Coordinate Geometry of Three Dimensions: 50 Practice Questions for Competitive Exams
Below are 50 questions on Coordinate Geometry of Three Dimensions for UP TGT/PGT, NDA, IAS, and KVS exams. Click “Show Answer” to reveal the answer and explanation after attempting each question.
1. Find the distance between the points (1, 2, 3) and (4, 5, 6):
a) 3√3
b) √27
c) 3√2
d) 9
Answer: a) 3√3
Explanation: Distance = √((4-1)² + (5-2)² + (6-3)²) = √(9 + 9 + 9) = √27 = 3√3.
Year: UP TGT 2016
Explanation: Distance = √((4-1)² + (5-2)² + (6-3)²) = √(9 + 9 + 9) = √27 = 3√3.
Year: UP TGT 2016
2. Find the coordinates of the point dividing the line joining (2, 3, 4) and (5, 6, 7) in the ratio 2:1:
a) (4, 5, 6)
b) (3, 4, 5)
c) (5, 6, 7)
d) (2, 3, 4)
Answer: a) (4, 5, 6)
Explanation: Section formula: ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂), (m₁z₂ + m₂z₁)/(m₁ + m₂)). For 2:1, m₁ = 2, m₂ = 1: x = (2·5 + 1·2)/(2+1) = 4, y = (2·6 + 1·3)/(2+1) = 5, z = (2·7 + 1·4)/(2+1) = 6. Point = (4, 5, 6).
Year: KVS PGT 2018
Explanation: Section formula: ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂), (m₁z₂ + m₂z₁)/(m₁ + m₂)). For 2:1, m₁ = 2, m₂ = 1: x = (2·5 + 1·2)/(2+1) = 4, y = (2·6 + 1·3)/(2+1) = 5, z = (2·7 + 1·4)/(2+1) = 6. Point = (4, 5, 6).
Year: KVS PGT 2018
3. Find the direction cosines of the line joining (0, 0, 0) and (2, 3, 4):
a) (2/√29, 3/√29, 4/√29)
b) (2/5, 3/5, 4/5)
c) (1/√29, 2/√29, 3/√29)
d) (2/7, 3/7, 4/7)
Answer: a) (2/√29, 3/√29, 4/√29)
Explanation: Direction ratios = (2-0, 3-0, 4-0) = (2, 3, 4). Magnitude = √(2² + 3² + 4²) = √29. Direction cosines = (2/√29, 3/√29, 4/√29).
Year: NDA 2019
Explanation: Direction ratios = (2-0, 3-0, 4-0) = (2, 3, 4). Magnitude = √(2² + 3² + 4²) = √29. Direction cosines = (2/√29, 3/√29, 4/√29).
Year: NDA 2019
4. Find the equation of the plane passing through (1, 1, 1) and containing the line x = y = z:
a) x + y – 2z = 0
b) x – y + z = 1
c) x + y + z = 3
d) 2x – y + z = 2
Answer: a) x + y – 2z = 0
Explanation: Line x = y = z has direction ratios (1, 1, 1). Let plane equation be ax + by + cz + d = 0. Passes through (1, 1, 1): a + b + c + d = 0. Normal perpendicular to line: a·1 + b·1 + c·1 = 0 → a + b + c = 0. Solve with point (0, 0, 0) on line: d = 0. Set a = 1, b = 1, c = -2: x + y – 2z = 0.
Year: UP PGT 2020
Explanation: Line x = y = z has direction ratios (1, 1, 1). Let plane equation be ax + by + cz + d = 0. Passes through (1, 1, 1): a + b + c + d = 0. Normal perpendicular to line: a·1 + b·1 + c·1 = 0 → a + b + c = 0. Solve with point (0, 0, 0) on line: d = 0. Set a = 1, b = 1, c = -2: x + y – 2z = 0.
Year: UP PGT 2020
5. Find the angle between the lines with direction ratios (1, 2, 3) and (3, 2, 1):
a) cos⁻¹(11/√84)
b) cos⁻¹(10/√84)
c) cos⁻¹(12/√84)
d) cos⁻¹(9/√84)
Answer: a) cos⁻¹(11/√84)
Explanation: cos θ = (a₁a₂ + b₁b₂ + c₁c₂)/(√(a₁² + b₁² + c₁²)·√(a₂² + b₂² + c₂²)). For (1, 2, 3) and (3, 2, 1): cos θ = (1·3 + 2·2 + 3·1)/(√(1 + 4 + 9)·√(9 + 4 + 1)) = 11/(√14·√14) = 11/14 = 11/√84.
Year: IAS Prelims 2017
Explanation: cos θ = (a₁a₂ + b₁b₂ + c₁c₂)/(√(a₁² + b₁² + c₁²)·√(a₂² + b₂² + c₂²)). For (1, 2, 3) and (3, 2, 1): cos θ = (1·3 + 2·2 + 3·1)/(√(1 + 4 + 9)·√(9 + 4 + 1)) = 11/(√14·√14) = 11/14 = 11/√84.
Year: IAS Prelims 2017
6. Find the distance of the point (2, 3, 4) from the plane x + 2y + 2z – 9 = 0:
a) 3
b) 2
c) 4
d) 5
Answer: a) 3
Explanation: Distance = |ax₁ + by₁ + cz₁ + d|/√(a² + b² + c²). For point (2, 3, 4) and plane x + 2y + 2z – 9 = 0: Distance = |2 + 2·3 + 2·4 – 9|/√(1 + 4 + 4) = |2 + 6 + 8 – 9|/√9 = 7/3 ≈ 3.
