Class 12 Mathematics Notes Chapter 11 (Three Dimensional Geometry) – Examplar Problems (English) Book

Alright class, let's get straight into the essential concepts of Three Dimensional Geometry from your NCERT Class 12 syllabus, focusing particularly on the types of problems and nuances highlighted in the Exemplar book, which are crucial for competitive government exams.

Chapter 11: Three Dimensional Geometry - Key Concepts & Formulas

1. Direction Cosines (DCs) and Direction Ratios (DRs) of a Line

• Direction Angles: If a directed line L passing through the origin makes angles α, β, and γ with the positive x, y, and z axes respectively, then α, β, γ are called direction angles.
• Direction Cosines (DCs): The cosines of the direction angles, i.e., l = cos α, m = cos β, n = cos γ, are called the direction cosines of the line L.
  • Key Property: l² + m² + n² = 1.
  • If P(x, y, z) is a point on line L such that OP = r, then x = lr, y = mr, z = nr.
  • DCs of the x-axis are (1, 0, 0), y-axis are (0, 1, 0), z-axis are (0, 0, 1).
• Direction Ratios (DRs): Any three numbers a, b, c proportional to the direction cosines l, m, n are called direction ratios.
  • l/a = m/b = n/c = k (for some constant k)
  • If a, b, c are DRs, then DCs are:
    l = ± a / √(a² + b² + c²),
    m = ± b / √(a² + b² + c²),
    n = ± c / √(a² + b² + c²)
    (The sign is chosen based on the direction of the line, often taken as positive).
• DRs of a line segment joining P(x₁, y₁, z₁) and Q(x₂, y₂, z₂): The DRs are (x₂ - x₁), (y₂ - y₁), (z₂ - z₁).

2. Equation of a Line in Space

• Equation of a line passing through a point A (position vector a) and parallel to a vector b:
  • Vector Form: r = a + λb (where r is the position vector of any point on the line, λ is a scalar parameter).
  • Cartesian Form: If A is (x₁, y₁, z₁) and DRs of the parallel vector b are (a, b, c), the equation is:
    (x - x₁) / a = (y - y₁) / b = (z - z₁) / c
    (Important Note: Ensure coefficients of x, y, z are 1 before reading DRs from this form).
• Equation of a line passing through two points A (position vector a) and B (position vector b):
  • Vector Form: r = a + λ(b - a)
  • Cartesian Form: If A is (x₁, y₁, z₁) and B is (x₂, y₂, z₂), the equation is:
    (x - x₁) / (x₂ - x₁) = (y - y₁) / (y₂ - y₁) = (z - z₁) / (z₂ - z₁)

3. Angle Between Two Lines

• Let the lines be r = a₁ + λb₁ and r = a₂ + μb₂. The angle θ between them is the angle between their parallel vectors b₁ and b₂.
  • Vector Form: cos θ = |(b₁ . b₂) / (|b₁| |b₂|)|
• Let the lines have DRs (a₁, b₁, c₁) and (a₂, b₂, c₂).
  • Cartesian Form (using DRs): cos θ = |(a₁a₂ + b₁b₂ + c₁c₂) / (√(a₁² + b₁² + c₁²) √(a₂² + b₂² + c₂²))|
• Using DCs (l₁, m₁, n₁) and (l₂, m₂, n₂): cos θ = |l₁l₂ + m₁m₂ + n₁n₂|
• Conditions:
  • Perpendicular Lines: b₁ . b₂ = 0 or a₁a₂ + b₁b₂ + c₁c₂ = 0.
  • Parallel Lines: b₁ = k b₂ (for some scalar k) or a₁/a₂ = b₁/b₂ = c₁/c₂.

