Class 12 Maths Chapter 11 Three Dimensional Geometry – Exercise 11.1 NCERT Solutions

Introduction

Exercise 11.1 introduces the basic concepts of three‑dimensional coordinate geometry. Students learn how to calculate distance between points, section formula, and midpoint formula in 3D space. This exercise builds the foundation for advanced 3D geometry topics such as equations of lines and planes.

Key Concepts

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

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

2. Midpoint Formula: Midpoint of $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$:

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

3. Section Formula: Point dividing line $PQ$ in ratio $m:n$:

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

Students Frequently Make Mistakes

• Forgetting to square differences in distance formula.

• Errors in applying section formula (mixing $m$ and $n$).

• Misplacing signs in coordinates.

• Skipping midpoint calculation steps.

• Confusing 2D formulas with 3D ones.

NCERT Questions with Step‑by‑Step Solutions (10)

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

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

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

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

Q3. Find midpoint of points $A(2,3,4)$ and $B(4,5,6)$.

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

Q4. Find midpoint of points $P(-1,2,3)$ and $Q(5,-2,7)$.

$$M=(\frac{-1+5}{2},\frac{2+(-2)}{2},\frac{3+7}{2})=(2,0,5)$$

Q5. Find coordinates of point dividing line joining $A(1,2,3)$ and $B(4,5,6)$ in ratio 1:2.

$$(\frac{1\cdot 4+2\cdot 1}{3},\frac{1\cdot 5+2\cdot 2}{3},\frac{1\cdot 6+2\cdot 3}{3})=(\frac{6}{3},\frac{9}{3},\frac{12}{3})=(2,3,4)$$

Q6. Find coordinates of point dividing line joining $P(2,3,4)$ and $Q(6,7,8)$ in ratio 3:1.

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

Q7. Find distance between points $A(0,0,0)$ and $B(1,1,1)$.

$$AB=\sqrt{(1-0{)}^2+(1-0{)}^2+(1-0{)}^2}=\sqrt{3}$$

Q8. Find midpoint of points $A(1,-1,2)$ and $B(-3,5,6)$.

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

Q9. Find coordinates of point dividing line joining $A(2,4,6)$ and $B(8,10,12)$ in ratio 2:3.

$$(\frac{2\cdot 8+3\cdot 2}{5},\frac{2\cdot 10+3\cdot 4}{5},\frac{2\cdot 12+3\cdot 6}{5})=(4,6,8)$$

Q10. Find distance between points $P(3,4,5)$ and $Q(-3,-4,-5)$.

$$PQ=\sqrt{(-3-3{)}^2+(-4-4{)}^2+(-5-5{)}^2}=\sqrt{36+64+100}=\sqrt{200}=10\sqrt{2}$$

FAQs (10)

FAQ1. What is distance formula in 3D? $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$.

FAQ2. What is midpoint formula in 3D? $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2})$.

FAQ3. What is section formula in 3D? $(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n},\frac{m{z}_2+n{z}_1}{m+n})$.

FAQ4. What is distance between $(1,2,3)$ and $(4,5,6)$? $3\sqrt{3}$.

FAQ5. What is midpoint of $(2,3,4)$ and $(4,5,6)$? $(3,4,5)$.

FAQ6. What is point dividing $(1,2,3)$ and $(4,5,6)$ in ratio 1:2? $(2,3,4)$.

FAQ7. What is distance between origin and $(1,1,1)$? $\sqrt{3}$.

FAQ8. What is midpoint of $(1,-1,2)$ and $(-3,5,6)$? $(-1,2,4)$.

FAQ9. What is point dividing $(2,4,6)$ and $(8,10,12)$ in ratio 2:3? $(4,6,8)$.

FAQ10. Why is Exercise 11.1 important? It builds foundation for 3D geometry concepts like lines and planes.

Conclusion

Exercise 11.1 has 10 solved questions and 10 FAQs that strengthen your understanding of distance, midpoint, and section formulas in 3D geometry. This begins the Three Dimensional Geometry chapter in Class 12 Maths.

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