Three Dimensional Geometry — Previous Year Questions (Class 12 Mathematics)
3D geometry extends coordinate systems to three dimensions. Master lines, planes, and distance formulas for problem solving.
TL;DR: 3D geometry extends coordinate systems to three dimensions. Master lines, planes, and distance formulas for problem solving.
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3D geometry extends coordinate systems to three dimensions. Master lines, planes, and distance formulas for problem solving.
Three Dimensional Geometry — Previous Year Questions with Solutions
Q (2023, 2 marks): Find the distance between the points A(1, 2, 3) and B(4, 5, 6).
Answer: Distance formula in 3D: d = √[(x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²]
A(1, 2, 3), B(4, 5, 6)
d = √[(4-1)² + (5-2)² + (6-3)²]
d = √[3² + 3² + 3²]
d = √[9 + 9 + 9]
d = √27 = 3√3
Final Answer: Distance = 3√3 units
Q (2022, 2 marks): Find the equation of the line passing through point (1, 2, 3) with direction ratios 2, 3, 4.
Answer: Equation of a line passing through point (x₀, y₀, z₀) with direction ratios (a, b, c):
(x - x₀)/a = (y - y₀)/b = (z - z₀)/c
With point (1, 2, 3) and direction ratios 2, 3, 4:
(x - 1)/2 = (y - 2)/3 = (z - 3)/4
Parametric form: x = 1 + 2t, y = 2 + 3t, z = 3 + 4t
Final Answer: (x - 1)/2 = (y - 2)/3 = (z - 3)/4
Q (2023, 2 marks): Find the equation of the plane passing through the point (1, 2, 3) with normal vector (2, 3, 4).
Answer: Equation of plane with normal vector (a, b, c) passing through point (x₀, y₀, z₀):
a(x - x₀) + b(y - y₀) + c(z - z₀) = 0
With point (1, 2, 3) and normal (2, 3, 4):
2(x - 1) + 3(y - 2) + 4(z - 3) = 0
2x - 2 + 3y - 6 + 4z - 12 = 0
2x + 3y + 4z - 20 = 0
Final Answer: 2x + 3y + 4z = 20
Q (2021, 3 marks): Find the angle between the lines with direction ratios (1, 2, 2) and (2, 2, 1).
Answer: Direction ratios: l₁ = (1, 2, 2), l₂ = (2, 2, 1)
Angle θ between two lines with direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂):
cos(θ) = |a₁a₂ + b₁b₂ + c₁c₂|/[√(a₁² + b₁² + c₁²) × √(a₂² + b₂² + c₂²)]
Dot product: a₁a₂ + b₁b₂ + c₁c₂ = (1)(2) + (2)(2) + (2)(1) = 2 + 4 + 2 = 8
Magnitude 1: √(1² + 2² + 2²) = √9 = 3
Magnitude 2: √(2² + 2² + 1²) = √9 = 3
cos(θ) = |8|/(3 × 3) = 8/9
θ = arccos(8/9) ≈ 26.39°
Final Answer: cos(θ) = 8/9, θ ≈ 26.39°
Q (2022, 3 marks): Find the perpendicular distance from the point (1, 1, 1) to the plane 2x + 3y + 4z = 5.
Answer: Distance from point (x₀, y₀, z₀) to plane ax + by + cz + d = 0:
Distance = |ax₀ + by₀ + cz₀ + d|/√(a² + b² + c²)
Point: (1, 1, 1)
Plane: 2x + 3y + 4z = 5, or 2x + 3y + 4z - 5 = 0
Distance = |2(1) + 3(1) + 4(1) - 5|/√(2² + 3² + 4²)
Distance = |2 + 3 + 4 - 5|/√(4 + 9 + 16)
Distance = |4|/√29 = 4/√29 = 4√29/29
Final Answer: Distance = 4/√29 = 4√29/29 units
Q (2023, 3 marks): Find the distance between the parallel planes 2x + 3y + 4z = 5 and 2x + 3y + 4z = 15.
Answer: Distance between parallel planes ax + by + cz = d₁ and ax + by + cz = d₂:
Distance = |d₁ - d₂|/√(a² + b² + c²)
Planes: 2x + 3y + 4z = 5 and 2x + 3y + 4z = 15
d₁ = 5, d₂ = 15
Distance = |5 - 15|/√(2² + 3² + 4²)
Distance = |-10|/√29 = 10/√29 = 10√29/29
Final Answer: Distance = 10/√29 = 10√29/29 units
Frequently Asked Questions
What is the difference between direction ratios and direction cosines?
Direction ratios (a, b, c) are any numbers proportional to the direction of a line. Direction cosines (l, m, n) are the cosines of angles the line makes with positive x, y, z axes. Relation: l = a/√(a² + b² + c²), m = b/√(a² + b² + c²), n = c/√(a² + b² + c²). Always l² + m² + n² = 1. Direction cosines are unique; direction ratios are not.
What is the equation of a line in vector form?
Vector equation of a line passing through point A with position vector a and direction vector b is: r = a + tb, where r is the position vector of any point on the line and t is a scalar parameter. In Cartesian form: (x - x₀)/l = (y - y₀)/m = (z - z₀)/n where (x₀, y₀, z₀) is a point and (l, m, n) are direction cosines.
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