4. Shortest Distance Between Two Lines

• Skew Lines: Lines that are neither parallel nor intersecting.
  • Lines: r = a₁ + λb₁ and r = a₂ + μb₂.
  • Shortest Distance (SD): SD = |((a₂ - a₁) . (b₁ x b₂)) / |b₁ x b₂||
  • Condition for Intersection: If the lines intersect, SD = 0, which implies (a₂ - a₁) . (b₁ x b₂) = 0.
  • Cartesian Form (Determinant): For lines (x-x₁)/a₁ = ... and (x-x₂)/a₂ = ..., SD is given by:
    SD = |det(A)| / √((a₁b₂-a₂b₁)² + (b₁c₂-b₂c₁)² + (c₁a₂-c₂a₁)²)
    where A = [[x₂-x₁, y₂-y₁, z₂-z₁], [a₁, b₁, c₁], [a₂, b₂, c₂]]. The vector form is generally easier to compute.
• Parallel Lines: Lines r = a₁ + λb and r = a₂ + μb.
  • Shortest Distance (SD): SD = |((a₂ - a₁) x b) / |b||

5. Equation of a Plane

• Equation of a plane in Normal Form:
  • Vector Form: r . n̂ = d (where n̂ is the unit vector normal to the plane from the origin, and d is the perpendicular distance of the plane from the origin, d ≥ 0).
  • Cartesian Form: lx + my + nz = d (where l, m, n are the DCs of the normal to the plane, and d is the distance from the origin).
• Equation of a plane perpendicular to a given vector N and passing through a point A (position vector a):
  • Vector Form: (r - a) . N = 0 or r . N = a . N (Here a . N is often written as q).
  • Cartesian Form: If A is (x₁, y₁, z₁) and DRs of the normal N are (A, B, C), the equation is:
    A(x - x₁) + B(y - y₁) + C(z - z₁) = 0
    This simplifies to the General Form: Ax + By + Cz + D = 0, where D = -(Ax₁ + By₁ + Cz₁). Here (A, B, C) are DRs of the normal.
• Equation of a plane passing through three non-collinear points A(a), B(b), C(c):
  • Vector Form: (r - a) . ((b - a) x (c - a)) = 0
  • Cartesian Form: If points are (x₁, y₁, z₁), (x₂, y₂, z₂), (x₃, y₃, z₃), the equation is:
    | (x - x₁) (y - y₁) (z - z₁) |
| (x₂ - x₁) (y₂ - y₁) (z₂ - z₁) | = 0
| (x₃ - x₁) (y₃ - y₁) (z₃ - z₁) |
• Intercept Form: Equation of a plane making intercepts a, b, c on the x, y, z axes respectively:
  x/a + y/b + z/c = 1

6. Coplanarity of Two Lines

• Two lines r = a₁ + λb₁ and r = a₂ + μb₂ are coplanar if the vector joining a point on each line (a₂ - a₁) is perpendicular to the vector normal to the plane containing both direction vectors (b₁ x b₂).
• Condition: (a₂ - a₁) . (b₁ x b₂) = 0. (This is the same condition as intersection for non-parallel lines).
• Cartesian Condition: Lines (x-x₁)/a₁ = ... and (x-x₂)/a₂ = ... are coplanar if:
  | (x₂ - x₁) (y₂ - y₁) (z₂ - z₁) |
| a₁ b₁ c₁ | = 0
| a₂ b₂ c₂ |

7. Angle Between Two Planes

• Let the planes be r . N₁ = d₁ and r . N₂ = d₂. The angle θ between the planes is the angle between their normals N₁ and N₂.
  • Vector Form: cos θ = |(N₁ . N₂) / (|N₁| |N₂|)|
• Let the planes be A₁x + B₁y + C₁z + D₁ = 0 and A₂x + B₂y + C₂z + D₂ = 0.
  • Cartesian Form: cos θ = |(A₁A₂ + B₁B₂ + C₁C₂) / (√(A₁² + B₁² + C₁²) √(A₂² + B₂² + C₂²))|
• Conditions:
  • Perpendicular Planes: N₁ . N₂ = 0 or A₁A₂ + B₁B₂ + C₁C₂ = 0.
  • Parallel Planes: N₁ = k N₂ or A₁/A₂ = B₁/B₂ = C₁/C₂.

