Class 9 Mathematics Chapter 11: Three Dimensional Geometry Important Questions & Answers

Three Dimensional Geometry (Chapter 11) bridges 2D coordinate geometry (Class 10) and advanced calculus. In the CBSE 2026-27 board exam pattern, this chapter typically carries 5–8 marks with questions spanning multiple difficulty levels. The 2024-25 rationalized syllabus emphasizes: (1) Direction cosines and direction ratios, (2) equations of a line in 3D (Cartesian and vector forms), and (3) equation of a plane. Examiners test both formula recall and conceptual problem-solving—for example, finding the angle between two lines, determining if points are collinear in 3D space, or finding the perpendicular distance from a point to a plane. Students who master direction cosines early gain an edge: these concepts appear in vector algebra, analytical geometry revision, and even in physics (mechanics, optics). Practising varied question types trains your brain to recognize which formula applies in each scenario, cutting down exam-room hesitation.

**Question 1:** The direction cosines of the line joining points A(1, 2, 3) and B(4, 5, 6) are proportional to: (a) 1 : 2 : 3 (b) 3 : 3 : 3 (c) 4 : 5 : 6 (d) (4−1) : (5−2) : (6−3) = 3 : 3 : 3 **Answer: (d) 3 : 3 : 3** — Direction ratios are the differences: (4−1, 5−2, 6−3) = (3, 3, 3) or simplified (1, 1, 1). **Question 2:** If l, m, n are direction cosines of a line, then l² + m² + n² = ? (a) 0 (b) 1 (c) 2 (d) 3 **Answer: (b) 1** — Fundamental property of direction cosines: the sum of squares always equals unity. **Question 3:** The direction cosines of the z-axis are: (a) (1, 0, 0) (b) (0, 1, 0) (c) (0, 0, 1) (d) (1, 1, 1) **Answer: (c) (0, 0, 1)** — The z-axis has no x or y component, only z. **Question 4:** A line passes through (2, −1, 3) and (5, 2, 9). Its direction ratios are: (a) 2 : −1 : 3 (b) 3 : 3 : 6 (c) 5 : 2 : 9 (d) 1 : 1 : 2 **Answer: (b) 3 : 3 : 6** — Direction ratios = (5−2) : (2−(−1)) : (9−3) = 3 : 3 : 6 or simplified 1 : 1 : 2. **Question 5:** The Cartesian equation of a plane with normal vector (1, 2, 3) passing through (0, 0, 0) is: (a) x + 2y + 3z = 0 (b) x + 2y + 3z = 5 (c) x − 2y + 3z = 1 (d) 2x + 3y + z = 0 **Answer: (a) x + 2y + 3z = 0** — Plane equation: a(x − x₀) + b(y − y₀) + c(z − z₀) = 0 gives 1(x) + 2(y) + 3(z) = 0.

**Question 1:** Find the direction ratios and direction cosines of the line joining A(1, 1, 1) and B(3, 5, 7). **Solution:** Direction ratios = (3−1, 5−1, 7−1) = (2, 4, 6) or simplified (1, 2, 3). Distance = √(1² + 2² + 3²) = √14. Direction cosines = (1/√14, 2/√14, 3/√14). **Question 2:** If a line has direction cosines l = 1/3, m = 2/3, find n. **Solution:** Using l² + m² + n² = 1: (1/3)² + (2/3)² + n² = 1 ⟹ 1/9 + 4/9 + n² = 1 ⟹ n² = 4/9 ⟹ n = ±2/3. **Question 3:** Write the equation of the plane with normal vector (1, −2, 3) passing through point P(1, 2, 3). **Solution:** Plane equation: 1(x − 1) − 2(y − 2) + 3(z − 3) = 0 ⟹ x − 2y + 3z − 1 + 4 − 9 = 0 ⟹ x − 2y + 3z = 6. **Question 4:** Are the points A(1, 2, 3), B(2, 4, 6), and C(3, 6, 9) collinear? Justify. **Solution:** Check if AB and AC are parallel. AB direction = (1, 2, 3), AC direction = (2, 4, 6) = 2(1, 2, 3). Since AC = 2·AB, the vectors are parallel, so points are **collinear**. **Question 5:** Find the equation of a plane passing through the origin and perpendicular to the line with direction ratios 2 : 3 : 4. **Solution:** If a plane is perpendicular to a line, the line's direction ratios are the plane's normal. Plane: 2x + 3y + 4z = 0 (passes through origin with normal (2, 3, 4)).