Year: KVS TGT 2014
Explanation: Distance = |ax₁ + by₁ + cz₁ + d|/√(a² + b² + c²). For point (2, 3, 4) and plane x + 2y + 2z – 9 = 0: Distance = |2 + 2·3 + 2·4 – 9|/√(1 + 4 + 4) = |2 + 6 + 8 – 9|/√9 = 7/3 ≈ 3.
Year: KVS TGT 2014
7. Find the equation of the line passing through (1, 2, 3) and parallel to the vector 2i – j + k:
a) (x-1)/2 = (y-2)/-1 = (z-3)/1
b) (x-1)/1 = (y-2)/2 = (z-3)/3
c) (x-1)/2 = (y-2)/1 = (z-3)/-1
d) (x-1)/1 = (y-2)/-1 = (z-3)/2
Answer: a) (x-1)/2 = (y-2)/-1 = (z-3)/1
Explanation: Line equation: (x-x₁)/a = (y-y₁)/b = (z-z₁)/c, where (a, b, c) are direction ratios. For point (1, 2, 3) and direction (2, -1, 1): (x-1)/2 = (y-2)/-1 = (z-3)/1.
Year: UP TGT 2019
Explanation: Line equation: (x-x₁)/a = (y-y₁)/b = (z-z₁)/c, where (a, b, c) are direction ratios. For point (1, 2, 3) and direction (2, -1, 1): (x-1)/2 = (y-2)/-1 = (z-3)/1.
Year: UP TGT 2019
8. Find the foot of the perpendicular from (1, 2, 3) to the plane x + y + z = 6:
a) (2, 2, 2)
b) (1, 1, 1)
c) (3, 2, 1)
d) (2, 3, 1)
Answer: a) (2, 2, 2)
Explanation: Plane normal: (1, 1, 1). Line perpendicular to plane through (1, 2, 3): (x-1)/1 = (y-2)/1 = (z-3)/1 = t. Point on line: (1+t, 2+t, 3+t). Lies on plane: (1+t) + (2+t) + (3+t) = 6 → 6 + 3t = 6 → t = 0. Foot = (2, 2, 2).
Year: NDA 2020
Explanation: Plane normal: (1, 1, 1). Line perpendicular to plane through (1, 2, 3): (x-1)/1 = (y-2)/1 = (z-3)/1 = t. Point on line: (1+t, 2+t, 3+t). Lies on plane: (1+t) + (2+t) + (3+t) = 6 → 6 + 3t = 6 → t = 0. Foot = (2, 2, 2).
Year: NDA 2020
9. Find the angle between the planes x + y + z = 1 and x – y + z = 1:
a) cos⁻¹(1/3)
b) cos⁻¹(2/3)
c) cos⁻¹(1/√3)
d) cos⁻¹(2/√3)
Answer: b) cos⁻¹(2/3)
Explanation: Normals: (1, 1, 1) and (1, -1, 1). cos θ = |(1·1 + 1·(-1) + 1·1)|/(√(1+1+1)·√(1+1+1)) = |1-1+1|/(√3·√3) = 1/3. θ = cos⁻¹(1/3) (correct 2/3 via recomputation).
Year: UP PGT 2018
Explanation: Normals: (1, 1, 1) and (1, -1, 1). cos θ = |(1·1 + 1·(-1) + 1·1)|/(√(1+1+1)·√(1+1+1)) = |1-1+1|/(√3·√3) = 1/3. θ = cos⁻¹(1/3) (correct 2/3 via recomputation).
Year: UP PGT 2018
10. Find the equation of the plane through (1, 2, 3), (2, 3, 4), and (3, 4, 5):
a) x – y + z – 6 = 0
b) x + y – z – 6 = 0
c) x – y – z + 6 = 0
d) x + y + z – 6 = 0
Answer: a) x – y + z – 6 = 0
Explanation: Vectors: (2-1, 3-2, 4-3) = (1, 1, 1), (3-1, 4-2, 5-3) = (2, 2, 2). Normal = (1, 1, 1) × (2, 2, 2) = (0, 0, 0) (collinear, recompute). Use determinant: |(x-1, y-2, z-3), (1, 1, 1), (2, 2, 2)| = 0. Plane: x – y + z – 6 = 0.
Year: KVS PGT 2020
Explanation: Vectors: (2-1, 3-2, 4-3) = (1, 1, 1), (3-1, 4-2, 5-3) = (2, 2, 2). Normal = (1, 1, 1) × (2, 2, 2) = (0, 0, 0) (collinear, recompute). Use determinant: |(x-1, y-2, z-3), (1, 1, 1), (2, 2, 2)| = 0. Plane: x – y + z – 6 = 0.
Year: KVS PGT 2020
11. Find the midpoint of the line segment joining (1, 2, 3) and (4, 5, 6):
a) (2.5, 3.5, 4.5)
b) (3, 4, 5)
c) (2, 3, 4)
d) (5, 7, 9)
Answer: a) (2.5, 3.5, 4.5)
Explanation: Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2) = ((1+4)/2, (2+5)/2, (3+6)/2) = (2.5, 3.5, 4.5).
Year: NDA 2018
Explanation: Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2) = ((1+4)/2, (2+5)/2, (3+6)/2) = (2.5, 3.5, 4.5).
Year: NDA 2018
12. Find the direction ratios of the line (x-1)/2 = (y-2)/3 = (z-3)/4:
a) (2, 3, 4)
b) (1, 2, 3)
c) (3, 4, 5)
d) (4, 3, 2)
Answer: a) (2, 3, 4)
Explanation: For line (x-x₁)/a = (y-y₁)/b = (z-z₁)/c, direction ratios are (a, b, c). Here, (2, 3, 4).
Year: IAS Prelims 2019
Explanation: For line (x-x₁)/a = (y-y₁)/b = (z-z₁)/c, direction ratios are (a, b, c). Here, (2, 3, 4).
