📐 CHAPTER 12: INTRODUCTION TO 3D GEOMETRY
Subject: Mathematics | Class: XI | Kendriya Vidyalaya Sangathan
📖 INTRODUCTION
In three-dimensional geometry, we use three mutually perpendicular axes: the x-axis, y-axis, and z-axis.
🔹 Key Concepts:
- Coordinate axes: x-axis, y-axis, z-axis (mutually perpendicular)
- Coordinate planes: xy-plane (z=0), yz-plane (x=0), zx-plane (y=0)
- Octants: 8 regions formed by the three coordinate planes
- Coordinates of a point: (x, y, z) - abscissa, ordinate, applicate
🔹 Points on Axes and Planes
- On x-axis: (x, 0, 0)
- On y-axis: (0, y, 0)
- On z-axis: (0, 0, z)
- On xy-plane: (x, y, 0)
- On yz-plane: (0, y, z)
- On zx-plane: (x, 0, z)
🔢 OCTANTS
The three coordinate planes divide space into eight octants.
| Octant | Signs (x, y, z) | Octant | Signs (x, y, z) |
|---|---|---|---|
| I | (+, +, +) | V | (+, +, -) |
| II | (-, +, +) | VI | (-, +, -) |
| III | (-, -, +) | VII | (-, -, -) |
| IV | (+, -, +) | VIII | (+, -, -) |
📏 DISTANCE FORMULA
The distance between two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) is:
PQ = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]
🔹 Distance from Origin
Distance of point P(x, y, z) from origin O(0, 0, 0):
OP = √(x² + y² + z²)
🔹 Distance from Coordinate Planes
- Distance from xy-plane = |z|
- Distance from yz-plane = |x|
- Distance from zx-plane = |y|
🔹 Distance from Axes
- Distance from x-axis = √(y² + z²)
- Distance from y-axis = √(x² + z²)
- Distance from z-axis = √(x² + y²)
📐 SECTION FORMULA
The coordinates of the point dividing the line segment joining P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) in the ratio m : n are:
((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n), (mz₂ + nz₁)/(m+n))
🔹 Midpoint Formula
((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2)
🔹 Centroid of a Triangle
((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3, (z₁+z₂+z₃)/3)
📝 MCQ QUESTIONS
Q1. The ratio in which the line joining (2, 4, 5) and (3, 5, -9) is divided by the yz-plane is:
(a) 2:3 (b) 3:2 (c) 2:-3 (d) 4:-3
Answer: (c) 2:-3
Solution: yz-plane ⇒ x = 0
Let ratio be k:1, x = (3k + 2)/(k+1) = 0 ⇒ 3k + 2 = 0 ⇒ k = -2/3
Ratio = -2/3:1 = 2:-3
Q2. The ratio in which the line joining the points (a, b, c) and (-a, -c, -b) is divided by the xy-plane is:
(a) a:b (b) b:c (c) c:a (d) c:b
Answer: (d) c:b
Q3. If P(0, 1, 2), Q(4, -2, 1) and O(0, 0, 0) are three points, then angle POQ is:
(a) π/6 (b) π/4 (c) π/3 (d) π/2
Answer: (d) π/2
Solution: OP = √(0+1+4) = √5
OQ = √(16+4+1) = √21
PQ = √(16+9+1) = √26
OP² + OQ² = 5 + 21 = 26 = PQ² ⇒ Right angle at O
Q4. If the extremities of the diagonal of a square are (1, -2, 3) and (2, -3, 5), then the length of the side is:
(a) √6 (b) √3 (c) √5 (d) √7
Answer: (b) √3
Solution: Diagonal = √[(2-1)² + (-3+2)² + (5-3)²] = √(1+1+4) = √6
Side = Diagonal/√2 = √6/√2 = √3
Q5. The points (5, -4, 2), (4, -3, 1), (7, 6, 4) and (8, -7, 5) are the vertices of:
(a) A rectangle (b) a square (c) a parallelogram (d) none of these
Answer: (a) A rectangle
Q6. In a three dimensional space, the equation x² - 5x + 6 = 0 represents:
(a) Points (b) Planes (c) Curves (d) Pair of Straight lines
Answer: (b) Planes
Solution: x² - 5x + 6 = 0 ⇒ (x-2)(x-3) = 0 ⇒ x = 2 or x = 3
These are planes parallel to yz-plane.
