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Introduction to Three Dimensional Geometry – Class 11 Mathematics notes, important questions and MCQs
NCERT Class 11 Mathematics, chapter 11. Notes, key terms, important questions with answers and MCQs.
Chapter overview
Points in space have three coordinates (x, y, z). Distance and section formulas extend naturally from two dimensions.
Coordinates
- Three mutually perpendicular axes divide space into 8 octants.
- A point on the x-axis is (x, 0, 0); on the xy-plane (x, y, 0).
- Signs of coordinates tell the octant.
Formulas
- Distance: √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²].
- Section formula (ratio m : n internally): [(mx₂ + nx₁)/(m + n), …].
- Mid-point: [(x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2]; centroid of a triangle: average of vertices.
Key terms
- Octant
- one of eight regions formed by the three coordinate planes
- Coordinate planes
- the xy-, yz- and zx-planes, on which z = 0, x = 0 and y = 0 respectively
- Section formula (3D)
- point dividing PQ in ratio m : n internally is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n), (mz₂ + nz₁)/(m + n))
Formulas
- d = √(Δx² + Δy² + Δz²)
Important questions with answers
Name the octant in which (−3, 1, −2) lies, and find its distance from the origin.2 marks
Signs (−, +, −): the octant XʹOYZʹ (the sixth octant in common numbering). Distance = √(9 + 1 + 4) = √14.
Show that (−2, 3, 5), (1, 2, 3) and (7, 0, −1) are collinear.3 marks
AB = √(9 + 1 + 4) = √14; BC = √(36 + 4 + 16) = √56 = 2√14; AC = √(81 + 9 + 36) = √126 = 3√14. Since AB + BC = AC, the points are collinear.
Find the coordinates of the point equidistant from (0, 0, 0), (2, 0, 0), (0, 4, 0) and (0, 0, 6), and also find the centroid of the triangle with vertices (2, 1, 3), (−1, 4, 0), (5, −2, 6).5 marks
Let P(x, y, z). From PO = PA: x² = (x − 2)², so x = 1. From PO = PB: y = 2. From PO = PC: z = 3. P = (1, 2, 3). Centroid = ((2 − 1 + 5)/3, (1 + 4 − 2)/3, (3 + 0 + 6)/3) = (2, 1, 3).
Quick revision
- 8 octants.
- On the yz-plane, x = 0.
- Centroid = average of vertices.
Exam tips
- Distance of (x, y, z) from the origin is √(x² + y² + z²).
- To check collinearity of three points in space, see whether the sum of two distances equals the third.
Practice MCQs
Number of octants in space is
- 4
- 6
- 8
- 12
Show answer
Answer: (c) 8. 2³.
A point on the z-axis has the form
- (x, 0, 0)
- (0, 0, z)
- (0, y, 0)
- (x, y, 0)
Show answer
Answer: (b) (0, 0, z). Only z.
Distance between (1, 2, 3) and (4, 6, 3) is
- 7
- 5
- 25
- √7
Show answer
Answer: (b) 5. √(9 + 16 + 0).
The point (−2, 3, −4) lies in the octant with signs
- (+, +, +)
- (+, −, +)
- (−, −, −)
- (−, +, −)
Show answer
Answer: (d) (−, +, −). Signs of coordinates.
The distance of (3, −4, 5) from the x-axis is
- √41
- 3
- 5√2
- 4
Show answer
Answer: (a) √41. √(y² + z²) = √(16 + 25).
Mid-point of (2, 4, 6) and (4, 8, 10) is
- (6, 12, 16)
- (3, 6, 8)
- (1, 2, 2)
- (2, 4, 4)
Show answer
Answer: (b) (3, 6, 8). Average.
Distance of (2, 3, 6) from the origin is
- 11
- 7
- 49
- 5
Show answer
Answer: (b) 7. √(4 + 9 + 36).
The point dividing the join of (1, −2, 3) and (3, 4, −5) in the ratio 2 : 3 internally is
- (2, 1, −1)
- (9/5, 2/5, −1/5)
- (7/5, 0, 1)
- (11/5, 2, −3)
Show answer
Answer: (b) (9/5, 2/5, −1/5). ((6 + 3)/5, (8 − 6)/5, (−10 + 9)/5).
All chapters of Class 11 Mathematics
Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.