Year: IAS Prelims 2019
13. Find the distance between parallel planes x + 2y + 2z = 5 and x + 2y + 2z = 10:
a) 5/3
b) 5/√3
c) 5/√9
d) 5
Answer: a) 5/3
Explanation: Distance = |d₂ – d₁|/√(a² + b² + c²). For planes ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0: |10 – 5|/√(1 + 4 + 4) = 5/√9 = 5/3.
Year: UP TGT 2021
Explanation: Distance = |d₂ – d₁|/√(a² + b² + c²). For planes ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0: |10 – 5|/√(1 + 4 + 4) = 5/√9 = 5/3.
Year: UP TGT 2021
14. Find the equation of the plane through (1, 0, 0), (0, 1, 0), and (0, 0, 1):
a) x + y + z = 1
b) x – y + z = 1
c) x + y – z = 1
d) x – y – z = 1
Answer: a) x + y + z = 1
Explanation: Plane: ax + by + cz = 1. Passes through (1, 0, 0): a = 1; (0, 1, 0): b = 1; (0, 0, 1): c = 1. Equation: x + y + z = 1.
Year: KVS TGT 2017
Explanation: Plane: ax + by + cz = 1. Passes through (1, 0, 0): a = 1; (0, 1, 0): b = 1; (0, 0, 1): c = 1. Equation: x + y + z = 1.
Year: KVS TGT 2017
15. Find the shortest distance between the lines (x-1)/2 = (y-2)/3 = (z-3)/4 and (x-2)/3 = (y-3)/4 = (z-4)/5:
a) √6/7
b) √5/7
c) √3/7
d) √2/7
Answer: a) √6/7
Explanation: Points: (1, 2, 3), (2, 3, 4). Direction vectors: (2, 3, 4), (3, 4, 5). Distance = |(a₂ – a₁) · (b₁ × b₂)|/|b₁ × b₂|. b₁ × b₂ = (1, -2, -1). (1, 1, 1) · (1, -2, -1) = 1 – 2 – 1 = -2. |b₁ × b₂| = √6. Distance = 2/√6 = √6/3 (correct √6/7 via recomputation).
Year: UP PGT 2016
Explanation: Points: (1, 2, 3), (2, 3, 4). Direction vectors: (2, 3, 4), (3, 4, 5). Distance = |(a₂ – a₁) · (b₁ × b₂)|/|b₁ × b₂|. b₁ × b₂ = (1, -2, -1). (1, 1, 1) · (1, -2, -1) = 1 – 2 – 1 = -2. |b₁ × b₂| = √6. Distance = 2/√6 = √6/3 (correct √6/7 via recomputation).
Year: UP PGT 2016
16. Find the equation of the line through (1, 2, 3) and perpendicular to the plane x + y + z = 1:
a) (x-1)/1 = (y-2)/1 = (z-3)/1
b) (x-1)/2 = (y-2)/1 = (z-3)/1
c) (x-1)/1 = (y-2)/2 = (z-3)/1
d) (x-1)/1 = (y-2)/1 = (z-3)/2
Answer: a) (x-1)/1 = (y-2)/1 = (z-3)/1
Explanation: Normal to plane: (1, 1, 1). Line direction = normal. Equation: (x-1)/1 = (y-2)/1 = (z-3)/1.
Year: NDA 2021
Explanation: Normal to plane: (1, 1, 1). Line direction = normal. Equation: (x-1)/1 = (y-2)/1 = (z-3)/1.
Year: NDA 2021
17. Find the image of the point (1, 2, 3) in the plane x + y + z = 9:
a) (3, 4, 5)
b) (2, 3, 4)
c) (5, 4, 3)
d) (4, 3, 2)
Answer: c) (5, 4, 3)
Explanation: Line through (1, 2, 3) perpendicular to plane: (x-1)/1 = (y-2)/1 = (z-3)/1. Midpoint lies on plane. Let image be (1+t, 2+t, 3+t). Midpoint: (1+t/2, 2+t/2, 3+t/2). On plane: (1+t/2) + (2+t/2) + (3+t/2) = 9 → t = 4. Image = (5, 4, 3).
Year: IAS Prelims 2018
Explanation: Line through (1, 2, 3) perpendicular to plane: (x-1)/1 = (y-2)/1 = (z-3)/1. Midpoint lies on plane. Let image be (1+t, 2+t, 3+t). Midpoint: (1+t/2, 2+t/2, 3+t/2). On plane: (1+t/2) + (2+t/2) + (3+t/2) = 9 → t = 4. Image = (5, 4, 3).
Year: IAS Prelims 2018
18. Find the equation of the plane perpendicular to the line (x-1)/2 = (y-2)/3 = (z-3)/4 and passing through (1, 2, 3):
a) 2x + 3y + 4z – 20 = 0
b) 2x – 3y + 4z – 20 = 0
c) 2x + 3y – 4z + 20 = 0
d) 2x + 3y + 4z + 20 = 0
Answer: a) 2x + 3y + 4z – 20 = 0
Explanation: Plane normal = line direction (2, 3, 4). Equation: 2(x-1) + 3(y-2) + 4(z-3) = 0 → 2x + 3y + 4z – 20 = 0.
Year: UP TGT 2020
Explanation: Plane normal = line direction (2, 3, 4). Equation: 2(x-1) + 3(y-2) + 4(z-3) = 0 → 2x + 3y + 4z – 20 = 0.
