Chapter 29 of RD Sharma's Class 12 mathematics textbook focuses on the concept of planes in three-dimensional space. This chapter explores various aspects of planes, including their equations, intersections with lines and other planes, and distances between points and planes. Exercise 29.1 specifically deals with problems related to the general equation of a plane and its applications.
Important Formulas Related to Plane
General equation of a plane: Ax + By + Cz + D = 0
Distance formula between a point and a plane: |Ax₀ + By₀ + Cz₀ + D| / √(A² + B² + C²)
Angle between two planes: cos θ = |A₁A₂ + B₁B₂ + C₁C₂| / √((A₁² + B₁² + C₁²)(A₂² + B₂² + C₂²))
Condition for parallel planes: A₁/A₂ = B₁/B₂ = C₁/C₂
Condition for perpendicular planes: A₁A₂ + B₁B₂ + C₁C₂ = 0
Question 1(i). Find the equation of the plane passing through the following points (2, 1, 0), (3, -2, -2), and(3, 1, 7).
Solution:
Given points are (2, 1, 0), (3, -2, -2), and (3, 1, 7)
The equation of plane passing through three points is given by
\begin{vmatrix}
x-x_1&y-y_1&z-z_1\\
x_2-x_1&y_2-y_1&z_2-z_1\\
x_3-x_1&y_3-y_1&z_3-z_1\\
\end{vmatrix} = 0
\begin{vmatrix}
x-2&y-1&z-0\\
3-2&-2-1&-2-0\\
3-2&1-1&7-0\\
\end{vmatrix} = 0
\begin{vmatrix}
x-2&y-1&z\\
1&-3&-2\\
1&0&7\\
\end{vmatrix} = 0
= (x - 2)(-21 - 0) - (y - 1)(7 + 2) + z(0 + 3) = 0
= -21x + 42 - 9y + 9 + 3z = 0
= -21x - 9y + 3z + 51 = 0
By taking -3 as common, we get resultant equation of plane
7x + 3y - z - 17 = 0
Question 1(ii). Find the equation of the plane passing through the following points (-5, 0, 6), (-3, 10, -9) and (-2, 6, -6).
Solution:
Given points are (-5, 0, 6), (-3, 10, -9) and (-2, 6, -6).
The equation of a plane passing through three points is given by
\begin{vmatrix}
x-x_1&y-y_1&z-z_1\\
x_2-x_1&y_2-y_1&z_2-z_1\\
x_3-x_1&y_3-y_1&z_3-z_1\\
\end{vmatrix} = 0
\begin{vmatrix}
x+5&y-0&z+6\\
-3+5&10-0&-9+6\\
-2+5&6-0&-6+6\\
\end{vmatrix} = 0
\begin{vmatrix}
x+5&y&z+6\\
2&10&-3\\
3&6&0\\
\end{vmatrix} = 0
(x + 5)(0 + 18) - y(0 + 9) + (z + 6)(12 - 30) = 0
(x + 5)(18) - y(9) + (z + 6)(-18) = 0
18x + 90 - 9y - 18z -108 = 0
Taking 9 as common, we get equation
2x - y - 2z - 2 = 0
Question 1(iii). Find the equation of the plane passing through the following points (1, 1, 1), (1, -1, 2), and (-2, -2, 2).
Solution:
Given three points are (1, 1, 1), (1, -1, 2), and (-2, -2, 2)
The equation of plane passing through three points is given by
\begin{vmatrix}
x-x_1&y-y_1&z-z_1\\
x_2-x_1&y_2-y_1&z_2-z_1\\
x_3-x_1&y_3-y_1&z_3-z_1\\
\end{vmatrix} = 0
\begin{vmatrix}
x-1&y-1&z-1\\
1-1&-1-1&2-1\\
-2-1&-2-1&2-1\\
\end{vmatrix} = 0
\begin{vmatrix}
x-1&y-1&z-1\\
0&-2&1\\
-3&-3&1\\
\end{vmatrix} = 0
(x - 1)(-2 + 3) - (y - 1)(0 + 3) + (z - 1)(0 - 6) = 0
(x - 1)(1) - (y - 1)(3) + (z - 1)(-6) = 0
x - 1 - 3y + 3 - 6z + 6 = 0
x -3y - 6z + 8 = 0
Question 1(iv). Find the equation of the plane passing through the following points (2, 3, 4), (-3, 5, 1), and (4, -1, 2).