**Question 1:** Show that the lines with direction ratios 1 : 2 : 3 and 2 : 4 : 6 are parallel. Also, write the Cartesian equation of a line with these direction ratios passing through (1, 1, 1). **Solution:** For parallel lines, direction ratios must be proportional. Here, (2, 4, 6) = 2(1, 2, 3), so they are **parallel**. Cartesian equation: (x − 1)/1 = (y − 1)/2 = (z − 1)/3. Simplifying: 2(x − 1) = (y − 1) and 3(y − 1) = 2(z − 1) give 2x − y = 1 and 3y − 2z = −2. **Question 2:** A plane has equation 2x − 3y + 4z = 12. Find the perpendicular distance from point P(1, −1, 2) to this plane. **Solution:** Formula: d = |ax₀ + by₀ + cz₀ − d| / √(a² + b² + c²). Here, d = |2(1) − 3(−1) + 4(2) − 12| / √(4 + 9 + 16) = |2 + 3 + 8 − 12| / √29 = |1| / √29 = **1/√29** or **√29/29** units. **Question 3:** Find the angle between two lines with direction ratios a₁ : b₁ : c₁ = 1 : 2 : 2 and a₂ : b₂ : c₂ = 2 : 3 : 6. **Solution:** cos θ = |a₁a₂ + b₁b₂ + c₁c₂| / (√(a₁² + b₁² + c₁²) × √(a₂² + b₂² + c₂²)) = |1(2) + 2(3) + 2(6)| / (√(1 + 4 + 4) × √(4 + 9 + 36)) = |2 + 6 + 12| / (3 × 7) = 20/21. θ = cos⁻¹(20/21) ≈ **9.59°**. **Question 4:** Write the equation of the plane passing through points A(1, 0, 0), B(0, 1, 0), and C(0, 0, 1). (Intercept form) **Solution:** Intercept form: x/a + y/b + z/c = 1. Here, a = 1, b = 1, c = 1. Plane equation: **x + y + z = 1**. Verify: A(1, 0, 0) ✓, B(0, 1, 0) ✓, C(0, 0, 1) ✓.

**Question 1:** Find the equation of the line passing through point (2, −1, 1) and parallel to the line (x − 1)/2 = (y + 1)/3 = (z − 2)/4. Also, verify that the point (6, 2, 5) lies on this line. **Full Solution:** The given line has direction ratios (2, 3, 4). Since our line is parallel, it has the same direction ratios. Passing through (2, −1, 1), the Cartesian form is: (x − 2)/2 = (y − (−1))/3 = (z − 1)/4 ⟹ **(x − 2)/2 = (y + 1)/3 = (z − 1)/4** To verify (6, 2, 5): (6 − 2)/2 = 4/2 = 2 (2 + 1)/3 = 3/3 = 1 (5 − 1)/4 = 4/4 = 1 Since 2 ≠ 1, the point **(6, 2, 5) does NOT lie on the line**. (If the question intended (6, 2, 3): (6−2)/2=2, (2+1)/3=1, (3−1)/4=0.5—still doesn't satisfy. Let's check (6, 2, 5) with a corrected line through (2, −1, 1) with ratios 2:3:4: At k=2, point is (2+4, −1+6, 1+8) = (6, 5, 9), not (6, 2, 5). Point verification fails as stated.) **Question 2:** Two planes are given: Π₁: x + 2y + 3z = 9 and Π₂: 2x + 4y + 6z = 15. Are these planes parallel? If they intersect, find the equation of the line of intersection. **Full Solution:** Normals: n₁ = (1, 2, 3) and n₂ = (2, 4, 6) = 2(1, 2, 3). Since n₂ = 2n₁, the normals are parallel, so **planes are parallel**. Check if they coincide: x + 2y + 3z = 9 vs. (2x + 4y + 6z)/2 = 15/2 = 7.5. Since 9 ≠ 7.5, the planes are **distinct and parallel** (no line of intersection). **Question 3:** Find the perpendicular distance from point P(1, 2, 3) to the plane x + y + z = 6. Also, find the foot of the perpendicular from P to the plane. **Full Solution:** Distance: d = |1 + 2 + 3 − 6| / √(1² + 1² + 1²) = |0| / √3 = **0**. Since distance = 0, point P(1, 2, 3) **lies on the plane** (verify: 1 + 2 + 3 = 6 ✓). Therefore, the foot of perpendicular is **P itself: (1, 2, 3)**. Alternative example: If P(2, 3, 4), distance = |2 + 3 + 4 − 6| / √3 = 3/√3 = √3 units. Foot of perpendicular: along normal (1, 1, 1), foot = P − d·(n/|n|) = (2, 3, 4) − √3·(1/√3, 1/√3, 1/√3) = (2, 3, 4) − (1, 1, 1) = **(1, 2, 3)**.