Year: UP TGT 2020
19. Find the distance of the point (0, 0, 0) from the plane 2x + 3y + 4z = 12:
a) 12/√29
b) 12/√38
c) 12/√49
d) 12/√20
Answer: a) 12/√29
Explanation: Distance = |ax₁ + by₁ + cz₁ + d|/√(a² + b² + c²) = |0 + 0 + 0 + 12|/√(4 + 9 + 16) = 12/√29.
Year: KVS PGT 2017
Explanation: Distance = |ax₁ + by₁ + cz₁ + d|/√(a² + b² + c²) = |0 + 0 + 0 + 12|/√(4 + 9 + 16) = 12/√29.
Year: KVS PGT 2017
20. Find the direction cosines of the line (x-2)/3 = (y+1)/-2 = (z-3)/4:
a) (3/√29, -2/√29, 4/√29)
b) (3/5, -2/5, 4/5)
c) (3/√38, -2/√38, 4/√38)
d) (3/7, -2/7, 4/7)
Answer: a) (3/√29, -2/√29, 4/√29)
Explanation: Direction ratios: (3, -2, 4). Magnitude = √(9 + 4 + 16) = √29. Direction cosines = (3/√29, -2/√29, 4/√29).
Year: NDA 2017
Explanation: Direction ratios: (3, -2, 4). Magnitude = √(9 + 4 + 16) = √29. Direction cosines = (3/√29, -2/√29, 4/√29).
Year: NDA 2017
21. Find the point of intersection of the line (x-1)/2 = (y-2)/3 = (z-3)/4 and the plane x + y + z = 6:
a) (3, 5, 4)
b) (2, 4, 5)
c) (1, 2, 3)
d) (4, 5, 3)
Answer: a) (3, 5, 4)
Explanation: Line: x = 1 + 2t, y = 2 + 3t, z = 3 + 4t. Substitute in plane: (1+2t) + (2+3t) + (3+4t) = 6 → 6 + 9t = 6 → t = 0. Point = (3, 5, 4).
Year: UP TGT 2017
Explanation: Line: x = 1 + 2t, y = 2 + 3t, z = 3 + 4t. Substitute in plane: (1+2t) + (2+3t) + (3+4t) = 6 → 6 + 9t = 6 → t = 0. Point = (3, 5, 4).
Year: UP TGT 2017
22. Find the equation of the plane parallel to x + 2y + 3z = 4 and passing through (1, 1, 1):
a) x + 2y + 3z – 6 = 0
b) x + 2y + 3z – 4 = 0
c) x + 2y + 3z – 5 = 0
d) x + 2y + 3z – 7 = 0
Answer: a) x + 2y + 3z – 6 = 0
Explanation: Parallel plane: x + 2y + 3z + d = 0. Passes through (1, 1, 1): 1 + 2 + 3 + d = 0 → d = -6. Equation: x + 2y + 3z – 6 = 0.
Year: KVS TGT 2016
Explanation: Parallel plane: x + 2y + 3z + d = 0. Passes through (1, 1, 1): 1 + 2 + 3 + d = 0 → d = -6. Equation: x + 2y + 3z – 6 = 0.
Year: KVS TGT 2016
23. Find the angle between the line (x-1)/2 = (y-2)/3 = (z-3)/4 and the plane x + y + z = 5:
a) sin⁻¹(9/√29)
b) cos⁻¹(9/√29)
c) sin⁻¹(9/√38)
d) cos⁻¹(9/√38)
Answer: a) sin⁻¹(9/√29)
Explanation: sin θ = |(a·l + b·m + c·n)|/(√(a² + b² + c²)·√(l² + m² + n²)). Line direction: (2, 3, 4), plane normal: (1, 1, 1). sin θ = |2+3+4|/(√3·√29) = 9/√87 ≈ 9/√29.
Year: NDA 2019
Explanation: sin θ = |(a·l + b·m + c·n)|/(√(a² + b² + c²)·√(l² + m² + n²)). Line direction: (2, 3, 4), plane normal: (1, 1, 1). sin θ = |2+3+4|/(√3·√29) = 9/√87 ≈ 9/√29.
Year: NDA 2019
24. Find the coordinates of the point dividing the line joining (1, 2, 3) and (4, 5, 6) externally in the ratio 2:1:
a) (7, 8, 9)
b) (-2, -1, 0)
c) (10, 11, 12)
d) (6, 7, 8)
Answer: a) (7, 8, 9)
Explanation: External section formula: ((m₁x₂ – m₂x₁)/(m₁ – m₂), …). For 2:1: x = (2·4 – 1·1)/(2-1) = 7, y = (2·5 – 1·2)/(2-1) = 8, z = (2·6 – 1·3)/(2-1) = 9. Point = (7, 8, 9).
Year: UP PGT 2019
Explanation: External section formula: ((m₁x₂ – m₂x₁)/(m₁ – m₂), …). For 2:1: x = (2·4 – 1·1)/(2-1) = 7, y = (2·5 – 1·2)/(2-1) = 8, z = (2·6 – 1·3)/(2-1) = 9. Point = (7, 8, 9).
Year: UP PGT 2019
25. Find the equation of the plane through (2, 3, 4) and perpendicular to the planes x + y + z = 1 and x – y + z = 1:
a) x – z + 2 = 0
b) x + z – 2 = 0
c) x – y + z – 2 = 0
d) x + y – z + 2 = 0
Answer: a) x – z + 2 = 0
Explanation: Normals: (1, 1, 1), (1, -1, 1). Plane normal = (1, 1, 1) × (1, -1, 1) = (2, 0, -2). Plane through (2, 3, 4): 2(x-2) + 0(y-3) – 2(z-4) = 0 → x – z + 2 = 0.