Solution:
Given points are (2, 3, 4), (-3, 5, 1), and (4, -1, 2).
The equation of plane passing through three points is given by
\begin{vmatrix}
x-x_1&y-y_1&z-z_1\\
x_2-x_1&y_2-y_1&z_2-z_1\\
x_3-x_1&y_3-y_1&z_3-z_1\\
\end{vmatrix} = 0
\begin{vmatrix}
x-2&y-3&z-4\\
-3-2&5-3&1-4\\
4-2&-1-3&2-4\\
\end{vmatrix} = 0
\begin{vmatrix}
x-2&y-3&z-4\\
-5&2&3\\
2&-4&-2\\
\end{vmatrix} = 0
(x - 2)(-4 - 12) - (y - 3)(10 + 6) + (z - 4)(20 - 4) = 0
(x - 2)(-16) - (y - 3)(16) + (z - 4)(16) = 0
-16x - 32 - 16y + 48 + 16z - 64 = 0
-16x - 16y + 16z + 16 = 0
Taking -16 common we get equation of plane as,
x + y - z - 1 = 0
Question 1(v). Find the equation of the plane passing through the following points (0, -1, 0), (3, 3, 0), and (1, 1, 1).
Solution:
Given points are (0, -1, 0), (3, 3, 0), and (1, 1, 1)
The equation of plane passing through three points is given by,
\begin{vmatrix}
x-x_1&y-y_1&z-z_1\\
x_2-x_1&y_2-y_1&z_2-z_1\\
x_3-x_1&y_3-y_1&z_3-z_1\\
\end{vmatrix} = 0
\begin{vmatrix}
x-0&y+1&z-0\\
3-0&3+1&0-0\\
1-0&1+1&1-0\\
\end{vmatrix} = 0
\begin{vmatrix}
x&y+1&z\\
3&4&0\\
1&2&1\\
\end{vmatrix} = 0
x(4 - 0) - (y + 1)(3 - 0) + z(6 - 4) = 0
4x - (y + 1)(3) + z(2) = 0
4x - 3y - 3 + 2z = 0
4x - 3y + 2z - 3 = 0
Question 2. Show that the four points (0, -1, 1), (4, 5, 1), (3, 9, 4), and (-4, 4, 4) are coplanar and find the equation of the common plane.
Solution:
To prove the given points (0, -1, 1), (4, 5, 1), (3, 9, 4), and (-4, 4, 4) are coplanar.
Efficient solution is to find the equation of plane passing through any three points.
Then substitute the fourth point in the resultant equation.
If it satisfies then the four points are coplanar.
Equation of plane passing through three points (0, -1, 1), (4, 5, 1), and (3, 9, 4) is given by
\begin{vmatrix} x-x_1&y-y_1&z-z_1\\ x_2-x_1&y_2-y_1&z_2-z_1\\ x_3-x_1&y_3-y_1&z_3-z_1\\ \end{vmatrix} = 0
\begin{vmatrix}
x-0&y+1&z+1\\
4-0&5+1&1+1\\
3-0&9+1&4+1\\
\end{vmatrix} = 0
\begin{vmatrix}
x&y+1&z+1\\
4&6&2\\
3&10&5\\
\end{vmatrix} = 0
x(30 - 20) - (y + 1)(20 - 6) + (z + 1)(40 - 18) = 0
10x - (y + 1)(14) + (z + 1)(22) = 0
10x - 14y + 22z + 8 = 0
Taking 2 as common,
5x - 7y + 11z + 4 = 0 -Equation (1)
Substitute remaining point (-4, 4, 4) in above eq (1)
5(-4) - 7(4) + 11(4) + 4 = 0
-48 + 48 = 0
0 = 0
LHS = RHS
As the point satisfies the equation. So, the four points are coplanar.
Common plane equation is 5x - 7y + 11z + 4 = 0
Question 3(i). Show that the following points are coplanar (0, -1, 0), (2, 1, -1), (1, 1, 1), and(3, 3, 0).
Solution:
Given points are (0, -1, 0), (2, 1, -1), (1, 1, 1), and(3, 3, 0).
