Class 11 — Mathematics (NCERT)

Introduction to Three Dimensional Geometry

Class 11

  • ✓Understand the concept of a three-dimensional coordinate system.
  • ✓Locate points in three-dimensional space using coordinates (x, y, z).
  • ✓Identify the coordinates of points lying on the coordinate axes and coordinate planes.
  • ✓Apply the distance formula to find the distance between two points in three-dimensional space.
  • ✓Solve problems involving the distance formula in various geometric contexts.

Key concepts

Coordinates in Space

In two-dimensional geometry, a point is located by two coordinates (x, y) relative to two mutually perpendicular axes. In three-dimensional geometry, we extend this idea by introducing a third axis, the z-axis, which is perpendicular to both the x-axis and the y-axis at their point of intersection (the origin). These three axes (x, y, z) are mutually perpendicular and intersect at a common point O, called the origin (0, 0, 0). They form a right-handed system. The position of any point P in space is uniquely determined by an ordered triplet of real numbers (x, y, z), called its coordinates.\nThe three coordinate axes divide the space into eight regions, called octants. The sign of the coordinates determines the octant.\nThe planes formed by pairs of axes are called coordinate planes:\n* XY-plane (or z=0 plane): Points on this plane have coordinates (x, y, 0).\n* YZ-plane (or x=0 plane): Points on this plane have coordinates (0, y, z).\n* ZX-plane (or y=0 plane): Points on this plane have coordinates (x, 0, z).\nPoints on the x-axis have coordinates (x, 0, 0).\nPoints on the y-axis have coordinates (0, y, 0).\nPoints on the z-axis have coordinates (0, 0, z).

Distance Formula

The distance between two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) in three-dimensional space is given by the formula, which is an extension of the two-dimensional distance formula. It is derived using the Pythagorean theorem twice.

d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

Key facts to remember

  • 1A point in three-dimensional space is represented by an ordered triplet (x, y, z).
  • 2The three coordinate axes (x, y, z) are mutually perpendicular and intersect at the origin O(0, 0, 0).
  • 3The coordinate planes are XY-plane (z=0), YZ-plane (x=0), and ZX-plane (y=0).
  • 4Points on the x-axis are (x, 0, 0); on the y-axis are (0, y, 0); on the z-axis are (0, 0, z).
  • 5The distance between two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) is given by d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²].
  • 6The distance of a point P(x, y, z) from the origin O(0, 0, 0) is √(x² + y² + z²).
  • 7The three coordinate planes divide the space into eight octants, determined by the signs of the coordinates.
  • 8Three points A, B, C are collinear if the sum of the distances of two pairs of points equals the distance of the third pair (e.g., AB + BC = AC).

Worked examples

Example 1

Find the coordinates of a point on the y-axis which is at a distance of 5 units from the origin.

ILet the point on the y-axis be P(0, y, 0).
IIThe origin is O(0, 0, 0).
IIIGiven that the distance OP = 5 units.
IVUsing the distance formula: OP = √[(0 - 0)² + (y - 0)² + (0 - 0)²]
V5 = √[0² + y² + 0²]
VI5 = √[y²]
VII5 = |y|
VIIITherefore, y = 5 or y = -5.
9The points are (0, 5, 0) and (0, -5, 0).

Answer

The coordinates of the points are (0, 5, 0) and (0, -5, 0).

Points on an axis can be in either positive or negative direction from the origin.

Example 2

Find the distance between the points P(1, -3, 4) and Q(-4, 1, 2).

ILet P(x₁, y₁, z₁) = (1, -3, 4) and Q(x₂, y₂, z₂) = (-4, 1, 2).
IIUsing the distance formula: PQ = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]
IIIPQ = √[(-4 - 1)² + (1 - (-3))² + (2 - 4)²]
IVPQ = √[(-5)² + (1 + 3)² + (-2)²]
VPQ = √[(-5)² + (4)² + (-2)²]
VIPQ = √[25 + 16 + 4]
VIIPQ = √[45]
VIIIPQ = √(9 × 5)
9PQ = 3√5 units.

Answer

The distance between the points is 3√5 units.

Always simplify the radical if possible.

Example 3

Show that the points A(1, 2, 3), B(2, 3, 4) and C(0, 1, 2) are collinear.

ITo show that three points A, B, C are collinear, we need to verify if the sum of the distances between two pairs of points is equal to the distance between the third pair. That is, AB + BC = AC or AB + AC = BC or BC + AC = AB.
IICalculate the distance AB:
IIIAB = √[(2 - 1)² + (3 - 2)² + (4 - 3)²]
IVAB = √[(1)² + (1)² + (1)²]
VAB = √[1 + 1 + 1]
VIAB = √3 units.
VIICalculate the distance BC:
VIIIBC = √[(0 - 2)² + (1 - 3)² + (2 - 4)²]
9BC = √[(-2)² + (-2)² + (-2)²]
10BC = √[4 + 4 + 4]
11BC = √12 = √(4 × 3) = 2√3 units.
12Calculate the distance AC:
13AC = √[(0 - 1)² + (1 - 2)² + (2 - 3)²]
14AC = √[(-1)² + (-1)² + (-1)²]
15AC = √[1 + 1 + 1]
16AC = √3 units.
17Check for collinearity:
18We observe that AB + AC = √3 + √3 = 2√3.
19Also, BC = 2√3.
20Since AB + AC = BC, the points A, B and C are collinear.
21Hence proved.

Answer

The points A(1, 2, 3), B(2, 3, 4) and C(0, 1, 2) are collinear.

For collinearity, the sum of the two smaller distances must equal the largest distance.

Common mistakes

  • ✗Confusing the signs of coordinates when determining the octant or performing calculations.
  • ✗Incorrectly applying the distance formula, especially with negative numbers (e.g., forgetting to square the negative difference).
  • ✗Not simplifying radical expressions for distances to their simplest form.
  • ✗Assuming collinearity without verifying the distance condition (AB + BC = AC).
  • ✗Mixing up coordinates, e.g., using (x₁, y₂, z₁) instead of (x₁, y₁, z₁) in the formula.

Exam tips

  • ★Clearly label your points as (x₁, y₁, z₁) and (x₂, y₂, z₂) before applying the distance formula to avoid errors.
  • ★Show all steps in calculations, especially when dealing with negative numbers, to minimise arithmetic mistakes.
  • ★Remember that distance is always a non-negative value.
  • ★Practice identifying the octant for various points to strengthen your understanding of 3D space.
  • ★For problems involving collinearity or types of triangles, calculate all three distances and then apply the relevant conditions carefully.

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