Class 12 Maths Chapter 11 Three Dimensional Geometry

Class 12 Maths Chapter 11 Three Dimensional Geometry – Exercise 11.4 NCERT Solutions

Introduction

Exercise 11.4 focuses on the equations of a plane in 3D geometry. Students learn how to represent planes in vector and Cartesian forms, derive equations using point and normal vector, and solve problems involving coplanarity and perpendicularity. This exercise is essential for understanding the geometry of planes in three‑dimensional space.

Key Concepts

Vector Equation of Plane: Plane through point $A (x_{1}, y_{1}, z_{1})$ and normal vector $\vec{n} = (a, b, c)$ :

r⃗⋅n⃗=a⃗⋅n⃗

where $\vec{a} = x_{1} \hat{i} + y_{1} \hat{j} + z_{1} \hat{k}$ .

Cartesian Equation of Plane:

a(x−x1)+b(y−y1)+c(z−z1)=0

ax+by+cz+d=0

Coplanarity Condition: Four points are coplanar if determinant of their position vectors = 0.
Perpendicularity Condition: If normals are perpendicular, dot product = 0.

Students Frequently Make Mistakes

Forgetting to substitute point coordinates correctly.
Mixing up vector and Cartesian forms.
Errors in identifying normal vector.
Skipping constant term $d$ in Cartesian equation.
Misinterpreting coplanarity condition.

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

Q1. Find equation of plane through point $A (1, 2, 3)$ and normal vector $(1, 1, 1)$ .

1(x−1)+1(y−2)+1(z−3)=0⇒x+y+z=6

Q2. Find equation of plane through point $P (2, - 1, 3)$ and normal vector $(2, 3, 1)$ .

2(x−2)+3(y+1)+1(z−3)=0⇒2x+3y+z=4

Q3. Find equation of plane through point $A (0, 0, 0)$ and normal vector $(1, 2, 3)$ .

x+2y+3z=0

Q4. Find equation of plane through point $A (1, 1, 1)$ and normal vector $(2, - 1, 3)$ .

2(x−1)−1(y−1)+3(z−1)=0⇒2x−y+3z=4

Q5. Find equation of plane through point $A (2, 3, 4)$ and normal vector $(1, - 2, 1)$ .

(x−2)−2(y−3)+(z−4)=0⇒x−2y+z=0

Q6. Find equation of plane through point $A (3, 4, 5)$ and normal vector $(2, 3, - 1)$ .

2(x−3)+3(y−4)−1(z−5)=0⇒2x+3y−z=20

Q7. Find equation of plane through point $A (1, - 2, 3)$ and normal vector $(0, 1, - 1)$ .

0(x−1)+1(y+2)−1(z−3)=0⇒y−z=1

Q8. Find equation of plane through point $A (2, 0, - 1)$ and normal vector $(3, 2, 1)$ .

3(x−2)+2(y−0)+1(z+1)=0⇒3x+2y+z=5

Q9. Find equation of plane through point $A (0, 1, 2)$ and normal vector $(1, - 1, 2)$ .

(x−0)−1(y−1)+2(z−2)=0⇒x−y+2z=3

Q10. Find equation of plane through point $A (1, 2, - 1)$ and normal vector $(2, 1, 3)$ .

2(x−1)+1(y−2)+3(z+1)=0⇒2x+y+3z=1

FAQs (10)

FAQ1. What is vector equation of plane? $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$ .

FAQ2. What is Cartesian equation of plane? $a x + b y + c z + d = 0$ .

FAQ3. What is normal vector of plane? Vector perpendicular to plane.

FAQ4. How to derive plane equation from point and normal? Substitute coordinates into $a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$ .

FAQ5. What is equation of plane through origin with normal $(1, 2, 3)$ ? $x + 2 y + 3 z = 0$ .

FAQ6. What is equation of plane through $(1, 1, 1)$ with normal $(2, - 1, 3)$ ? $2 x - y + 3 z = 4$ .

FAQ7. What is coplanarity condition? Determinant of position vectors = 0.

FAQ8. What is perpendicularity condition for planes? Dot product of normals = 0.

FAQ9. What is equation of plane through $(2, 0, - 1)$ with normal $(3, 2, 1)$ ? $3 x + 2 y + z = 5$ .

FAQ10. Why is Exercise 11.4 important? It builds foundation for equations of planes in 3D geometry.

Conclusion

Exercise 11.4 has 10 solved questions and 10 FAQs that strengthen your understanding of equations of planes in 3D geometry. This builds the foundation for advanced study of planes and their applications in Class 12 Maths.

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