RD Sharma Class 12 Ex 29.1 Solutions Chapter 29 The Plane

Here we provide RD Sharma Class 12 Ex 29.1 Solutions Chapter 29 The Plane for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 29.1 Solutions Chapter 29 The Plane book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter29
Exercise29.1
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 29.1 Solutions Chapter 29 The Plane

Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed

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.

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.

Leave a Comment

Your email address will not be published.