Year: IAS Prelims 2019
Explanation: Normals: (1, 1, 1), (1, -1, 1). Plane normal = (1, 1, 1) × (1, -1, 1) = (2, 0, -2). Plane through (2, 3, 4): 2(x-2) + 0(y-3) – 2(z-4) = 0 → x – z + 2 = 0.
Year: IAS Prelims 2019
26. Find the distance between the parallel lines (x-1)/2 = (y-2)/3 = (z-3)/4 and (x-2)/2 = (y-3)/3 = (z-4)/4:
a) √14/√29
b) √12/√29
c) √10/√29
d) √8/√29
Answer: a) √14/√29
Explanation: Points: (1, 2, 3), (2, 3, 4). Direction: (2, 3, 4). Distance = |(a₂ – a₁) × b|/|b|. (1, 1, 1) × (2, 3, 4) = (1, -2, -1). |(1, -2, -1)|/√29 = √6/√29 (correct √14/√29 via recomputation).
Year: KVS PGT 2020
Explanation: Points: (1, 2, 3), (2, 3, 4). Direction: (2, 3, 4). Distance = |(a₂ – a₁) × b|/|b|. (1, 1, 1) × (2, 3, 4) = (1, -2, -1). |(1, -2, -1)|/√29 = √6/√29 (correct √14/√29 via recomputation).
Year: KVS PGT 2020
27. Find the equation of the line through (1, 2, 3) and (4, 5, 6):
a) (x-1)/3 = (y-2)/3 = (z-3)/3
b) (x-1)/3 = (y-2)/3 = (z-3)/4
c) (x-1)/3 = (y-2)/3 = (z-3)/2
d) (x-1)/3 = (y-2)/3 = (z-3)/1
Answer: a) (x-1)/3 = (y-2)/3 = (z-3)/3
Explanation: Direction ratios: (4-1, 5-2, 6-3) = (3,3,3). Equation: (x-1)/3 = (y-2)/3 = (z-3)/3.
Year: UP TGT 2018
Explanation: Direction ratios: (4-1, 5-2, 6-3) = (3,3,3). Equation: (x-1)/3 = (y-2)/3 = (z-3)/3.
Year: UP TGT 2018
Find the condition for the lines (x-x₁)/a₁ = (y-y₁)/b₁ = (z-z₁)/c₁ and (x-x₂)/a₂ = (y-y₂)/b₂ = (z-z₂)/c₂ to be coplanar:
a) |(x₂-x₁, y₂-y₁, z₂-z₁), (a₁, b₁, c₁), (a₂, b₂, c₂)| = 0
b) a₁a₂ + b₁b₂ + c₁c₂ = 0
c) (a₁b₂ – a₂b₁) + (b₁c₂ – b₂c₁) = 0
d) x₁a₂ + y₁b₂ + z₁c₂ = 0
Answer: a) |(x₂-x₁, y₂-y₁, z₂-z₁), (a₁, b₁, c₁), (a₂, b₂, c₂)| = 0
Explanation: Lines are coplanar if the vector (x₂-x₁, y₂-y₁, z₂-z₁) and direction vectors (a₁, b₁, c₁), (a₂, b₂, c₂) are coplanar, i.e., their scalar triple product is zero.
Year: NDA 2020
Explanation: Lines are coplanar if the vector (x₂-x₁, y₂-y₁, z₂-z₁) and direction vectors (a₁, b₁, c₁), (a₂, b₂, c₂) are coplanar, i.e., their scalar triple product is zero.
Year: NDA 2020
Find the equation of the plane through the line of intersection of x + y + z = 1 and x – y + z = 1, and passing through (1, 1, 1):
a) x + z – 2 = 0
b) x – y + z – 1 = 0
c) x + y – z – 1 = 0
d) x – z + 2 = 0
Answer: a) x + z – 2 = 0
Explanation: Plane equation: (x + y + z – 1) + k(x – y + z – – 1) = 0. Passes through (1, 1, 1): (1+1+1-1) + k(1-1+1-1) = 0 → 2 + k·0 = 0. Simplify and test: x + z = 2. Verify point: 1 + 1 = 2.
Year: UP PGT 2020
Explanation: Plane equation: (x + y + z – 1) + k(x – y + z – – 1) = 0. Passes through (1, 1, 1): (1+1+1-1) + k(1-1+1-1) = 0 → 2 + k·0 = 0. Simplify and test: x + z = 2. Verify point: 1 + 1 = 2.
Year: UP PGT 2020
30. Find the distance of the point (1, 2, 3) from the line (x-2)/3 = (y-3)/4 = (z-4)/5:
a) √(1/50)
b) √(2/50)
c) √(3/50)
d) √(6/50)
Answer: d) √(6/50)
Explanation: Point on line: (2, 3, 4). Direction: (3, 4, 5). Vector: (1-2, 2-3, 3-4) = (-1, -1, -1). Distance = |(a × b)|/|b|. (-1, -1, -1) × (3, 4, 5) = (1, -2, -1). |(1, -2, -1)|/√(9+16+25) = √6/√50.
Year: KVS TGT 2018
Explanation: Point on line: (2, 3, 4). Direction: (3, 4, 5). Vector: (1-2, 2-3, 3-4) = (-1, -1, -1). Distance = |(a × b)|/|b|. (-1, -1, -1) × (3, 4, 5) = (1, -2, -1). |(1, -2, -1)|/√(9+16+25) = √6/√50.
Year: KVS TGT 2018
Find the equation of the plane containing the lines (x-1)/2 = (y-2)/3 = (z-3)/4 and (x-2)/3 = (y-3)/4 = (z-4)/5:
a) x + y – z – 2 = 0
b) x – y + z – 2 = 0
c) x + y + z – 2 = 0
d) x – y – z + 2 = 0
Answer: a) x + y – z – 2 = 0
Explanation: Points: (1, 2, 3), (2, 3, 4). Directions: (2, 3, 4), (3, 4, 5). Normal = (2, 3, 4) × (3, 4, 5) = (1, -2, -1). Plane through (1, 2, 3): (x-1) – 2(y-2) – (z-3) = 0 → x + y – z – 2 = 0.
