Exercises of Class 12 Maths Chapter 11 – Three Dimensional Geometry

In Class 12 Mathematics, Chapter 11 - Three Dimensional Geometry, is a fundamental segment of the curriculum that expands students' understanding of geometry from two dimensions to three. This chapter introduces concepts that are crucial for higher studies in mathematics, engineering, and various fields of science. This article will delve into the core exercises and topics of Chapter 11, providing a comprehensive overview to help students excel in their understanding and application of three-dimensional geometry.

Introduction to Three Dimensional Geometry

Three-dimensional geometry, or 3D geometry, deals with the study of shapes and figures in a three-dimensional space. Unlike two-dimensional geometry, which deals with shapes on a flat plane, 3D geometry involves depth, in addition to length and width. This chapter equips students with the tools to work with points, lines, planes, and various geometric shapes in three-dimensional space.

Key Concepts in Three Dimensional Geometry

Before diving into the exercises, it's crucial to understand the key concepts covered in Chapter 11:

1. Coordinate System in 3D Space:

   • Cartesian Coordinates: In three-dimensional space, each point is represented by an ordered triplet (x, y, z), where x, y, and z are coordinates along the X, Y, and Z axes, respectively.
   • Distance Formula: The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$

2. Equation of a Plane:

   • General Form: The equation of a plane in 3D space can be expressed as $ax + by + cz = d$, where (a, b, c) is a normal vector perpendicular to the plane.
   • Plane Through Three Points: To find the equation of a plane passing through three given points, use the determinant method or vector algebra.

3. Line and Plane Intersection:

   • Intersection of Two Lines: To determine if two lines intersect, solve their parametric equations simultaneously.
   • Intersection of a Line and a Plane: Substitute the parametric equations of the line into the plane's equation to find the point of intersection.

4. Angle Between Two Planes:

   • Formula: The angle θ between two planes with normal vectors $\mathbf{n}_1$ and $\mathbf{n}_2$ is given by: $$\cos \theta = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{\|\mathbf{n}_1\| \|\mathbf{n}_2\|}$$

5. Distance Between Two Parallel Planes:

   • Formula: For two parallel planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$, the distance between them is: $$\text{Distance} = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}}$$

6. Skew Lines:

   • Definition: Skew lines are non-intersecting and non-parallel lines that do not lie in the same plane.
   • Shortest Distance: The shortest distance between two skew lines can be found using vector methods.

Detailed Solution of Key Exercises

Let's explore some representative exercises from Chapter 11 to understand how to apply these concepts effectively.

Exercise 11.1: Distance Between Two Points

Question: Find the distance between the points $A(2, 3, 4)$ and $B(-1, -2, 1)$.

Solution: Using the distance formula:

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

Substitute the coordinates:

$$d = \sqrt{((-1) - 2)^2 + ((-2) - 3)^2 + (1 - 4)^2}$$ $$d = \sqrt{(-3)^2 + (-5)^2 + (-3)^2}$$ $$d = \sqrt{9 + 25 + 9}$$ $$d = \sqrt{43}$$

Thus, the distance between the points is $\sqrt{43}$ units.

Exercise 11.2: Equation of a Plane

Question: Find the equation of the plane passing through the points $A(1, 2, 3)$, $B(4, 5, 6)$, and $C(7, 8, 9)$.

Solution: To find the equation of the plane through three points, first find two vectors in the plane:

$$\vec{AB} = (4 - 1, 5 - 2, 6 - 3) = (3, 3, 3)$$ $$\vec{AC} = (7 - 1, 8 - 2, 9 - 3) = (6, 6, 6)$$

The vectors $\vec{AB}$ and $\vec{AC}$ are parallel, implying the points are collinear, and hence do not define a unique plane. Therefore, the points are not sufficient to define a unique plane.

Exercise 11.3: Intersection of Line and Plane

Question: Find the point of intersection of the line given by $x = 2 + t$, $y = 3 - 2t$, $z = 4 + 3t$ with the plane $x - y + z = 7$.

Solution: Substitute the parametric equations of the line into the plane equation:

$$(2 + t) - (3 - 2t) + (4 + 3t) = 7$$ $$2 + t - 3 + 2t + 4 + 3t = 7$$ $$6t + 3 = 7$$ $$6t = 4$$ $$t = \frac{2}{3}$$

Substitute $t = \frac{2}{3}$ into the parametric equations:

$$x = 2 + \frac{2}{3} = \frac{8}{3}$$ $$y = 3 - 2 \cdot \frac{2}{3} = \frac{5}{3}$$ $$z = 4 + 3 \cdot \frac{2}{3} = 5$$

Thus, the point of intersection is $\left(\frac{8}{3}, \frac{5}{3}, 5\right)$.

Exercise 11.4: Angle Between Two Planes

Question: Find the angle between the planes $2x - 3y + z = 4$ and $x + 2y - 2z = 5$.

Solution: First, find the normal vectors of the planes:

$$\mathbf{n}_1 = (2, -3, 1)$$ $$\mathbf{n}_2 = (1, 2, -2)$$

The angle θ between the planes is given by:

$$\cos \theta = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{\|\mathbf{n}_1\| \|\mathbf{n}_2\|}$$

Calculate the dot product:

$$\mathbf{n}_1 \cdot \mathbf{n}_2 = (2 \cdot 1) + (-3 \cdot 2) + (1 \cdot -2) = 2 - 6 - 2 = -6$$

Calculate the magnitudes:

$$\|\mathbf{n}_1\| = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{14}$$ $$\|\mathbf{n}_2\| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9}$$ $$\cos \theta = \frac{-6}{\sqrt{14} \cdot 3} = -\frac{2}{\sqrt{14}}$$

Thus,

$$\theta = \cos^{-1}\left(-\frac{2}{\sqrt{14}}\right)$$

Exercise 11.5: Distance Between Two Parallel Planes

Question: Find the distance between the planes $3x - 4y + 5z = 6$ and $3x - 4y + 5z = 12$.

Solution: For two parallel planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$, the distance between them is given by:

$$\text{Distance} = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}}$$

Substitute the values:

$$\text{Distance} = \frac{|12 - 6|}{\sqrt{3^2 + (-4)^2 + 5^2}} = \frac{6}{\sqrt{50}} = \frac{6}{5\sqrt{2}} = \frac{3\sqrt{2}}{5}$$

Thus, the distance between the planes is $\frac{3\sqrt{2}}{5}$ units.

Conclusion

Understanding three-dimensional geometry is essential for solving complex problems in mathematics and science. By mastering the exercises of Class 12 Maths Chapter 11, students can gain proficiency in analyzing and solving problems involving 3D shapes, lines, and planes. The exercises provided in this chapter not only reinforce key concepts but also prepare students for more advanced studies in geometry and related fields.

Whether it's calculating distances, finding equations of planes, or determining angles and intersections, a thorough grasp of these concepts will help students excel in their academic pursuits and beyond. Keep practicing these exercises to build a solid foundation in three-dimensional geometry.