Class 11 Maths Chapter 11 Three Dimensional Geometry – Exercise 11.1 NCERT Solutions

Introduction

Exercise 11.1 introduces the basics of three‑dimensional coordinate geometry. You will learn how to represent points in 3D space, calculate distances between points, and apply formulas for section and midpoint. This exercise builds the foundation for vectors and advanced geometry.

Key Concepts

1. Coordinates in 3D: A point in space is represented as $(x,y,z)$.

2. Distance Formula in 3D: For points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$:

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

3. Midpoint Formula in 3D:

$$M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2})$$

4. Section Formula in 3D: Point dividing line joining $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ in ratio $m:n$:

$$P=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n},\frac{m{z}_2+n{z}_1}{m+n})$$

NCERT Questions with Solutions (10)

Q1. Find distance between points (1,2,3) and (4,5,6).

$$d=\sqrt{(4-1{)}^2+(5-2{)}^2+(6-3{)}^2}=\sqrt{9+9+9}=\sqrt{27}=3\sqrt{3}$$

Q2. Find distance between points (2,-1,3) and (-2,2,1).

$$d=\sqrt{(-2-2{)}^2+(2+1{)}^2+(1-3{)}^2}=\sqrt{16+9+4}=\sqrt{29}$$

Q3. Find midpoint of line joining (1,2,3) and (5,6,7).

$$M=(\frac{1+5}{2},\frac{2+6}{2},\frac{3+7}{2})=(3,4,5)$$

Q4. Find midpoint of line joining (-1,4,2) and (3,-2,6).

$$M=(\frac{-1+3}{2},\frac{4+(-2)}{2},\frac{2+6}{2})=(1,1,4)$$

Q5. Find point dividing line joining (2,3,4) and (6,7,8) in ratio 1:3.

$$P=(\frac{1\cdot 6+3\cdot 2}{4},\frac{1\cdot 7+3\cdot 3}{4},\frac{1\cdot 8+3\cdot 4}{4})=(\frac{12}{4},\frac{16}{4},\frac{20}{4})=(3,4,5)$$

Q6. Find point dividing line joining (1,-2,3) and (3,2,-1) in ratio 2:1.

$$P=(\frac{2\cdot 3+1\cdot 1}{3},\frac{2\cdot 2+1\cdot (-2)}{3},\frac{2\cdot (-1)+1\cdot 3}{3})=(\frac{7}{3},\frac{2}{3},\frac{1}{3})$$

Q7. Find distance between points (0,0,0) and (2,2,1).

$$d=\sqrt{(2-0{)}^2+(2-0{)}^2+(1-0{)}^2}=\sqrt{4+4+1}=\sqrt{9}=3$$

Q8. Find distance between points (1,1,1) and (-1,-1,-1).

$$d=\sqrt{(-1-1{)}^2+(-1-1{)}^2+(-1-1{)}^2}=\sqrt{4+4+4}=\sqrt{12}=2\sqrt{3}$$

Q9. Find midpoint of line joining (2,4,6) and (8,10,12).

$$M=(\frac{2+8}{2},\frac{4+10}{2},\frac{6+12}{2})=(5,7,9)$$

Q10. Find point dividing line joining (0,0,0) and (4,6,8) in ratio 1:2.

$$P=(\frac{1\cdot 4+2\cdot 0}{3},\frac{1\cdot 6+2\cdot 0}{3},\frac{1\cdot 8+2\cdot 0}{3})=(\frac{4}{3},2,\frac{8}{3})$$

FAQs (10)

1. What is 3D coordinate geometry? Study of points, lines, and planes in three dimensions.

2. What is distance formula in 3D? $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$.

3. What is midpoint formula in 3D? $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2})$.

4. What is section formula in 3D? Point dividing line in ratio $m:n$.

5. What is coordinate of origin in 3D? (0,0,0).

6. What is x‑axis in 3D? Line along x direction with y=z=0.

7. What is y‑axis in 3D? Line along y direction with x=z=0.

8. What is z‑axis in 3D? Line along z direction with x=y=0.

9. Can distance be negative? No, always non‑negative.

10. Why is this exercise important? It builds foundation for vectors and advanced 3D geometry.

Conclusion

Exercise 11.1 has 10 solved questions and 10 FAQs that strengthen your understanding of three‑dimensional coordinate geometry basics. This builds the foundation for vectors and advanced geometry in Class 11 Maths.

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