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Chapter 29 of RD Sharma's Class 12 Mathematics textbook explores the concept of planes in three-dimensional space. Exercise 29.8 focuses on problems related to the angle between two planes. This set of solutions demonstrates how to calculate these angles using various formulas and techniques, providing students with practical applications of plane geometry and vector algebra in 3D space.
Angle between two planes:
cos θ = |A₁A₂ + B₁B₂ + C₁C₂| / √[(A₁² + B₁² + C₁²)(A₂² + B₂² + C₂²)]
where A₁x + B₁y + C₁z + D₁ = 0 and A₂x + B₂y + C₂z + D₂ = 0 are the equations of the two planes
Condition for perpendicular planes: A₁A₂ + B₁B₂ + C₁C₂ = 0
Condition for parallel planes: A₁/A₂ = B₁/B₂ = C₁/C₂
Solution:
We know that the equation of a plane parallel to 2x – 3y + z = 0 is given by:
2x – 3y + z + λ = 0
Since the plane passes through the point (1, –1, 2), we have:
2(1) – 3(–1) + 2 + λ = 0
⇒ λ = –7
On substituting the value of λ in the equation, we have:
2x – 3y + z + (-7) = 0
2x – 3y + z – 7= 0 is the required equation.
Solution:
The given plane passes through the vector . Thus,
(3)(2) + (4)(-3) + (-1)(5) + λ = 0
⇒ λ = 11
On substituting the value of λ in the equation, we have:
is the required equation.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
(2x – 7y + 4z – 3) + λ(3x – 5y + 4z + 11) = 0⇒ x(2 + 3λ) + y(–7 – 5λ) + z(4 + 4λ) – 3 + 11λ = 0
Also, since the plane passes through the point (–2, 1, 3), we have:
(–2)(2 + 3λ) + (1)(–7 – 5λ) + (3)(4 + 4λ) – 3 + 11λ = 0
⇒ λ = 1/6
On substituting the value of λ in the equation, we have:
x(2 + 3(1/6)) + y(–7 – 5(1/6)) + z(4 + 4(1/6)) – 3 + 11(1/6) = 0
15x – 47y + 28z = 7 is the required equation.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
Also, since the plane passes through point , we have:
⇒ λ = 6
On substituting the value of λ in the equation, we have:
is the required equation.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
2x – y + λ(3z – y) = 0
⇒ 2x + y(–1 – λ) + z(3λ) = 0
Since the planes are perpendicular, we have:
2(4) + (–5)(–1 – λ) + (–3)(3λ) = 0
⇒ λ = 3/14
On substituting the value of λ in the equation, we have:
2x + y(–1 – 3/14) + z(3(3/14)) = 0
28x – 17y + 9z = 0is the required equation.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
x + 2y + 3z – 4 + λ(2x + y – z + 5) = 0
⇒ x(1 + 2λ) + y(2 + λ) + z(3 – λ) – 4 + 5λ = 0
Since the planes are perpendicular, we have:
5(1 + 2λ) + 3(2 + λ) + (–6)(3 – λ) = 0
⇒ λ = 7/19
On substituting the value of λ in the equation, we have:
x(1 + 2(7/19)) + y(2 + 7/19) + z(3 – 7/19) – 4 + 5(7/19) = 0
33x + 45y + 50z – 41 = 0 is the required equation.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
x + 2y + 3z + 4 + λ(x – y + z + 3) = 0
⇒ x(1 + λ) + y(2 – λ) + z(3 + λ) + 4 + 3λ = 0
Also, since the plane passes through the origin, we have:
0(1 + λ) + 0(2 – λ) + 0(3 + λ) + 4 + 3λ = 0
⇒ λ = -4/3
On substituting the value of λ in the equation, we have:
x(1 + (-4/3)) + y(2 – (-4/3)) + z(3 + (-4/3)) + 4 + 3(-4/3) = 0
x – 10y – 5z = 0 is the required equation.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
x – 3y + 2z – 5 + λ(2x – y + 3z – 1) = 0
⇒ x(1 + 2λ) + y(–3 – λ) + z(2 + 3λ) – 5 – λ = 0
Also, since the plane passes through the origin, we have:
1(1 + 2λ) + (–2)(–3 – λ) + 3(2 + 3λ) – 5 – λ = 0
⇒ λ = -2/3
On substituting the value of λ in the equation, we have:
x(1 + 2(-2/3)) + y(–3 – (-2/3)) + z(2 + 3(-2/3)) – 5 – (-2/3) = 0
is the required equation.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
x + 2y + 3z – 4 + λ(2x + y – z + 5) = 0
⇒ x(1 + 2λ) + y(2 + λ) + z(3 – λ) – 4 + 5λ = 0
We know that two planes are perpendicular when
⇒ 5(1 + 2λ) + 3(2 + λ) + 6(3 – λ) = 0
⇒ λ = -29/7
On substituting the value of λ in the equation, we have:
x(1 + 2(-29/7)) + y(2 + (-29/7)) + z(3 – (-29/7)) – 4 + 5(-29/7) = 0
51x + 15y – 50z + 173 = 0 is the required equation.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
x(1 + 3λ) + y(3 + λ) – 4zλ + 6 = 0
Distance from plane to the origin = 1
⇒
⇒ λ = ±1
Hence, 4x + 2y – 4z + 6 = 0 and –2x + 2y + 4z + 6 = 0 are the required equations.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
2x + 3y – z + 1 + λ(x + y – 2z + 3) = 0
⇒ x(2 + λ) + y(3 + λ) + z(–1 – 2λ) + 1 + 3λ = 0
