT
Source
11th
Standard
▶ Watch Before You Read
3D Geometry — Full Chapter Video Lecture
1
Coordinate Axes & Planes in 3D
📌 The 3D Coordinate System
To locate a point in 3-dimensional space, three mutually perpendicular lines are required — the x-axis, y-axis, and z-axis — all meeting at the origin O.
The three axes form the rectangular coordinate system (also called Cartesian 3D system).
Three coordinate planes: XY-plane (z=0), YZ-plane (x=0), ZX-plane (y=0)
The three planes divide all of space into eight octants.
Any point P(x, y, z): x = distance from YZ-plane, y = distance from ZX-plane, z = distance from XY-plane
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X-Axis
Points on x-axis: (x, 0, 0) x is positive toward viewer, negative away
↕️
Y-Axis
Points on y-axis: (0, y, 0) y is positive to right, negative to left
⬆️
Z-Axis
Points on z-axis: (0, 0, z) z is positive upward, negative downward
🖼️
IMAGE SPACE — 3D Coordinate System
Add diagram: three mutually perpendicular axes X, Y, Z meeting at origin O, showing positive and negative directions, and the three coordinate planes
🗺️ Sign Convention in Eight Octants
The signs of (x, y, z) determine the octant. Octant I = all positive, and so on:
I
(+, +, +)
II
(–, +, +)
III
(–, –, +)
IV
(+, –, +)
V
(+, +, –)
VI
(–, +, –)
VII
(–, –, –)
VIII
(+, –, –)
💡Special points: Origin = (0,0,0) | On XY-plane: z=0 | On YZ-plane: x=0 | On ZX-plane: y=0. Point (x,y,0) always lies on XY-plane.
📝 MCQ Practice
Coordinate System — Test Yourself
1 In which octant does the point (–3, 1, 2) lie?
A Octant I
B Octant II
C Octant III
D Octant VI
Solution: Point (–3, 1, 2): x = –, y = +, z = +
From the octant table: (–, +, +) → Octant II ✓
2 If P is (2, 4, 5) in 3D space, what are the coordinates of the point F on the ZX-plane (where y = 0)?
A (0, 4, 5)
B (2, 0, 5)
C (2, 4, 0)
D (0, 0, 5)
Solution: F lies on the ZX-plane (y = 0).
P = (2, 4, 5) → Set y = 0 → F = (2, 0, 5) ✓
3 A point lies on the y-axis. Its coordinates must be of the form:
A (x, 0, 0)
B (0, y, 0)
C (0, 0, z)
D (x, y, 0)
Solution: On the y-axis, x = 0 and z = 0.
Only y-coordinate varies → form is (0, y, 0) ✓
(x-axis: (x,0,0) | z-axis: (0,0,z))
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2
Distance Formula in 3D
📌 Distance Between Two Points in Space
The distance formula in 3D is a natural extension of the 2D formula. For points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂), the derivation uses a rectangular parallelepiped with PQ as diagonal:
PQ² = PA² + AQ² (right angle at A), and AQ² = AN² + NQ²
This gives PQ² = (x₂–x₁)² + (y₂–y₁)² + (z₂–z₁)²
Distance from origin to Q(x, y, z): OQ = √(x² + y² + z²)
Collinear test: PQ + QR = PR (or any permutation summing to longest distance)
🎯 Distance Formula in 3D
PQ = √[(x₂–x₁)² + (y₂–y₁)² + (z₂–z₁)²]
Distance from origin: OQ = √(x² + y² + z²)
Collinear condition: PQ + QR = PR
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IMAGE SPACE — Distance derivation diagram
Add diagram: rectangular parallelepiped with vertices P and Q, showing PA, AN, NQ as the three perpendicular components
💡 Worked Examples from NCERT
Distance P(1,–3,4) to Q(–4,1,2): √[(–4–1)²+(1+3)²+(2–4)²] = √[25+16+4] = √45 = 3√5 units
If a point R divides the line joining P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) in ratio m:n internally, its coordinates are:
🎯 Section Formula (Internal Division)
R = ( (mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n), (mz₂+nz₁)/(m+n) )
Midpoint (m = n): M = ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2)
Centroid of triangle with vertices A, B, C:
G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3, (z₁+z₂+z₃)/3)
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IMAGE SPACE — Section formula diagram in 3D
Add diagram: line segment from P(x₁,y₁,z₁) to Q(x₂,y₂,z₂) with point R dividing it in ratio m:n, showing the 3D parallelepiped construction
💡 Key Applications
Collinear points: If PQ + QR = PR, then P, Q, R are collinear.
Equidistant point: If PA = PB, set up the distance equations and simplify to get the locus.
Centroid: Average of all three coordinates of the three vertices.
Parallelogram: Opposite sides equal (AB = CD and BC = AD) OR diagonals bisect each other (same midpoint).
Right angle test: If AB² + BC² = AC², then angle B is 90°.
📐Locus in 3D: For "all points equidistant from A and B" — set PA = PB, square both sides, expand and simplify to get a plane equation (linear in x, y, z).
📝 MCQ Practice
Section Formula & Applications — Test Yourself
7 The centroid of triangle ABC is (1,1,1). If A = (3,–5,7) and B = (–1,7,–6), find the z-coordinate of C.