Class 11 Maths Chapter 11 Three Dimensional Geometry – Exercise 11.3 NCERT Solutions

Introduction

Exercise 11.3 introduces the equations of a line in 3D geometry. Students learn both vector form and Cartesian form of a line, along with conversion between them. This exercise is crucial for understanding how to represent lines in space and solve problems involving their intersection, parallelism, and angles.

Key Concepts

1. Vector Equation of a Line: If a line passes through point $A(x_1,y_1,z_1)$ and is parallel to vector $\vec{b}=⟨l,m,n⟩$, then

$$\vec{r}=\vec{a}+\lambda \vec{b},\lambda \in \mathbb{R}$$

2. Cartesian Equation of a Line:

$$\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$$

3. Conversion:

   • From vector to Cartesian: Compare coefficients.

   • From Cartesian to vector: Identify point and direction ratios.

Common Mistakes

• Forgetting that parameter $\lambda$ can take all real values.

• Confusing direction ratios with coordinates of a point.

• Misapplying conversion between vector and Cartesian forms.

• Mixing up line equations with plane equations.

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

Q1. Find vector equation of line through (1,2,3) and parallel to vector $⟨2,3,4⟩$.

$$\vec{r}=⟨1,2,3⟩+\lambda ⟨2,3,4⟩$$

Q2. Find Cartesian equation of line through (1,2,3) and parallel to vector $⟨2,3,4⟩$.

$$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$$

Q3. Find vector equation of line through (2,−1,3) and parallel to vector $⟨1,2,-1⟩$.

$$\vec{r}=⟨2,-1,3⟩+\lambda ⟨1,2,-1⟩$$

Q4. Find Cartesian equation of line through (2,−1,3) and parallel to vector $⟨1,2,-1⟩$.

$$\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-3}{-1}$$

Q5. Write vector equation of line through (0,0,0) and parallel to $⟨1,1,1⟩$.

$$\vec{r}=\lambda ⟨1,1,1⟩$$

Q6. Write Cartesian equation of line through (0,0,0) and parallel to $⟨1,1,1⟩$.

$$\frac{x}{1}=\frac{y}{1}=\frac{z}{1}$$

Q7. Find vector equation of line through (3,−2,1) and parallel to $⟨2,-1,3⟩$.

$$\vec{r}=⟨3,-2,1⟩+\lambda ⟨2,-1,3⟩$$

Q8. Find Cartesian equation of line through (3,−2,1) and parallel to $⟨2,-1,3⟩$.

$$\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z-1}{3}$$

Q9. Write vector equation of line through (1,1,1) and parallel to $⟨0,1,-1⟩$.

$$\vec{r}=⟨1,1,1⟩+\lambda ⟨0,1,-1⟩$$

Q10. Write Cartesian equation of line through (1,1,1) and parallel to $⟨0,1,-1⟩$.

$$\frac{x-1}{0}=\frac{y-1}{1}=\frac{z-1}{-1}$$

(Note: First ratio undefined, so line is parallel to y‑z plane.)

FAQs (10)

FAQ1. What is vector equation of a line? $\vec{r}=\vec{a}+\lambda \vec{b}$.

FAQ2. What is Cartesian equation of a line? $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$.

FAQ3. What are direction ratios? Numbers proportional to direction cosines of a line.

FAQ4. How to convert vector form to Cartesian form? Compare coefficients of $\lambda$.

FAQ5. How to convert Cartesian form to vector form? Take point and direction ratios.

FAQ6. What is parameter $\lambda$? Real number representing position on line.

FAQ7. Can two different equations represent same line? Yes, if they are equivalent.

FAQ8. What if denominator in Cartesian form is zero? That coordinate remains constant.

FAQ9. Why is Exercise 11.3 important? It builds foundation for line‑plane problems.

FAQ10. What is difference between line and plane equations? Line has one parameter, plane has one linear relation.

Conclusion

Exercise 11.3 covers equations of lines in 3D geometry in both vector and Cartesian forms. With solved examples and FAQs, students gain clarity on representing lines in space and converting between forms.

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New Syllabus-Class 11 Maths Chapter 11 Three Dimensional Geometry – Exercise 11.3 NCERT Solutions