Three Dimensional Geometry — Class 12 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Three Dimensional Geometry" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Three Dimensional Geometry" — 8 important questions with detailed answers for CBSE…
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NCERT Solutions for Class 12 Mathematics — Three Dimensional Geometry. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Three Dimensional Geometry — All 8 Questions Solved
Q1. Find the distance of the point (1, 2, 3) from the origin.
To find the distance from P(1, 2, 3) to origin O(0, 0, 0):
Step 1: Use the distance formula.
For points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂):
d = √[(x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²]
Step 2: Substitute values.
Here: P(1, 2, 3) and O(0, 0, 0)
d = √[(0-1)² + (0-2)² + (0-3)²]
= √[1 + 4 + 9]
= √14
Final Answer: Distance = √14 units
Q2. Find the equation of the plane passing through points A(1, 2, 3), B(2, 3, 4), and C(3, 1, 2).
To find the equation of plane through A, B, C:
Step 1: Find vectors AB and AC.
AB = (2-1, 3-2, 4-3) = (1, 1, 1)
AC = (3-1, 1-2, 2-3) = (2, -1, -1)
Step 2: Find the normal vector n = AB × AC.
n = |î ĵ k̂ |
|1 1 1 |
|2 -1 -1 |
= î[1×(-1) - 1×(-1)] - ĵ[1×(-1) - 1×2] + k̂[1×(-1) - 1×2]
= î(-1 + 1) - ĵ(-1 - 2) + k̂(-1 - 2)
= 0î + 3ĵ - 3k̂
= (0, 3, -3)
Simplify: n = (0, 1, -1)
Step 3: Use point-normal form.
Equation: 0(x - 1) + 1(y - 2) - 1(z - 3) = 0
0 + y - 2 - z + 3 = 0
y - z + 1 = 0
Step 4: Verify with points.
A(1, 2, 3): 2 - 3 + 1 = 0 ✓
B(2, 3, 4): 3 - 4 + 1 = 0 ✓
C(3, 1, 2): 1 - 2 + 1 = 0 ✓
Final Answer: y - z + 1 = 0
Q3. Find the foot of perpendicular from point P(1, 3, 2) to the plane x + 2y - z = 5.
To find the foot of perpendicular from P(1, 3, 2) to plane x + 2y - z = 5:
Step 1: Find the equation of the line perpendicular to the plane.
Normal to plane: n = (1, 2, -1)
Line through P parallel to n:
(x, y, z) = (1, 3, 2) + t(1, 2, -1)
x = 1 + t
y = 3 + 2t
z = 2 - t
Step 2: Find intersection with the plane.
Substitute the parametric equations into plane equation:
(1 + t) + 2(3 + 2t) - (2 - t) = 5
1 + t + 6 + 4t - 2 + t = 5
5 + 6t = 5
6t = 0
t = 0
Step 3: Find the foot.
x = 1 + 0 = 1
y = 3 + 0 = 3
z = 2 - 0 = 2
Foot of perpendicular = (1, 3, 2)
Note: The point P itself lies on the plane. Verify: 1 + 2(3) - 2 = 1 + 6 - 2 = 5 ✓
Final Answer: Foot of perpendicular is (1, 3, 2)
Q4. Find the distance between the parallel planes x + 2y - 2z = 3 and x + 2y - 2z = 9.
To find distance between parallel planes ax + by + cz = d₁ and ax + by + cz = d₂:
Step 1: Verify planes are parallel.
Both planes: x + 2y - 2z = d
Normals are identical: n = (1, 2, -2)
Planes are parallel ✓
Step 2: Use distance formula.
For parallel planes ax + by + cz = d₁ and ax + by + cz = d₂:
distance = |d₁ - d₂|/√(a² + b² + c²)
Step 3: Substitute values.
Here: d₁ = 3, d₂ = 9
a = 1, b = 2, c = -2
distance = |3 - 9|/√(1² + 2² + (-2)²)
= |-6|/√(1 + 4 + 4)
= 6/√9
= 6/3
= 2
Final Answer: Distance = 2 units
Q5. Find the equation of the line passing through A(1, 2, 3) and B(4, 5, 6).
To find the equation of line through A(1, 2, 3) and B(4, 5, 6):
Step 1: Find direction vector.
d = B - A = (4-1, 5-2, 6-3) = (3, 3, 3)
Simplify: d = (1, 1, 1)
Step 2: Write the parametric form.
