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.
Important Formulas Related to Plane
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
Class 12 RD Sharma Mathematics Solutions - Exercise 29.8
Question 1. Find the equation of the plane which is parallel to 2x – 3y + z = 0 and passes through the point (1, –1, 2).
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.
Question 2. Find the equation of the plane through (3, 4, –1) which is parallel to the plane \vec{r}.(2\hat{i}-3\hat{j}+5\hat{k})+2=0.
Solution:
The given plane passes through the vector
3\hat{i}+4\hat{j}-\hat{k} . Thus,
(3\hat{i}+4\hat{j}-\hat{k}).(2\hat{i}-3\hat{j}+5\hat{k})+ λ=0. (3)(2) + (4)(-3) + (-1)(5) + λ = 0
⇒ λ = 11
On substituting the value of λ in the equation, we have:
\vec{r}.(2\hat{i}-3\hat{j}+5\hat{k})+11=0 is the required equation.
Question 3. Find the equation of the plane passing through the line of intersection of the planes 2x – 7y + 4z – 3 = 0 and 3x – 5y + 4z + 11 = 0 and the point (–2, 1, 3).
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.
Question 4. Find the equation of the plane passing through the point 2\hat{i}+\hat{j}-\hat{k} and passing through the line of intersection of the planes \vec{r}.(\hat{i}+3\hat{j}-\hat{k})=0 and \vec{r}.(\hat{j}+\hat{k})=0.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
\vec{r}.[(\hat{i}+3\hat{j}-\hat{k})+λ(\hat{j}+2\hat{k})]=0 Also, since the plane passes through point
2\hat{i}+\hat{j}-\hat{k} , we have:
(2\hat{i}+\hat{j}-\hat{k})(\hat{i}+3\hat{j}-\hat{k})+λ(2\hat{i}+\hat{j}-\hat{k})(\hat{j}+2\hat{k})=0 ⇒ λ = 6
On substituting the value of λ in the equation, we have:
\vec{r}.(\hat{i}+9\hat{j}+11\hat{k})=0 is the required equation.
Question 5. Find the equation of the plane passing through the intersection of 2x – y = 0 and 3z – y = 0 and perpendicular to 4x + 5y – 3z = 8.
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 = 0 is the required equation.
Question 6. Find the equation of the plane which contains the line of intersection of planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0 and is perpendicular to the plane 5x + 3y – 6z + 8 = 0.
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.
Question 7. Find the equation of the plane passing through the line of intersection of the planes x + 2y + 3z + 4 = 0 and x – y + z + 3 = 0 and passing through the origin.
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.
Question 8. Find the vector equation in scalar product form of the plane containing the line of intersection of the planes x – 3y + 2z – 5 = 0 and 2x – y + 3z – 1 = 0 and passing through (1, –2, 3).
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
\vec{r}.(\hat{i}+7\hat{j})+13=0 is the required equation.
Question 9. Find the equation of the plane which contains the line of intersection of planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0 and is perpendicular to the plane 5x + 3y + 6z + 8 = 0.
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
a_1a_2+b_1b_2+c_1c_2=0. ⇒ 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.
Question 10. Find the equation of the plane passing through the line of intersection of the planes \vec{r}.(\hat{i}+3\hat{j})+6=0 and \vec{r}.(3\hat{i}-\hat{j}-4\hat{k})=0 and which is at a unit distance from the origin.
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
⇒
|\frac{6}{\sqrt{(1+3λ )^2+(3-λ)^2+(4λ)^2}}|=1 ⇒ λ = ±1
Hence, 4x + 2y – 4z + 6 = 0 and –2x + 2y + 4z + 6 = 0 are the required equations.
Question 11. Find the equation of the plane passing through the line of intersection of the planes 2x + 3y – z + 1 = 0 and x + y – 2z + 3 = 0 and perpendicular to the plane 3x – 2y – z – 4 = 0.
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
a_1a_2+b_1b_2+c_1c_2=0. ⇒ 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.
Question 12. Find the equation of the plane that contains the line of intersection of the planes \vec{r}.(\hat{i}+2\hat{j}+3\hat{k})-4=0 and \vec{r}.(2\hat{i}+\hat{j}-\hat{k})+5=0 and which is perpendicular to the plane \vec{r}.(5\hat{i}+3\hat{j}-6\hat{k})+8=0.
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
\vec{r}.(\hat{i}+2\hat{j}+3\hat{k})-4+λ(\vec{r}.(2\hat{i}+\hat{j}-\hat{k})+5)=0 ⇒
\vec{r}.[(\hat{i}+2\hat{j}+3\hat{k})+λ(2\hat{i}+\hat{j}-\hat{k})]-4+5λ=0 We know that two planes are perpendicular if
\vec{n_1}.\vec{n_2}=0 ⇒
[(\hat{i}+2\hat{j}+3\hat{k})+λ(2\hat{i}+\hat{j}-\hat{k})](5\hat{i}+3\hat{j}-6\hat{k})=0 ⇒ 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.
Question 13. Find the vector equation of the plane passing through the intersection of planes \vec{r}.(\hat{i}+\hat{j}+\hat{k})=6 and \vec{r}.(2\hat{i}+3\hat{j}+4\hat{k})=5 and the point (1, 1, 1).
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
\vec{r}.(\hat{i}+\hat{j}+\hat{k})-6+λ(\vec{r}.(2\hat{i}+3\hat{j}+4\hat{k})-5)=0 ⇒ 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)
\vec{r}.(20\hat{i}+23\hat{j}+26\hat{k})=69 is the required equation.
Question 14. Find the equation of the plane passing through the intersection of the planes \vec{r}.(2\hat{i}+\hat{j}+3\hat{k})=7 and \vec{r}.(2\hat{i}+5\hat{j}+3\hat{k})=9 and the point (2, 1, 3).
Solution:
The equation of the plane passing through the line of intersection of the given planes is:
\vec{r}.(2\hat{i}+\hat{j}+3\hat{k})-7+λ(\vec{r}.(2\hat{i}+5\hat{j}+3\hat{k})-9)=0 ⇒
\vec{r}.[(2+2λ)\hat{i}+(1+5λ)\hat{j}+(3+3λ)\hat{k}]-7-9λ=0 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:
\vec{r}.(2\hat{i}-13\hat{j}+3\hat{k})=0 is the required equation.
Question 15. Find the equation of the plane passing through the intersection of the planes 3x – y + 2z = 4 and x + y + z = 2 and the point (2, 2, 1).
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.
Question 16. Find the vector equation of the plane through the line of intersection of the planes x + 2y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.
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
a_1a_2+b_1b_2+c_1c_2=0. ⇒ 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.
Question 17. Find the equation of the plane passing through (a, b, c) and parallel to the plane \vec{r}.(\hat{i}+\hat{j}+\hat{k})=2.
Solution:
Equation of the family of planes parallel to the given plane =
\vec{r}.(\hat{i}+\hat{j}+\hat{k})=d 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:
\vec{r}(\hat{i}+\hat{j}+\hat{k})=a+b+c Hence, x + y + z = a + b + c is the required equation of the plane.
Practice Problems on 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.
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Conclusion
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