Q7. Let (3, 4, -1) and (-1, 2, 3) be the end points of a diameter of a sphere, then the radius of the sphere is equal to:
(a) 2 (b) 3 (c) 6 (d) 7
Answer: (b) 3
Solution: Diameter = √[(3+1)² + (4-2)² + (-1-3)²] = √(16+4+16) = √36 = 6
Radius = 3
Q8. XOZ-Plane divides the join of (2, 3, 1) and (6, 7, 1) in the ratio:
(a) 3:7 (b) 2:7 (c) -3:7 (d) -2:7
Answer: (c) -3:7
Q9. What is the locus of a point for which y = 0, z = 0?
(a) x-axis (b) y-axis (c) z-axis (d) yz-plane
Answer: (a) x-axis
Q10. The coordinates of the foot of the perpendicular drawn from the point P(3, 4, 5) on the yz-plane are:
(a) (3, 4, 0) (b) (0, 4, 5) (c) (3, 0, 5) (d) (3, 0, 0)
Answer: (b) (0, 4, 5)
Q11. The coordinates of the foot of the perpendicular from a point P(6, 7, 8) on x-axis are:
(a) (6, 0, 0) (b) (0, 7, 0) (c) (0, 0, 8) (d) (0, 7, 8)
Answer: (a) (6, 0, 0)
Q12. The perpendicular distance of the point P(6, 7, 8) from xy-plane is:
(a) 8 (b) 7 (c) 6 (d) 10
Answer: (a) 8
Q13. The length of the perpendicular drawn from P(3, 4, 5) on y-axis is:
(a) 10 (b) √34 (c) √113 (d) 5√2
Answer: (b) √34
Solution: Distance from y-axis = √(x² + z²) = √(9 + 25) = √34
Q14. The perpendicular distance of the point P(3, 3, 4) from the x-axis is:
(a) 3√2 (b) 5 (c) 3 (d) 4
Answer: (b) 5
Solution: Distance from x-axis = √(y² + z²) = √(9 + 16) = 5
Q15. The length of the perpendicular drawn from the point P(a, b, c) from z-axis is:
(a) √(a² + b²) (b) √(b² + c²) (c) √(a² + c²) (d) √(a² + b² + c²)
Answer: (a) √(a² + b²)
Q16. A plane is parallel to the yz-plane, so it is perpendicular to the:
(a) x-axis (b) y-axis (c) z-axis (d) none of these
Answer: (a) x-axis
Q17. The point (-2, -3, -4) lies in the:
(a) first octant (b) seventh octant (c) second octant (d) eighth octant
Answer: (b) seventh octant
Solution: (-, -, -) is the seventh octant.
Q18. If the distance between the points (a, 0, 1) and (0, 1, 2) is √27, then the value of a is:
(a) 5 (b) ±5 (c) -5 (d) none of these
Answer: (b) ±5
Solution: √[(a-0)² + (0-1)² + (1-2)²] = √27
a² + 1 + 1 = 27 ⇒ a² = 25 ⇒ a = ±5
Q19. x-axis is the intersection of the planes:
(a) xy and xz (b) yz and zx (c) xy and yz (d) none of these
Answer: (a) xy and xz
Q20. Equation of y-axis is considered as:
(a) x = 0, y = 0 (b) y = 0, z = 0 (c) z = 0, x = 0 (d) none of these
Answer: (c) z = 0, x = 0
📚 CASE BASED QUESTIONS
Q21. A three dimensional cartesian coordinate system is formed by a point called the origin (denoted by O) and a basis consisting of three mutually perpendicular vectors. These vectors define the three coordinate axes x, y and z axes. They are also known as abscissa, ordinate and applicate axis, respectively. The coordinate planes divide the space into OCTANTS.
(i) In which octant does the point (3, -2, -5) lie?