(x, y, z) = (1, 2, 3) + t(1, 1, 1)
x = 1 + t
y = 2 + t
z = 3 + t
Step 3: Write the symmetric form.
(x - 1)/1 = (y - 2)/1 = (z - 3)/1
Simplified: x - 1 = y - 2 = z - 3
Step 4: Verify with points.
At t = 0: (1, 2, 3) = A ✓
At t = 3: (1+3, 2+3, 3+3) = (4, 5, 6) = B ✓
Final Answer: x - 1 = y - 2 = z - 3 or (x, y, z) = (1, 2, 3) + t(1, 1, 1)
Q6. Find the angle between the planes x + 2y - z = 7 and 3x - 2y - z = 11.
To find the angle between two planes:
Step 1: Extract normal vectors.
Plane 1: x + 2y - z = 7, normal n₁ = (1, 2, -1)
Plane 2: 3x - 2y - z = 11, normal n₂ = (3, -2, -1)
Step 2: Use the formula.
cos θ = |n₁ · n₂|/(|n₁| × |n₂|)
Step 3: Calculate dot product.
n₁ · n₂ = (1)(3) + (2)(-2) + (-1)(-1)
= 3 - 4 + 1
= 0
Step 4: Find magnitudes.
|n₁| = √(1² + 2² + (-1)²) = √6
|n₂| = √(3² + (-2)² + (-1)²) = √14
Step 5: Calculate angle.
cos θ = |0|/(√6 × √14) = 0
θ = cos⁻¹(0) = π/2 = 90°
Final Answer: The planes are perpendicular; angle = 90° or π/2 radians
Q7. Find the distance of the point (1, 0, 1) from the line (x, y, z) = (1, 1, 1) + t(1, -1, 0).
To find distance from point P(1, 0, 1) to line:
Step 1: Identify line parameters.
Point on line: A(1, 1, 1)
Direction vector: d = (1, -1, 0)
Step 2: Find vector AP.
AP = P - A = (1-1, 0-1, 1-1) = (0, -1, 0)
Step 3: Use distance formula.
distance = |AP × d|/|d|
Step 4: Calculate cross product AP × d.
AP × d = |î ĵ k̂ |
|0 -1 0 |
|1 -1 0 |
= î[(-1)(0) - (0)(-1)] - ĵ[(0)(0) - (0)(1)] + k̂[(0)(-1) - (-1)(1)]
= î(0) - ĵ(0) + k̂(0 + 1)
= k̂ = (0, 0, 1)
Step 5: Calculate magnitudes.
|AP × d| = √(0² + 0² + 1²) = 1
|d| = √(1² + (-1)² + 0²) = √2
Step 6: Find distance.
distance = 1/√2 = √2/2
Final Answer: Distance = √2/2 units or 1/√2 units
Q8. Find the equation of the plane perpendicular to the line (x-1)/2 = (y+1)/3 = (z-2)/-1 and passing through (1, -1, 2).
To find the plane perpendicular to the given line:
Step 1: Find the normal to the plane.
The line has direction vector d = (2, 3, -1)
For a plane perpendicular to the line, the normal n = d = (2, 3, -1)
Step 2: Use point-normal form.
The plane passes through P(1, -1, 2) with normal n = (2, 3, -1)
Equation: 2(x - 1) + 3(y - (-1)) + (-1)(z - 2) = 0
2(x - 1) + 3(y + 1) - (z - 2) = 0
2x - 2 + 3y + 3 - z + 2 = 0
2x + 3y - z + 3 = 0
Step 3: Verify with given point.
At P(1, -1, 2): 2(1) + 3(-1) - 2 + 3 = 2 - 3 - 2 + 3 = 0 ✓
Final Answer: 2x + 3y - z + 3 = 0
Three Dimensional Geometry — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. If a line makes angles 90°, 135°, 45° with the positive x, y, and z-axes respectively, what are its direction cosines?
- A. (0, -1/√2, 1/√2)
- B. (0, 1/√2, -1/√2)
- C. (1, -1/√2, 1/√2)
- D. (0, -1, 1)
Answer: A. (0, -1/√2, 1/√2) — The direction cosines are given by cos α, cos β, cos γ. Here, α=90°, β=135°, γ=45°. So, cos 90°=0, cos 135°=-1/√2, cos 45°=1/√2.
MCQ 2. What are the direction ratios of the line passing through the points A(2, 3, -1) and B(3, -1, 2)?