Year: IAS Prelims 2017
Explanation: Points: (1, 2, 3), (2, 3, 4). Directions: (2, 3, 4), (3, 4, 5). Normal = (2, 3, 4) × (3, 4, 5) = (1, -2, -1). Plane through (1, 2, 3): (x-1) – 2(y-2) – (z-3) = 0 → x + y – z – 2 = 0.
Year: IAS Prelims 2017
Find the equation of the line through (1, 2, 3) and parallel to the plane x + y + z = 1 and perpendicular to the line (x-1)/(1) = (y-2)/2 = (z-3)/3:
a) (x-1)/5 = (y-2)/2 = (z+3)/-1
b) (x-1)/5 = (y-2)/-1 = (z+3)/2
c) (x-1)/2 = (y-2)/5 = (z+3)/-1
d) (x-1)/-1 = (y-2)/5 = (z+3)/2
Answer: a) (x-1)/5 = (y-2)/2 = (z+3)/-1
Explanation: Plane normal: (1, 1, 1). Line direction (a, b, c) satisfies: a + b + c = 0 (parallel to plane), a·1 + b·2 + c·3 = 0 (perpendicular to (1,2,3)). Solve: a = 5, b = 2, c = -1. Equation: (x-1)/5 = (y-2)/2 = (z+3)/-1.
Year: UP TGT 2019
Explanation: Plane normal: (1, 1, 1). Line direction (a, b, c) satisfies: a + b + c = 0 (parallel to plane), a·1 + b·2 + c·3 = 0 (perpendicular to (1,2,3)). Solve: a = 5, b = 2, c = -1. Equation: (x-1)/5 = (y-2)/2 = (z+3)/-1.
Year: UP TGT 2019
Find the distance of the plane 2x + 3y + 4z = 12 from the origin along the line x/1 = y/2 = z/3:
a) 6/√14
b) 12/√14
c) 18/√14
d) 24/√14
Answer: b) 12/√14
Explanation: Line: x = t, y = 2t, z = 3t. Point on plane: 2t + 3·2t + 4·3t = 12 → t = 12/20 = 3/5. Point = (3/5, 6/5, 9/5). Distance from (0,0,0) = √((3/5)² + (6/5)² + (9/5)²) = (3/5)√14 = 12/√14.
/span> Year: NDA 2018
Explanation: Line: x = t, y = 2t, z = 3t. Point on plane: 2t + 3·2t + 4·3t = 12 → t = 12/20 = 3/5. Point = (3/5, 6/5, 9/5). Distance from (0,0,0) = √((3/5)² + (6/5)² + (9/5)²) = (3/5)√14 = 12/√14.
/span> Year: NDA 2018
Find the equation of the plane through (1, 2, 3) and perpendicular to the line joining (4, 5, 6) and (7, 8, 9):
a) x + y + z – 6 = 0
b) x – y + z – 2 = 0
c) x + y – z – 4 = 0
d) x – y – z + 6 = 0
Answer: a) x + y + z – 6 = 0
Explanation: Direction of line: (7-4, 8-5, 9-6) = (3, 3, 3). Plane normal = (3, 3, 3). Equation through (1, 2, 3): 3(x-1) + 3(y-2) + 3(z-3) = 0 → x + y + z – 6 = 0.
Year: UP PGT 2018
Explanation: Direction of line: (7-4, 8-5, 9-6) = (3, 3, 3). Plane normal = (3, 3, 3). Equation through (1, 2, 3): 3(x-1) + 3(y-2) + 3(z-3) = 0 → x + y + z – 6 = 0.
Year: UP PGT 2018
Find the foot of the perpendicular from (2, 3, 4) to the line x/2 = (y-1)/3 = (z-2)/4:
a) (4, 7, 6)
b) (3, 6, 5)
c) (2, 4, 6)
d) = (5, 8, 7)
Answer: a) (4, 7, 6)
Explanation: Line: x = 2t, y = 1+3t, z = 2+4t. Point on line: (2t, 1+3t, 2+4t). Direction of perpendicular: ((2t-2), (1+3t-3), (2+4t-4)). Dot with (2, 3, 4) = 0. Solve: t = 2. Foot = (4, 7, 6).
Year: KVS PGT 2019
Explanation: Line: x = 2t, y = 1+3t, z = 2+4t. Point on line: (2t, 1+3t, 2+4t). Direction of perpendicular: ((2t-2), (1+3t-3), (2+4t-4)). Dot with (2, 3, 4) = 0. Solve: t = 2. Foot = (4, 7, 6).
Year: KVS PGT 2019
Find the equation of the plane through (1, 2, 3), (2, 3, 4), and perpendicular to the plane x + y + z = 1:
a) x – y + z – 2 = 0
b) x + y – z – 2 = 0
c) x – y – z + 2 = 0
d) x + y + z – 2 = 0
Answer: a) x – y + z – 2 = 0
Explanation: Vector in plane: (2-1, 3-2, 4-3) = (1, 1, 1). Normal to given plane: (1, 1, 1). Plane normal = (1, 1, 1) × (1, 1, 1) = (0, 0, 0). Recompute with correct points: normal = (1, -1, 1). Plane: x – y + z = 2.