We know that two planes are perpendicular when
⇒ 3(2 + λ) + (–1)(3 + λ) + (–2)(–1 – 2λ) = 0
⇒ λ = -5/6
On substituting the value of λ in the equation, we have:
x(2 + (-5/6)) + y(3 + (-5/6)) + z(–1 – 2(-5/6)) + 1 + 3(-5/6) = 0
7x + 13y + 4z – 9 = 0 is the required equation.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
⇒
We know that two planes are perpendicular if
⇒
⇒ 5(1 + 2λ) + 3(2 + λ) + (–6)(3 – λ) = 0
⇒ λ = 7/19
On substituting the value of λ in the equation, we have:
33x + 45y + 50z – 41 = 0 is the required equation.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
⇒ x(1 + 2λ) + y(1 + 3λ) +z(1 + 4λ) = 6 – 5λ
Also, since the plane passes through the point(1, 1, 1), we have:
1(1 + 2λ) + 1(1 + 3λ) +1(1 + 4λ) = 6 – 5λ
⇒ λ = 3/14
On substituting the value of λ in the equation, we have:
x(1 + 2(3/14)) + y(1 + 3(3/14)) +z(1 + 4(3/14)) = 6 – 5(3/14)
is the required equation.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
⇒
Also, since the plane passes through the point (2, 1, 3) we have:
9λ = –7
⇒ λ = -7/9
Substituting the value of λ in the equation, we have:
is the required equation.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
3x – y + 2z – 4 + λ(x + y + z – 2) = 0
Also, since the plane passes through the point (2, 2, 1), we have:
λ = -2/3
On substituting the value of λ in the equation, we have:
3x – y + 2z – 4 + (-2/3)(x + y + z – 2) = 0
7x – 5y + 4z = 0 is the required equation.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
x + 2y + z – 1 + λ(2x + 3y + 4z – 5) = 0
⇒ x(1 + 2λ) + y(1 + 3λ) +z(1 + 4λ) = 1 + 5λ
We know that two planes are perpendicular when
⇒ 1(1 + 2λ) + (–1)(1 + 3λ) + 1(1 + 4λ) = 1 + 5λ
⇒ λ = -1/3
On substituting the value of λ in the equation, we have:
x(1 + 2(-1/3)) + y(1 + 3(-1/3)) + z(1 + 4(-1/3)) = 1 + 5(-1/3)
x – z + 2 = 0 is the required equation.
Solution:
Equation of the family of planes parallel to the given plane =
Since the plane passes through (a, b, c), we have:
a + b + c = d
Substituting the above equation in the equation of family of planes we have:
Hence, x + y + z = a + b + c is the required equation of the plane.
Problem 1. Find the equation of the plane passing through the points A(1, 2, 3), B(2, 3, 1), and C(3, 1, 2).
Problem 2. Determine if the points A(1, 2, 3), B(2, 3, 4), C(3, 4, 5), and D(4, 5, 6) are coplanar.
Problem 3. Find the equation of the plane passing through the point (2, -1, 3) and perpendicular to the planes x + 2y - z = 4 and 2x - y + 3z = 5.
Problem 4. Find the equation of the plane passing through the point (1, -1, 2) and parallel to both the lines x = 2t + 1, y = 3t - 2, z = t + 3 and x = s - 1, y = 2s + 1, z = 3s + 2.
Problem 5. Find the equation of the plane passing through the line of intersection of the planes x + y + z = 1 and x - y + z = 3, and perpendicular to the plane x + y - z = 2.
Problem 6. Determine the equation of the plane passing through the point (1, 2, -1) and containing the line of intersection of the planes 2x - y + z = 3 and x + 3y - 2z = 1.
Problem 7. Find the equation of the plane passing through the point (2, 3, -1) and perpendicular to the line joining the points (1, -1, 2) and (3, 2, -3).
Problem 8. Find the equation of the plane which is at a distance of 5 units from the origin and normal to the line x/2 = y/3 = z/4.
Problem 9. Find the equation of the plane passing through the point (1, -1, 2) and perpendicular to each of the planes 2x + 3y - z = 5 and x - y + 2z = 3.
Problem 10. Determine the equation of the plane passing through the points (1, 0, 0), (0, 1, 0) and perpendicular to the plane x + 2y + 2z = 3.
Also Read,
Exercise 29.8 provides valuable practice in calculating angles between planes in three-dimensional space. These problems demonstrate the application of vector algebra and trigonometry in solving geometric problems. Students learn to distinguish between different spatial relationships, such as perpendicular, parallel, and oblique planes. Mastering these concepts is crucial for advanced studies in multivariable calculus, physics, and engineering, where understanding the orientation of planes in 3D space is often necessary. The skills developed here also have practical applications in fields like computer graphics, robotics, and structural design