- A. (1, -4, 3)
- B. (1, 2, 1)
- C. (5, 2, 1)
- D. (-1, 4, -3)
Answer: A. (1, -4, 3) — The direction ratios of a line passing through (x1, y1, z1) and (x2, y2, z2) are (x2-x1, y2-y1, z2-z1). So, (3-2, -1-3, 2-(-1)) = (1, -4, 3).
MCQ 3. The vector equation of a line passing through the point (1, 2, -4) and parallel to the vector (2î + 3ĵ - 5k̂) is:
- A. r⃗ = (î + 2ĵ - 4k̂) + λ(2î + 3ĵ - 5k̂)
- B. r⃗ = (2î + 3ĵ - 5k̂) + λ(î + 2ĵ - 4k̂)
- C. r⃗ = (î + 2ĵ - 4k̂) - λ(2î + 3ĵ - 5k̂)
- D. r⃗ = (î + 2ĵ - 4k̂) + λ(-2î - 3ĵ + 5k̂)
Answer: A. r⃗ = (î + 2ĵ - 4k̂) + λ(2î + 3ĵ - 5k̂) — The vector equation of a line passing through a point with position vector a⃗ and parallel to vector b⃗ is given by r⃗ = a⃗ + λb⃗. Here a⃗ = î + 2ĵ - 4k̂ and b⃗ = 2î + 3ĵ - 5k̂.
MCQ 4. The Cartesian equation of a line passing through (1, 2, 3) and having direction ratios (2, -1, 4) is:
- A. (x-1)/2 = (y-2)/(-1) = (z-3)/4
- B. (x+1)/2 = (y+2)/(-1) = (z+3)/4
- C. (x-2)/1 = (y+1)/2 = (z-4)/3
- D. (x-1)/2 = (y-2)/1 = (z-3)/4
Answer: A. (x-1)/2 = (y-2)/(-1) = (z-3)/4 — The Cartesian equation of a line passing through (x1, y1, z1) with direction ratios a, b, c is (x-x1)/a = (y-y1)/b = (z-z1)/c. Substituting the given values gives (x-1)/2 = (y-2)/(-1) = (z-3)/4.
MCQ 5. The angle between the lines (x-1)/2 = (y-2)/2 = (z-3)/1 and (x-5)/4 = (y-3)/1 = (z-1)/8 is:
- A. cos⁻¹(14/27)
- B. cos⁻¹(18/27)
- C. cos⁻¹(14/sqrt(81))
- D. cos⁻¹(18/sqrt(81))
Answer: B. cos⁻¹(18/27) — The direction ratios of the first line are (2, 2, 1) and of the second line are (4, 1, 8). The cosine of the angle θ between them is given by |a₁.a₂ + b₁.b₂ + c₁.c₂| / (√(a₁²+b₁²+c₁²) * √(a₂²+b₂²+c₂²)). This calculates to |2*4 + 2*1 + 1*8| / (√(2²+2²+1²) * √(4²+1²+8²)) = |8+2+8| / (√9 * √81) = 18 / (3*9) = 18/27.
MCQ 6. The shortest distance between the lines r⃗ = (î + ĵ) + λ(2î - ĵ + k̂) and r⃗ = (2î + ĵ - k̂) + μ(3î - 5ĵ + 2k̂) is:
- A. 10/√59 units
- B. 10/√59
- C. 1/√59
- D. 10/sqrt(59)
Answer: A. 10/√59 units — For lines r⃗ = a₁⃗ + λb₁⃗ and r⃗ = a₂⃗ + μb₂⃗, the shortest distance is |(a₂⃗ - a₁⃗) ⋅ (b₁⃗ × b₂⃗)| / |b₁⃗ × b₂⃗|. Calculation yields |(î - k̂) ⋅ (3î - ĵ - 7k̂)| / |3î - ĵ - 7k̂| = |3+7| / √59 = 10/√59 units.
MCQ 7. The shortest distance between the parallel lines r⃗ = (î + ĵ) + λ(2î + ĵ + 2k̂) and r⃗ = (2î + 3ĵ + k̂) + μ(2î + ĵ + 2k̂) is:
- A. √2 units
- B. 2 units
- C. 3 units
- D. √3 units
Answer: A. √2 units — For parallel lines r⃗ = a₁⃗ + λb⃗ and r⃗ = a₂⃗ + μb⃗, the shortest distance is |(b⃗ × (a₂⃗ - a₁⃗))| / |b⃗|. Here, a₁⃗ = î + ĵ, a₂⃗ = 2î + 3ĵ + k̂, b⃗ = 2î + ĵ + 2k̂. This results in a distance of √2 units.