(a) II (b) VI (c) VII (d) IV
Answer: (c) VII
Solution: (+, -, -) is the seventh octant.
(ii) If a point lies on z-axis, then the coordinates of the point are:
(a) (x, y, z) (b) (x, y, 0) (c) (0, y, 0) (d) (0, 0, z)
Answer: (d) (0, 0, z)
(iii) The locus of a point for which x = 0 is:
(a) xy-plane (b) yz-plane (c) zx-plane (d) none of these
Answer: (b) yz-plane
(iv) A plane is parallel to yz-plane, so it is perpendicular to:
(a) x-axis (b) y-axis (c) z-axis (d) none of these
Answer: (a) x-axis
Q22. The coordinates of a point P in the space are (a, b, c).
(i) Distance of point P from xy-plane is:
(a) |a| (b) |b| (c) |c| (d) √(a² + b²)
Answer: (c) |c|
(ii) The coordinates of foot of perpendicular drawn from point P to yz-plane are:
(a) (0, a, b) (b) (0, a, c) (c) (0, b, a) (d) (0, a, b)
Answer: (a) (0, a, b) [Wait: yz-plane means x=0, so (0, b, c)]
Actually, yz-plane has x=0, so point is (0, b, c).
(iii) The coordinates of foot of perpendicular drawn from point P to z-axis are:
(a) (0, 0, √(b²+c²)) (b) (0, 0, c) (c) (0, 0, √(a²+b²)) (d) (0, 0, √(c²+a²))
Answer: (b) (0, 0, c)
(iv) Distance between point P and its image in zx-plane is:
(a) 2|b| (b) 2|a| (c) 2|c| (d) 2|a + c|
Answer: (a) 2|b|
Solution: Image in zx-plane (y=0): (a, -b, c)
Distance = 2|b|
📝 VERY SHORT ANSWER TYPE QUESTIONS
Q23. Write the distance of the point P(2, 3, 5) from the xy-plane.
Answer: Distance = |z| = 5
Q24. Write the distance of the point P(3, 4, 5) from z-axis.
Answer: Distance from z-axis = √(x² + y²) = √(9 + 16) = 5
Q25. If the distance between the points P(a, 2, 1) and Q(1, -1, -1) is 5 units, find the value of a.
Answer: √[(a-1)² + (2+1)² + (1+1)²] = 5
(a-1)² + 9 + 4 = 25 ⇒ (a-1)² = 12 ⇒ a = 1 ± 2√3
Q26. The coordinates of the midpoints of sides AB, BC and CA of ABC are D(1, 2, -3), E(3, 0, 1) and F(-1, 1, -4) respectively. Write the coordinates of its centroid.
Answer: Centroid = (D + E + F)/3 = ((1+3-1)/3, (2+0+1)/3, (-3+1-4)/3)
= (3/3, 3/3, -6/3) = (1, 1, -2)
Q27. Write the coordinates of third vertex of a triangle having centroid at the origin and two vertices are (3, -5, 7) and (3, 0, 1).
Answer: (x₁+x₂+x₃)/3 = 0 ⇒ x₃ = -6
(y₁+y₂+y₃)/3 = 0 ⇒ -5 + 0 + y₃ = 0 ⇒ y₃ = 5
(z₁+z₂+z₃)/3 = 0 ⇒ 7 + 1 + z₃ = 0 ⇒ z₃ = -8
Third vertex = (-6, 5, -8)
Q28. What is the locus of a point (x, y, z) for y = 0, z = 0?
Answer: x-axis
Q29. Find the point on y-axis which is at a distance of √10 units from the point (1, 2, 3).
Answer: Point on y-axis: (0, y, 0)
√[(0-1)² + (y-2)² + (0-3)²] = √10
1 + (y-2)² + 9 = 10 ⇒ (y-2)² = 0 ⇒ y = 2
Point = (0, 2, 0)
Q30. If the origin is the centroid of a triangle ABC having vertices A(a, 1, 3), B(-2, b, -5) and C(4, 7, c). Find the values of a, b, c.
