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Class 12 RD Sharma Solutions – Chapter 29 The Plane – Exercise 29.1

Last Updated : 25 Feb, 2021
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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.



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