8. Distance of a Point from a Plane

• Distance of point P (position vector a) from the plane r . N = q:
  • Vector Form: Distance = |(a . N - q) / |N||
• Distance of point P(x₁, y₁, z₁) from the plane Ax + By + Cz + D = 0:
  • Cartesian Form: Distance = |(Ax₁ + By₁ + Cz₁ + D) / √(A² + B² + C²)|
• Distance between two parallel planes Ax + By + Cz + D₁ = 0 and Ax + By + Cz + D₂ = 0:
  • Distance = |(D₁ - D₂) / √(A² + B² + C²)|

9. Angle Between a Line and a Plane

• Let the line be r = a + λb and the plane be r . N = q.
• Let φ be the angle between the line and the plane. Let θ be the angle between the line and the normal to the plane (b and N). Then φ = 90° - θ.
• cos θ = |(b . N) / (|b| |N|)|
• Therefore, sin φ = cos(90° - φ) = cos θ = |(b . N) / (|b| |N|)|
• Cartesian Form: Line (x-x₁)/a = ..., Plane Ax + By + Cz + D = 0. DRs of line are (a, b, c), DRs of normal are (A, B, C).
  • sin φ = |(aA + bB + cC) / (√(a² + b² + c²) √(A² + B² + C²))|
• Conditions:
  • Line Parallel to Plane: The line is perpendicular to the normal. b . N = 0 or aA + bB + cC = 0.
  • Line Perpendicular to Plane: The line is parallel to the normal. b = k N or a/A = b/B = c/C.

10. Plane Passing Through the Intersection of Two Planes

• The equation of any plane passing through the intersection of planes P₁: r . N₁ = d₁ and P₂: r . N₂ = d₂ is given by:
  r . (N₁ + λN₂) = d₁ + λd₂ (where λ is a scalar parameter).
• Cartesian Form: For planes A₁x + B₁y + C₁z + D₁ = 0 and A₂x + B₂y + C₂z + D₂ = 0, the equation of a plane passing through their intersection is:
  (A₁x + B₁y + C₁z + D₁) + λ(A₂x + B₂y + C₂z + D₂) = 0
  The value of λ is found using an additional condition (e.g., the plane passes through a given point).

11. Intersection of a Line and a Plane

• To find the point of intersection of a line r = a + λb (or (x-x₁)/a = ... = λ) and a plane r . N = q (or Ax + By + Cz + D = 0):
  1. Write the coordinates of a general point on the line in terms of λ: (x₁ + aλ, y₁ + bλ, z₁ + cλ).
  2. Substitute these coordinates into the equation of the plane.
  3. Solve the resulting linear equation for λ.
  4. Substitute the value of λ back into the coordinates of the general point to get the intersection point.

Multiple Choice Questions (MCQs)

1. If the direction cosines of a line are (k, k, k), then:
   (A) k > 0
   (B) 0 < k < 1
   (C) k = 1
   (D) k = 1/√3 or k = -1/√3

2. The distance of the plane r . (2î - 3ĵ + 6k̂) + 14 = 0 from the origin is:
   (A) 14
   (B) 7
   (C) 2
   (D) 49

3. The angle between the lines (x-2)/1 = (y+1)/-2 = (z)/3 and (x)/-2 = (y-1)/4 = (z-2)/-6 is:
   (A) 0°
   (B) 90°
   (C) 180°
   (D) 60°

4. The vector equation of the line passing through the point (1, 2, -4) and perpendicular to the two lines (x-8)/3 = (y+19)/-16 = (z-10)/7 and (x-15)/3 = (y-29)/8 = (z-5)/-5 is:
   (A) r = (î + 2ĵ - 4k̂) + λ(2î + 3ĵ + 6k̂)
   (B) r = (î + 2ĵ - 4k̂) + λ(2î - 3ĵ + 6k̂)
   (C) r = (î + 2ĵ - 4k̂) + λ(3î - 16ĵ + 7k̂)
   (D) r = (î + 2ĵ - 4k̂) + λ(3î + 8ĵ - 5k̂)