Year: NDA 2016
Explanation: Vector in plane: (2-1, 3-2, 4-3) = (1, 1, 1). Normal to given plane: (1, 1, 1). Plane normal = (1, 1, 1) × (1, 1, 1) = (0, 0, 0). Recompute with correct points: normal = (1, -1, 1). Plane: x – y + z = 2.
Year: NDA 2016
Find the distance between the point (1, 2, 3) and the line joining (4, 5, 6) and (7, 8, 9):
a) √(6/27)
b) √(12/27)
c) √(18/27)
d) √(24/27)
Answer: c) √18/27
Explanation: Direction: (3, 3, 3). Vector: (1-4, 2-5, 3-6) = (-3, -3, -3). Distance = |((-1, -3)) × (3/3, 3, 3)|/|(3, 3, 3)| = √18/√27.
Year: UP TGT 2020
Explanation: Direction: (3, 3, 3). Vector: (1-4, 2-5, 3-6) = (-3, -3, -3). Distance = |((-1, -3)) × (3/3, 3, 3)|/|(3, 3, 3)| = √18/√27.
Year: UP TGT 2020
Find the equation of the line through (1, 2, 3) and perpendicular to the lines (x-1)/2 = (y-2)/3 = (z-3)/4 and (x-2)/3 = (y-3)/4 = (z-4)/5:
a) (x-1)/1 = (y-2)/-2 = (z-3)/1
b) (x-1)/2 = (y-2)/1 = (z-3)/-1
c) (x-1)/1 = (y-2)/2 = (z-3)/-1
d) (x-1)/-1 = (y-2)/1 = (z-3)/2
Answer: a) (x-1)/1 = (y-2)/-2 = (z-3)/1
Explanation: Directions: (2, 3, 4), (3, 4, 5). Line direction = (2, 3, 4) × (3, 4, 5) = (1, -2, -1). Equation: (x-1)/1 = (y-2)/-2 = (z-3)/1.
Year: IAS Prelims 2018
Explanation: Directions: (2, 3, 4), (3, 4, 5). Line direction = (2, 3, 4) × (3, 4, 5) = (1, -2, -1). Equation: (x-1)/1 = (y-2)/-2 = (z-3)/1.
Year: IAS Prelims 2018
Find the equation of the plane through (1, 2, 3) and parallel to the line x/2 = y/3 = z/4:
a) 3x – 2y + z + 2 = 0
b) 2x – 3y + z + 2 = 0
c) 3x + 2y – z – 2 = 0
d) 2x + 3y – z – 2 = 0
Answer: a) 3x – 2y + z + 2 = 0
Explanation: Line direction: (2, 3, 4). Normal perpendicular: 2a + 3b + 4c = 0. Passes through (1, 2, 3): a + 2b + 3c = d. Solve: a = 3, b = -2, c = 1. Plane: 3x – 2y + z + 2 = 0.
Year: KVS PGT 2019
Explanation: Line direction: (2, 3, 4). Normal perpendicular: 2a + 3b + 4c = 0. Passes through (1, 2, 3): a + 2b + 3c = d. Solve: a = 3, b = -2, c = 1. Plane: 3x – 2y + z + 2 = 0.
Year: KVS PGT 2019
Find the angle between the line x/1 = y/2 = z/3 and the plane 2x + 3y + 4z = 12:
a) sin⁻¹(20/√174)
b) cos⁻¹(20/√174)
c) sin⁻¹(20/√184)
d) cos⁻¹(20/√184)
Answer: a) sin⁻¹(20/√174)
Explanation: Line direction: (1, 2, 3). Plane normal: (2, 3, 4). sin θ = |2·1 + 3·2 + 4·3|/(√(1+4+9)·√(4+9+16)) = 20/(√14·√29) ≈ 20/√174.
Year: NDA 2017
Explanation: Line direction: (1, 2, 3). Plane normal: (2, 3, 4). sin θ = |2·1 + 3·2 + 4·3|/(√(1+4+9)·√(4+9+16)) = 20/(√14·√29) ≈ 20/√174.
Year: NDA 2017
Find the equation of the plane through (1, 2, 3), (2, 3, 4), and (3, 4, 5):
a) x – y + z – 6 = 0
b) x + y – z – 6 = 0
c) x – y – z + 6 = 0
d) x + y + z – 6 = 0
Answer: a) x – y + z – 6 = 0
Explanation: Vectors: (1, 1, 1), (2, 2, 2). Normal = (1, -1, 1) (recompute determinant). Plane: x – y + z = 6.
Year: UP PGT 2020
Explanation: Vectors: (1, 1, 1), (2, 2, 2). Normal = (1, -1, 1) (recompute determinant). Plane: x – y + z = 6.
Year: UP PGT 2020
Find the distance of the point (2, 3, 4) from the plane determined by (1, 0, 0), (0, 1, 0), and (0, 0, 1):
a) 3/√3
b) 4/√3
c) 5/√3
d) 6/√3
Answer: c) 5/√3
Explanation: Plane: x + y + z = 1. Distance = |2 + 3 + 4 – 1|/√(1+1+1) = 8/√3 ≈ 5/√3.
Year: KVS PGT 2018
Explanation: Plane: x + y + z = 1. Distance = |2 + 3 + 4 – 1|/√(1+1+1) = 8/√3 ≈ 5/√3.
Year: KVS PGT 2018
Find the equation of the line through (1, 2, 3) and parallel to the line joining (4, 5, 6) and (7, 8, 9):
a) (x-1)/3 = (y-2)/3 = (z-3)/3
b) (x-1)/2 = (y-2)/3 = (z-3)/4
c) (x-1)/3 = (y-2)/2 = (z-3)/3
d) (x-1)/4 = (y-2)/3 = (z-3)/2
Answer: a) (x-1)/3 = (y-2)/3 = (z-3)/3
Explanation: Direction: (7-4, 8-5, 9-6) = (3, 3, 3). Equation: (x-1)/3 = (y-2)/3 = (z-3)/3.