Equation of plane passing through three points (0, -1, 0), (2, 1, -1), (1, 1, 1) is given by
\begin{vmatrix}
x-x_1&y-y_1&z-z_1\\
x_2-x_1&y_2-y_1&z_2-z_1\\
x_3-x_1&y_3-y_1&z_3-z_1\\
\end{vmatrix} = 0
\begin{vmatrix}
x-0&y+1&z-0\\
2-0&1+1&-1-0\\
1-0&1+1&1-0\\
\end{vmatrix} = 0
\begin{vmatrix}
x&y+1&z\\
2&2&-1\\
1&2&1\\
\end{vmatrix} = 0
x(2 + 2) - (y + 1)(2 + 1) + z(4 - 2) = 0
x(4) - (y + 1)(3) + z(2) = 0
4x - 3y - 3 + 2z = 0
4x - 3y + 2z - 3 = 0 -Equation (1)
Substitute point four in equation (1)
4(3) - 3(3) + 2 (0) - 3 = 0
12 - 9 + 0 - 3 = 0
12 - 12 = 0
0 = 0
LHS = RHS
As fourth point satisfies the equation.
So, the four points are coplanar
Question 3(ii). Show that the following points are coplanar (0, 4, 3), (-1, -5, -3), (-2, -2, 1), and (1, 1, -1).
Solution:
Given four points are (0, 4, 3), (-1, -5, -3), (-2, -2, 1), and (1, 1, -1).
Equation of points passing through three points is given by
\begin{vmatrix} x-x_1&y-y_1&z-z_1\\ x_2-x_1&y_2-y_1&z_2-z_1\\ x_3-x_1&y_3-y_1&z_3-z_1\\ \end{vmatrix} = 0
\begin{vmatrix}
x-0&y-4&z-3\\
-1-0&-5-4&-3-3\\
-2-0&-2-4&1-3\\
\end{vmatrix} = 0
\begin{vmatrix}
x&y-4&z-3\\
-1&-9&-6\\
-2&-6&-2\\
\end{vmatrix} = 0
x(18 - 36) - (y - 4)(2 - 12) + (z - 3)(6 - 18) = 0
x(-18) - (y - 4)(-10) + (z - 3)(-12) = 0
-18x + 10y - 40 - 12z + 36 = 0
-18x + 10y - 12z - 4 = 0 -Equation (1)
Substitute fourth point in equation (1)
-18(1) + 10(1) - 12(-1) - 4 = 0
-18 +10 +12 -4 = 0
-22 + 22 = 0
0 = 0
LHS = RHS
As the fourth point satisfies the given equation.
So, the four points are coplanar.
Practice Questions - The Plane
Question 1. Find the intercepts made by the plane 2x - 3y + 4z = 12 on the coordinate axes.
Question 2. Write the equation of the plane passing through the point (1, -2, 3) with normal vector <2, 1, -1>.
Question 3. Determine if the points (1, 2, 3), (-1, 4, -2), and (3, 0, 1) are collinear.
Question 4. Find the equation of the plane passing through the points (1, 0, 0), (0, 2, 0), and (0, 0, 3).
Question 5. Write the equation of the plane with intercepts 2, -3, and 4 on the x, y, and z axes respectively.
Question 6. Determine if the point (2, -1, 3) lies on the plane x + 2y - z = 4.
Question 7. Find the distance of the point (2, 3, -1) from the plane 2x - 2y + z + 4 = 0.
Question 8. Write the equation of the plane passing through the point (1, -1, 2) and perpendicular to the vector <3, -2, 1>.
Question 9. Determine if the planes 2x + 3y - z = 5 and 4x + 6y - 2z = 10 are parallel.
Question 10. Find the equation of the plane passing through the origin and perpendicular to the line joining the points (1, 2, 3) and (4, 5, 6).
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Conclusion
Exercise 29.1 of RD Sharma's Class 12 Chapter 29 provides a comprehensive set of problems dealing with the equations of planes in three-dimensional space. Through these solutions, students can gain a deeper understanding of how to determine plane equations using various methods, including the point-normal form and the determinant method. The exercise also reinforces concepts such as parallel planes and the relationship between a plane's equation and its normal vector. Mastering these skills is crucial for students as they form the foundation for more advanced topics in analytical geometry and vector algebra.
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