5. The shortest distance between the lines r = (î + ĵ) + λ(2î - ĵ + k̂) and r = (2î + ĵ - k̂) + μ(3î - 5ĵ + 2k̂) is:
   (A) 0
   (B) 10/√59
   (C) √59/10
   (D) 5/√59

6. The equation of the plane passing through the point (1, 1, 1) and perpendicular to the line (x-1)/2 = (y+1)/3 = (z)/-1 is:
   (A) 2x + 3y - z = 4
   (B) x + y + z = 3
   (C) 2x + 3y - z = 0
   (D) x + y - z = 1

7. The coordinates of the foot of the perpendicular drawn from the point (2, -1, 5) to the line (x-11)/10 = (y+2)/-4 = (z+8)/-11 are:
   (A) (1, 2, 3)
   (B) (1, -6, 3)
   (C) (-1, 6, -3)
   (D) (1, 2, -3)

8. The angle between the planes 2x - y + z = 6 and x + y + 2z = 7 is:
   (A) π/6
   (B) π/4
   (C) π/3
   (D) π/2

9. The distance between the parallel planes 2x - 2y + z + 3 = 0 and 4x - 4y + 2z + 18 = 0 is:
   (A) 3
   (B) 6
   (C) 2
   (D) 4

10. The lines (x-1)/2 = (y-2)/3 = (z-3)/4 and (x-1)/3 = (y-3)/4 = (z-k)/5 are coplanar if k equals:
    (A) 3
    (B) 4
    (C) 5
    (D) 6

Answers to MCQs:

1. (D) [Hint: Use l² + m² + n² = 1, so k² + k² + k² = 1 => 3k² = 1]
2. (C) [Hint: Equation is r . N = -14. Distance = |-14| / |N|. |N| = √(2² + (-3)² + 6²) = √49 = 7. Distance = 14/7 = 2]
3. (C) [Hint: DRs are (1, -2, 3) and (-2, 4, -6). Notice that -2 * (1, -2, 3) = (-2, 4, -6). The lines are parallel. Since the directions are opposite, the angle can be considered 180° or 0° depending on interpretation, but parallel implies 0° or 180°. Often 0° is used for parallel vectors, but since the DRs point opposite ways, 180° is appropriate here. Let's recheck standard convention. Parallel vectors b1 = k b2. If k>0, angle is 0. If k<0, angle is 180. Here k=-2. So 180°.]
4. (A) [Hint: The line is perpendicular to both given lines, so its direction vector b is parallel to the cross product of the direction vectors of the given lines: b₁ = (3, -16, 7) and b₂ = (3, 8, -5). b₁ x b₂ = (80-56)î - (-15-21)ĵ + (24 - (-48))k̂ = 24î + 36ĵ + 72k̂. The DRs can be simplified to (2, 3, 6). Line is r = (î + 2ĵ - 4k̂) + λ(2î + 3ĵ + 6k̂).]
5. (B) [Hint: a₁ = (1, 1, 0), b₁ = (2, -1, 1). a₂ = (2, 1, -1), b₂ = (3, -5, 2). a₂ - a₁ = (1, 0, -1). b₁ x b₂ = ((-1)(2) - (1)(-5))î - ((2)(2) - (1)(3))ĵ + ((2)(-5) - (-1)(3))k̂ = 3î - ĵ - 7k̂. |b₁ x b₂| = √(9 + 1 + 49) = √59. (a₂ - a₁) . (b₁ x b₂) = (1)(3) + (0)(-1) + (-1)(-7) = 3 + 7 = 10. SD = |10 / √59| = 10/√59.]
6. (A) [Hint: The normal to the plane has DRs (2, 3, -1). The plane passes through (1, 1, 1). Equation: 2(x-1) + 3(y-1) - 1(z-1) = 0 => 2x - 2 + 3y - 3 - z + 1 = 0 => 2x + 3y - z - 4 = 0 or 2x + 3y - z = 4.]
7. (D) [Hint: General point on the line is (11 + 10λ, -2 - 4λ, -8 - 11λ). Let this be Q. Let P = (2, -1, 5). DRs of PQ are (9 + 10λ, -1 - 4λ, -13 - 11λ). Since PQ is perpendicular to the line, the dot product of their DRs is 0: 10(9 + 10λ) - 4(-1 - 4λ) - 11(-13 - 11λ) = 0. 90 + 100λ + 4 + 16λ + 143 + 121λ = 0. 237λ + 237 = 0 => λ = -1. Foot Q = (11 - 10, -2 + 4, -8 + 11) = (1, 2, 3). Let me recheck the calculation. 10(9+10λ) = 90+100λ. -4(-1-4λ) = 4+16λ. -11(-13-11λ) = 143+121λ. Sum = (100+16+121)λ + (90+4+143) = 237λ + 237. So λ=-1. Point is (11+10(-1), -2-4(-1), -8-11(-1)) = (11-10, -2+4, -8+11) = (1, 2, 3). My previous answer D was (1, 2, -3). Let's recheck the options. (A) is (1, 2, 3). Okay, the calculation leads to (A). Let me re-verify the question and calculation once more. Point (2, -1, 5). Line (x-11)/10 = (y+2)/-4 = (z+8)/-11. General point Q (11+10λ, -2-4λ, -8-11λ). P(2, -1, 5). PQ DRs (11+10λ-2, -2-4λ-(-1), -8-11λ-5) = (9+10λ, -1-4λ, -13-11λ). Line DRs (10, -4, -11). Dot product: 10(9+10λ) -4(-1-4λ) -11(-13-11λ) = 0. 90+100λ + 4+16λ + 143+121λ = 0. 237λ + 237 = 0. λ = -1. Foot Q = (11+10(-1), -2-4(-1), -8-11(-1)) = (1, 2, 3). The correct answer is (A). I will correct the answer key.
8. (C) [Hint: Normals N₁ = (2, -1, 1), N₂ = (1, 1, 2). N₁ . N₂ = 2(1) + (-1)(1) + 1(2) = 2 - 1 + 2 = 3. |N₁| = √(4+1+1) = √6. |N₂| = √(1+1+4) = √6. cos θ = |3 / (√6 √6)| = 3/6 = 1/2. θ = π/3.]
9. (C) [Hint: Rewrite second plane as 2x - 2y + z + 9 = 0 (dividing by 2). Now planes are 2x - 2y + z + 3 = 0 and 2x - 2y + z + 9 = 0. D₁ = 3, D₂ = 9. A=2, B=-2, C=1. Distance = |3 - 9| / √(2² + (-2)² + 1²) = |-6| / √(4+4+1) = 6 / √9 = 6/3 = 2.]
10. (C) [Hint: Use the coplanarity condition. x₁=1, y₁=2, z₁=3. a₁=2, b₁=3, c₁=4. x₂=1, y₂=3, z₂=k. a₂=3, b₂=4, c₂=5. Determinant:
    | (1-1) (3-2) (k-3) |
| 2 3 4 | = 0
| 3 4 5 |
| 0 1 (k-3) |
| 2 3 4 | = 0
| 3 4 5 |
    Expand along R1: 0(...) - 1(2*5 - 4*3) + (k-3)(2*4 - 3*3) = 0.
    -1(10 - 12) + (k-3)(8 - 9) = 0.
    -1(-2) + (k-3)(-1) = 0.
    2 - k + 3 = 0.
    5 - k = 0 => k = 5.]

(Corrected Answer for Q7): (A)

Study these concepts thoroughly, paying close attention to the different forms of equations and the conditions for parallel/perpendicular lines and planes. Practice problems from the Exemplar book to master the application of these formulas, especially for shortest distance, coplanarity, and intersection problems. Good luck!