Class 11 Maths Chapter 11 – Introduction to 3D Geometry (NCERT Solutions with Formulas & Examples)

Introduction

In two‑dimensional geometry, we locate a point using two coordinates (x, y). But in real life, objects exist in space, so we need three coordinates (x, y, z). This chapter introduces the basics of three‑dimensional geometry: coordinate axes, coordinate planes, octants, distance formula, collinearity, and equations of loci. Let’s go step by step with formulas and solved NCERT examples.

Key Formulas

• Coordinates of a point in space: P(x, y, z)

• Distance between two points:

$$PQ=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$$

• Distance from origin to point (x, y, z):

$$\sqrt{x^2+y^2+z^2}$$

• Condition for collinearity: PQ + QR = PR

• Centroid of triangle with vertices (x1, y1, z1), (x2, y2, z2), (x3, y3, z3):

$$(\frac{x1+x2+x3}{3},\frac{y1+y2+y3}{3},\frac{z1+z2+z3}{3})$$

 Solved NCERT Examples (Step by Step)

Example 1

If P = (2,4,5), find coordinates of F in Fig 11.3. Solution: Distance along OY = 0 → F = (2,0,5).

Example 2

Find octants of (-3,1,2) and (-3,1,-2). Solution: From sign table: (-3,1,2) → II octant, (-3,1,-2) → VI octant.

Example 3

Find distance between P(1,-3,4) and Q(-4,1,2). Solution: PQ = √[(-4-1)² + (1+3)² + (2-4)²] = √(25 + 16 + 4) = √45 = 3√5 units.

Example 4

Show P(-2,3,5), Q(1,2,3), R(7,0,-1) are collinear. Solution: PQ = √[(1+2)² + (2-3)² + (3-5)²] = √14 QR = √[(7-1)² + (0-2)² + (-1-3)²] = √56 = 2√14 PR = √[(7+2)² + (0-3)² + (-1-5)²] = √126 = 3√14 Since PQ + QR = PR, points are collinear.

Example 5

Are A(3,6,9), B(10,20,30), C(25,-41,5) vertices of a right triangle? Solution: AB² = 686, BC² = 4571, CA² = 2709. Check Pythagoras: CA² + AB² ≠ BC². So not a right triangle.

Example 9

Centroid of triangle ABC is (1,1,1). A(3,-5,7), B(-1,7,-6). Find C. Solution: x = 1, y = 1, z = 2. So C = (1,1,2).

(Continue similarly for all NCERT examples in Exercises 11.1, 11.2, and Miscellaneous.)

 15 FAQs with Step‑by‑Step Solutions (Compose Format)

Q1. A point is on x-axis. What are its y and z coordinates? Answer: On x-axis → coordinates (x,0,0). So y=0, z=0.

Q2. A point is in XZ-plane. What is its y-coordinate? Answer: In XZ-plane → y=0.

Q3. Find octant of (4,-2,3). Answer: x=+, y=-, z=+. From table → IV octant.

Q4. Fill blank: x-axis and y-axis determine which plane? Answer: XY-plane.

Q5. Fill blank: Coordinates of points in XY-plane are of form? Answer: (x,y,0).

Q6. Fill blank: Coordinate planes divide space into how many parts? Answer: 8 octants.

Q7. Find distance between (2,3,5) and (4,3,1). Answer: PQ = √[(4-2)² + (3-3)² + (1-5)²] = √(4+0+16) = √20 = 2√5.

Q8. Show (-2,3,5), (1,2,3), (7,0,-1) are collinear. Answer: PQ=√14, QR=2√14, PR=3√14 → PQ+QR=PR → collinear.

Q9. Verify (0,7,-10), (1,6,-6), (4,9,-6) form isosceles triangle. Answer: Calculate distances: AB=√17, BC=√10, AC=√29. Since AB=AC, triangle is isosceles.

Q10. Verify (0,7,10), (-1,6,6), (-4,9,6) form right triangle. Answer: Distances: AB²+BC²=AC² → satisfies Pythagoras → right triangle.

Q11. Show (-1,2,1),(1,-2,5),(4,-7,8),(2,-3,4) form parallelogram. Answer: Opposite sides equal → parallelogram.

Q12. Find equation of set of points equidistant from (1,2,3) and (3,2,-1). Answer: (x-1)²+(y-2)²+(z-3)² = (x-3)²+(y-2)²+(z+1)² → simplifies to 4x+4z-8=0.

Q13. Find equation of set of points P with PA+PB=10, A(4,0,0), B(-4,0,0). Answer: Condition → ellipse equation: √[(x-4)²+y²+z²] + √[(x+4)²+y²+z²] = 10.

Q14. Find coordinates of fourth vertex of parallelogram with A(3,-1,2), B(1,2,-4), C(-1,1,2). Answer: D = A + C - B = (3,-1,2)+(-1,1,2)-(1,2,-4) = (1,-2,8).

Q15. Find coordinates of C if centroid of triangle is origin, A(2a,2,6), B(-4,3b,-10), R(8,14,2c). Answer: (2a-4+8)/3=0 → a=-2; (2+3b+14)/3=0 → b=-16/3; (6-10+2c)/3=0 → c=2.

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