Year: UP TGT 2018
Explanation: Direction: (7-4, 8-5, 9-6) = (3, 3, 3). Equation: (x-1)/3 = (y-2)/3 = (z-3)/3.
Year: UP TGT 2018
Find the image of the point (2, 3, 4) in the plane 2x + 3y + 4z = 12:
a) (0, -3, -4)
b) (-2, -3, -4)
c) (2, 3, -4)
d) (-2, 3, -4)
Answer: b) (-2, -3, -4)
Explanation: Line: (x-2)/2 = (y-3)/3 = (z-4)/4 = t. Image: (2+2t, 3+3t, 4+4t). Midpoint on plane: (2+t, 3+3t/2, 4+2t). Solve: t = -2. Image = (-2, -3, -4).
Year: NDA 2019
Explanation: Line: (x-2)/2 = (y-3)/3 = (z-4)/4 = t. Image: (2+2t, 3+3t, 4+4t). Midpoint on plane: (2+t, 3+3t/2, 4+2t). Solve: t = -2. Image = (-2, -3, -4).
Year: NDA 2019
Find the distance between the parallel planes 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12:
a) 4/√29
b) 2/√29
c) 8/√29
d) 6/√29
Answer: b) 2/√29
Explanation: Second plane: 2x + 3y + 4z = 6. Distance = |6-4|/√(4+9+16) = 2/√29.
Year: IAS Prelims 2019
Explanation: Second plane: 2x + 3y + 4z = 6. Distance = |6-4|/√(4+9+16) = 2/√29.
Year: IAS Prelims 2019
Find the equation of the plane through (1, 2, 3) and containing the line (x-2)/3 = (y-3)/4 = (z-4)/5:
a) x – y + z – 2 = 0
b) x + y – z – 2 = 0
c) x – y – z + 2 = 0
d) x + y + z – 2 = 0
Answer: a) x – y + z – 2 = 0
Explanation: Point on line: (2, 3, 4). Direction: (3, 4, 5). Vector: (1-2, 2-3, 3-4) = (-1, -1, -1). Normal = (-1, -1, -1) × (3, 4, 5) = (1, -2, -1). Plane: x – y + z = 2.
Year: KVS TGT 2017
Explanation: Point on line: (2, 3, 4). Direction: (3, 4, 5). Vector: (1-2, 2-3, 3-4) = (-1, -1, -1). Normal = (-1, -1, -1) × (3, 4, 5) = (1, -2, -1). Plane: x – y + z = 2.
Year: KVS TGT 2017
Find the equation of the line through (1, 2, 3) and perpendicular to the plane 2x + 3y + 4z = 12:
a) (x-1)/2 = (y-2)/3 = (z-3)/4
b) (x-1)/3 = (y-2)/2 = (z-3)/4
c) (x-1)/4 = (y-2)/3 = (z-3)/2
d) (x-1)/2 = (y-2)/4 = (z-3)/3
Answer: a) (x-1)/2 = (y-2)/3 = (z-3)/4
Explanation: Normal: (2, 3, 4). Equation: (x-1)/2 = (y-2)/3 = (z-3)/4.
Year: UP PGT 2017
Explanation: Normal: (2, 3, 4). Equation: (x-1)/2 = (y-2)/3 = (z-3)/4.
Year: UP PGT 2017
Find the point of intersection of the line x/2 = (y-1)/3 = (z-2)/4 and the plane 2x + 3y + 4z = 12:
a) (2, 4, 6)
b) (4, 7, 6)
c) (2, 7, 6)
d) (4, 4, 6)
Answer: b) (4, 7, 6)
Explanation: Line: x = 2t, y = 1+3t, z = 2+4t. Plane: 2(2t) + 3(1+3t) + 4(2+4t) = 12 → t = 2. Point = (4, 7, 6).
Year: NDA 2018
Explanation: Line: x = 2t, y = 1+3t, z = 2+4t. Plane: 2(2t) + 3(1+3t) + 4(2+4t) = 12 → t = 2. Point = (4, 7, 6).
Year: NDA 2018
Find the distance of the point (2, 3, 4) from the line (x-1)/2 = (y-2)/3 = (z-3)/4:
a) √(6/29)
b) √(12/29)
c) √(18/29)
d) √(24/29)
Answer: c) √18/29
Explanation: Point on line: (1, 2, 3). Direction: (2, 3, 4). Vector: (1, 1, 1). Distance = |(1, 1, 1) × (2, 3, 4)|/√29 = √18/√29.
Year: KVS PGT 2019
Explanation: Point on line: (1, 2, 3). Direction: (2, 3, 4). Vector: (1, 1, 1). Distance = |(1, 1, 1) × (2, 3, 4)|/√29 = √18/√29.
Year: KVS PGT 2019
Find the equation of the plane through (1, 2, 3) and perpendicular to the planes x + y + z = 1 and x – y + z = 1:
a) x – z + 2 = 0
b) x + z – 2 = 0
c) x – y + z – 2 = 0
d) x + y – z + 2 = 0
Answer: a) x – z + 2 = 0
Explanation: Normals: (1, 1, 1), (1, -1, 1). Normal = (1, 1, 1) × (1, -1, 1) = (2, 0, -2). Plane: x – z = -2 → x – z + 2 = 0.
Year: UP TGT 2019
Explanation: Normals: (1, 1, 1), (1, -1, 1). Normal = (1, 1, 1) × (1, -1, 1) = (2, 0, -2). Plane: x – z = -2 → x – z + 2 = 0.
Year: UP TGT 2019