MCQ 8. The equation of a plane at a distance of 5 units from the origin and having normal vector 2î - 3ĵ + 6k̂ is:
- A. (2/7)x - (3/7)y + (6/7)z = 5
- B. 2x - 3y + 6z = 5
- C. (2/7)x - (3/7)y + (6/7)z = 35
- D. 2x - 3y + 6z = 35
Answer: A. (2/7)x - (3/7)y + (6/7)z = 5 — The unit normal vector n̂ is (2î - 3ĵ + 6k̂) / √(2²+(-3)²+6²) = (2/7)î - (3/7)ĵ + (6/7)k̂. The equation of the plane is r⃗ ⋅ n̂ = d, so x(2/7) + y(-3/7) + z(6/7) = 5.
MCQ 9. The Cartesian equation of a plane passing through the point (2, -1, 3) and perpendicular to the vector 3î + 2ĵ - k̂ is:
- A. 3x + 2y - z - 1 = 0
- B. 3x + 2y - z + 1 = 0
- C. 3x + 2y - z - 6 = 0
- D. 2x - y + 3z - 11 = 0
Answer: A. 3x + 2y - z - 1 = 0 — The equation of a plane through (x₁, y₁, z₁) perpendicular to (a, b, c) is a(x-x₁) + b(y-y₁) + c(z-z₁) = 0. Substituting (2, -1, 3) and (3, 2, -1) gives 3(x-2) + 2(y+1) - 1(z-3) = 0, which simplifies to 3x + 2y - z - 1 = 0.
MCQ 10. The equation of the plane passing through the points (0, 0, 0), (1, 1, 0), and (1, 0, 1) is:
- A. x + y - z = 0
- B. x - y + z = 0
- C. -x + y + z = 0
- D. x + y + z = 0
Answer: C. -x + y + z = 0 — Let the equation be Ax + By + Cz + D = 0. Passing through (0,0,0) implies D=0. Using (1,1,0) gives A+B=0 (B=-A). Using (1,0,1) gives A+C=0 (C=-A). Substituting these into Ax+By+Cz=0 yields Ax-Ay-Az=0, or x-y-z=0, which is equivalent to -x+y+z=0.
MCQ 11. The intercepts made by the plane 3x - 4y + 2z = 12 on the coordinate axes are:
- A. x-intercept = 4, y-intercept = -3, z-intercept = 6
- B. x-intercept = 3, y-intercept = -4, z-intercept = 2
- C. x-intercept = -4, y-intercept = 3, z-intercept = -6
- D. x-intercept = 12, y-intercept = 12, z-intercept = 12
Answer: A. x-intercept = 4, y-intercept = -3, z-intercept = 6 — To get the intercept form x/a + y/b + z/c = 1, divide the equation 3x - 4y + 2z = 12 by 12. This gives x/4 + y/(-3) + z/6 = 1, so the intercepts are 4, -3, and 6 respectively.
MCQ 12. The equation of the plane passing through the intersection of the planes x + y + z = 6 and 2x + 3y + 4z = -5, and through the point (1, 1, 1) is:
- A. 20x + 23y + 26z - 69 = 0
- B. 20x + 23y + 26z + 69 = 0
- C. x + y + z = 3
- D. 2x + 3y + 4z = 9
Answer: A. 20x + 23y + 26z - 69 = 0 — The equation of the plane is (x + y + z - 6) + λ(2x + 3y + 4z + 5) = 0. Substituting (1, 1, 1) gives (-3) + λ(14) = 0, so λ = 3/14. Plugging λ back results in 20x + 23y + 26z - 69 = 0.
MCQ 13. The distance of the point (2, 3, -5) from the plane x + 2y - 2z = 8 is:
- A. 10/3 units
- B. 10 units
- C. 9/3 units
- D. 1 units
Answer: A. 10/3 units — The distance of a point (x₁, y₁, z₁) from Ax + By + Cz + D = 0 is |Ax₁ + By₁ + Cz₁ + D| / √(A² + B² + C²). Substituting (2, 3, -5) into x + 2y - 2z - 8 = 0 gives |1(2) + 2(3) - 2(-5) - 8| / √(1²+2²+(-2)²) = |2+6+10-8| / 3 = 10/3 units.