Answer: (a-2+4)/3 = 0 ⇒ a = -2
(1+b+7)/3 = 0 ⇒ b = -8
(3-5+c)/3 = 0 ⇒ c = 2
a = -2, b = -8, c = 2
Q31. Find the distance of the point (a, b, c) from zx-plane.
Answer: Distance from zx-plane = |y| = |b|
Q32. If L and M are the feet of the perpendiculars from P(3, 4, 5) on xy and yz-planes, then find LM.
Answer: L = (3, 4, 0), M = (0, 4, 5)
LM = √[(3-0)² + (4-4)² + (0-5)²] = √(9 + 0 + 25) = √34
Q33. Find the value of β, so that distance between the points (7, 1, -3) and (4, 5, β) is 13.
Answer: √[(7-4)² + (1-5)² + (-3-β)²] = 13
9 + 16 + (β+3)² = 169 ⇒ (β+3)² = 144 ⇒ β+3 = ±12
β = 9 or -15
Q34. Are the points A(3, 6, 9), B(10, 20, 30) and C(25, -41, 5) vertices of a right angled triangle? Justify your answer.
Answer: AB = √(49+196+441) = √686
BC = √(225+3721+625) = √4571
AC = √(484+2209+16) = √2709
Not a right triangle.
Q35. Find the midpoint of (1, 4, 6) and (5, 8, 10).
Answer: ((1+5)/2, (4+8)/2, (6+10)/2) = (3, 6, 8)
Q36. Find the distance between the points A(3, 4, 5) and origin (0, 0, 0).
Answer: OA = √(9 + 16 + 25) = √50 = 5√2
Q37. A point is on x-axis, what are its y-coordinate and z-coordinate?
Answer: y = 0, z = 0
Q38. A point is in the xz-plane, what can you say about its y-coordinate?
Answer: y = 0
Q39. Name the octant in which point (7, -1, 5) lies.
Answer: (+, -, +) = IV octant
Q40. Find the coordinates of a point on the y-axis which is at a distance of 5√2 from the point P(3, -2, 5).
Answer: Point on y-axis: (0, y, 0)
√[(0-3)² + (y+2)² + (0-5)²] = 5√2
9 + (y+2)² + 25 = 50 ⇒ (y+2)² = 16 ⇒ y+2 = ±4
y = 2 or y = -6
Points: (0, 2, 0) or (0, -6, 0)
Q41. Write the plane containing the x-axis and y-axis.
Answer: xy-plane (z = 0)
Q42. Find the distance of the point P(3, 4, 5) from yz-plane.
Answer: Distance from yz-plane = |x| = 3
📝 SHORT ANSWER TYPE QUESTIONS
Q43. Show that the points (a, b, c), (b, c, a) and (c, a, b) are vertices of an equilateral triangle.
Answer: Let P(a,b,c), Q(b,c,a), R(c,a,b)
PQ = √[(b-a)² + (c-b)² + (a-c)²]
QR = √[(c-b)² + (a-c)² + (b-a)²] = same
RP = √[(a-c)² + (b-a)² + (c-b)²] = same
All sides equal ⇒ Equilateral triangle.
Q44. Using distance formula prove that points A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8) are collinear.
Answer: AB = √(1+16+49) = √66
BC = √(4+64+196) = √264 = 2√66
AC = √(1+16+49) = √66
Since AB + AC = BC, points are collinear.
Q45. Find the point on y-axis which is equidistant from the points (3, 1, 2) and (5, 5, 2).
Answer: Point on y-axis: (0, y, 0)
Distance to (3,1,2): √(9 + (y-1)² + 4) = √((y-1)² + 13)
Distance to (5,5,2): √(25 + (y-5)² + 4) = √((y-5)² + 29)
(y-1)² + 13 = (y-5)² + 29
y² - 2y + 1 + 13 = y² - 10y + 25 + 29
-2y + 14 = -10y + 54 ⇒ 8y = 40 ⇒ y = 5
Point = (0, 5, 0)
Q46. Find the point on z-axis which is at a distance √21 from the point (1, 2, 3).