**Case Study: 3D Architecture Problem** An architect designs a roof structure with three corner points: A(0, 0, 0), B(4, 0, 0), and C(2, 2√3, 0). A support beam runs from vertex D(2, √3, h) perpendicular to the triangular base ABC. (a) Find the direction ratios of line AD. [1 mark] (b) Show that triangle ABC is equilateral and find its area. [2 marks] (c) If the support beam (line from D perpendicular to plane ABC) has direction ratio (0, 0, 1), find the equation of the plane containing A, B, C. [2 marks] **Solution:** (a) Direction ratios of AD: (2 − 0, √3 − 0, h − 0) = **(2, √3, h)**. (b) AB = √(16) = 4, BC = √((2−4)² + (2√3)²) = √(4 + 12) = 4, AC = √(4 + 12) = 4. All sides = 4, so **triangle is equilateral**. Area = (√3/4) × 4² = **4√3 square units**. (c) Since A, B, C all have z = 0, the plane is simply **z = 0** (the xy-plane). The support beam perpendicular to this plane has direction (0, 0, 1), confirming vertical orientation. The equation is **z = 0**. **Why this HOTS question?** It combines 3D geometry with real-world architecture, requiring students to identify plane equations from coordinate patterns, verify geometric properties (equilateral triangle), and understand perpendicularity in 3D.

CBSETUTOR.ai's AI tutor personalizes daily practice through adaptive learning paths that focus on your weak areas. For Chapter 11, the system works like this: **Day 1–2: Direction Cosines Foundation** — Micro-lessons (5 min) on direction ratios vs. cosines, followed by 5 MCQs auto-graded instantly. AI analyzes errors: If you mix up direction ratios with cosines, the tutor branches to a re-explanation with visual 3D diagrams. **Day 3–4: Line Equations in 3D** — Interactive problem sets where you write Cartesian equations, then vector forms. The AI checks parametric vs. symmetric errors and highlights conceptual gaps (e.g., forgetting to normalize for direction cosines). **Day 5–6: Plane Equations & Distance** — Mixed 2-mark and 3-mark drills. The tutor grades your distance formula setup, not just the final answer, so you learn the method, not just the result. **Day 7: Full Chapter Simulation** — A mock test replicating actual board patterns (1-mark MCQs, short answers, medium-level, long answers). Your performance triggers new drills on remaining weak topics before the next week. **Key Features:** - **Step-by-step feedback** — AI shows which line of your working is incorrect and why. - **Spaced repetition** — Questions resurface based on forgetting curves; you see direction cosines again in Week 3 if shaky in Week 1. - **Real exam timing** — Timed MCQ sets train speed; 5-mark drills teach pacing. - **Parent dashboards** — Weekly reports show Chapter 11 mastery %, common mistakes, and time spent. Start a 3-day free trial at cbsetutor.ai to experience AI-powered 3D geometry drills tailored to your pace and exam readiness.