MCQ 14. The angle between the planes 2x - y + z = 6 and x + y + 2z = 7 is:
- A. cos⁻¹(1/3)
- B. cos⁻¹(√3/2)
- C. cos⁻¹(2/3)
- D. cos⁻¹(1/2)
Answer: D. cos⁻¹(1/2) — The angle between two planes is the angle between their normal vectors, n₁⃗ = 2î - ĵ + k̂ and n₂⃗ = î + ĵ + 2k̂. cos θ = |n₁⃗ ⋅ n₂⃗| / (|n₁⃗| |n₂⃗|) = |2-1+2| / (√6 * √6) = 3/6 = 1/2. So, θ = cos⁻¹(1/2).
MCQ 15. The angle between the line (x+1)/2 = y/3 = (z-3)/6 and the plane 10x + 2y - 11z = 3 is:
- A. sin⁻¹(8/21)
- B. sin⁻¹(1/21)
- C. cos⁻¹(8/21)
- D. sin⁻¹(√3/2)
Answer: A. sin⁻¹(8/21) — The angle θ between a line (direction vector b⃗) and a plane (normal vector n⃗) is given by sin θ = |b⃗ ⋅ n⃗| / (|b⃗| |n⃗|). Here, b⃗ = 2î + 3ĵ + 6k̂ (|b⃗|=7) and n⃗ = 10î + 2ĵ - 11k̂ (|n⃗|=15). sin θ = |20+6-66| / (7*15) = |-40| / 105 = 40/105 = 8/21.
MCQ 16. The lines (x-1)/k = (y-2)/3 = (z-3)/4 and (x-2)/1 = (y-3)/(-4) = (z-5)/2 are perpendicular if k is:
- A. 2
- B. 3
- C. 4
- D. 5
Answer: C. 4 — Two lines with direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂) are perpendicular if a₁a₂ + b₁b₂ + c₁c₂ = 0. So, k(1) + 3(-4) + 4(2) = 0. This simplifies to k - 12 + 8 = 0, which means k = 4.
MCQ 17. For what value of p are the lines r⃗ = (î + 2ĵ + 3k̂) + λ(-3î + 2pĵ + 2k̂) and r⃗ = (î + 5ĵ + 6k̂) + μ(6î - 4ĵ - 4k̂) parallel?
- A. p = -1
- B. p = 1
- C. p = -2
- D. p = 2
Answer: B. p = 1 — Two lines are parallel if their direction vectors are proportional. So, (-3î + 2pĵ + 2k̂) = k(6î - 4ĵ - 4k̂). Comparing coefficients: -3 = 6k (k=-1/2), 2p = -4k (2p = -4(-1/2) = 2, so p=1), and 2 = -4k (k=-1/2). All conditions are consistent for p=1.
MCQ 18. Which of the following sets of points are collinear?
- A. (1,2,3), (2,3,4), (3,4,5)
- B. (1,0,0), (0,1,0), (0,0,1)
- C. (1,1,1), (1,2,3), (2,1,3)
- D. (0,0,0), (1,1,0), (2,2,1)
Answer: A. (1,2,3), (2,3,4), (3,4,5) — Points are collinear if the direction ratios between any two pairs of points are proportional. For (1,2,3), (2,3,4), (3,4,5), the direction ratios for (2,3,4)-(1,2,3) are (1,1,1) and for (3,4,5)-(2,3,4) are also (1,1,1), showing they are collinear.
MCQ 19. The lines (x-1)/2 = (y-2)/3 = (z-3)/4 and (x-2)/3 = (y-3)/4 = (z-4)/5 are:
- A. Coplanar
- B. Skew
- C. Perpendicular
- D. Parallel
Answer: A. Coplanar — Two lines r⃗ = a₁⃗ + λb₁⃗ and r⃗ = a₂⃗ + μb₂⃗ are coplanar if (a₂⃗ - a₁⃗) ⋅ (b₁⃗ × b₂⃗) = 0. Here, a₂⃗ - a₁⃗ = î + ĵ + k̂ and b₁⃗ × b₂⃗ = -î + 2ĵ - k̂. Their dot product is (1)(-1) + (1)(2) + (1)(-1) = 0, indicating they are coplanar.
MCQ 20. The foot of the perpendicular from the origin (0, 0, 0) to the plane x + y + z = 3 is:
- A. (1, 1, 1)
- B. (1, -1, 1)
- C. (0, 0, 0)
- D. (1/3, 1/3, 1/3)
Answer: A. (1, 1, 1) — The line from the origin perpendicular to the plane x+y+z=3 is x/1 = y/1 = z/1 = λ. Substituting a point (λ,λ,λ) into the plane equation (λ+λ+λ=3) gives λ=1. So the foot of the perpendicular is (1,1,1).
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.