Answer: Point on z-axis: (0, 0, z)
√[(0-1)² + (0-2)² + (z-3)²] = √21
1 + 4 + (z-3)² = 21 ⇒ (z-3)² = 16 ⇒ z-3 = ±4
z = 7 or z = -1
Points: (0, 0, 7) or (0, 0, -1)
Q47. Show that the points (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of an isosceles right angled triangle.
Answer: Let A(0,7,10), B(-1,6,6), C(-4,9,6)
AB = √(1+1+16) = √18 = 3√2
BC = √(9+9+0) = √18 = 3√2
AC = √(16+4+16) = √36 = 6
AB² + BC² = 18 + 18 = 36 = AC²
Isosceles right triangle.
Q48. Three vertices of a parallelogram ABCD are A(3, -1, 2), B(1, 2, -4) and C(-1, 1, 2). Find the coordinates of the fourth vertex.
Answer: In parallelogram, diagonals bisect each other.
Midpoint of AC = ((3-1)/2, (-1+1)/2, (2+2)/2) = (1, 0, 2)
Midpoint of BD = ((1+x)/2, (2+y)/2, (-4+z)/2)
(1+x)/2 = 1 ⇒ x = 1
(2+y)/2 = 0 ⇒ y = -2
(-4+z)/2 = 2 ⇒ z = 8
D = (1, -2, 8)
Q49. Find the equation of the set of points P such that its distance from the points A(3, 4, -5) and B(-2, 1, 4) are equal.
Answer: PA = PB
(x-3)² + (y-4)² + (z+5)² = (x+2)² + (y-1)² + (z-4)²
x²-6x+9 + y²-8y+16 + z²+10z+25 = x²+4x+4 + y²-2y+1 + z²-8z+16
-6x-8y+10z+50 = 4x-2y-8z+21
-10x-6y+18z+29 = 0
10x + 6y - 18z - 29 = 0
Q50. Using section formula, prove that the three points (-4, 6, 10), (2, 4, 6) and (14, 0, -2) are collinear.
Answer: Let A(-4,6,10), B(2,4,6), C(14,0,-2)
Check if B divides AC in some ratio:
Let AB:BC = m:n
x: (m(14)+n(-4))/(m+n) = 2
14m - 4n = 2m + 2n ⇒ 12m = 6n ⇒ m:n = 1:2
y: (m(0)+n(6))/(m+n) = 4
6n = 4m + 4n ⇒ 2n = 4m ⇒ m:n = 1:2
z: (m(-2)+n(10))/(m+n) = 6
-2m + 10n = 6m + 6n ⇒ 4n = 8m ⇒ m:n = 1:2
Hence collinear.
✅ ANSWERS - MCQs
| Q.No. | Answer | Q.No. | Answer |
|---|---|---|---|
| 1 | (c) | 11 | (a) |
| 2 | (d) | 12 | (a) |
| 3 | (d) | 13 | (b) |
| 4 | (b) | 14 | (b) |
| 5 | (a) | 15 | (a) |
| 6 | (b) | 16 | (a) |
| 7 | (b) | 17 | (b) |
| 8 | (c) | 18 | (b) |
| 9 | (a) | 19 | (a) |
| 10 | (b) | 20 | (c) |
📌 KEY FORMULAS SUMMARY
| Formula | Expression |
|---|---|
| Distance between two points | √[(x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²] |
| Distance from origin | √(x² + y² + z²) |
| Distance from xy-plane | |z| |
| Distance from yz-plane | |x| |
| Distance from zx-plane | |y| |
| Distance from x-axis | √(y² + z²) |
| Distance from y-axis | √(x² + z²) |
| Distance from z-axis | √(x² + y²) |
| Section Formula (internal) | ((mx₂+nx₁)/(m+n), ...) |
| Midpoint Formula | ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2) |
| Centroid of Triangle | ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3, (z₁+z₂+z₃)/3) |
🔹 Important Points:
- xy-plane: z = 0
- yz-plane: x = 0
- zx-plane: y = 0
- Octants: 8 regions with sign combinations
- On axes: two coordinates are zero
📚 Kendriya Vidyalaya Sangathan | Class XI Mathematics
Chapter 12: Introduction to 3D Geometry | Student Support Material