RD Sharma Class 12 Exercise 28.1 The Plane Solutions Maths - Download PDF Free Online

The Plane Excercise: 28.1

The Plane exercise 28.1 question 1(i)

Answer:-

The required equation of the plane is $7x+3y-z=17$
Hint:- Use equation of the plane $(x_1,y_1,z_1),(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$
Given:-$(2,1,0),(3,-2,-2)$ and $(3,1,7)$
Solution:- We know that, the equation of the plane passing through three non-collinear points $(x_1,y_1,z_1),(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$ is
$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$
$\begin{array}{rl}& \Rightarrow \left|\begin{array}{ccc}x-2& y-1& z-0\\ 3-2& -2-1& -2-0\\ 3-2& 1-1& 7-0\end{array}\right|=0\\ \\ & \Rightarrow \left|\begin{array}{ccc}x-2& y-1& z\\ 1& -3& -2\\ 1& 0& 7\end{array}\right|=0\end{array}$
$\begin{array}{c}(x-2)(-21)-(y-1)(9)+z(3)=0\\ \\ \therefore 7x+3y-z=17\end{array}$

The Plane exercise 28.1 question 1(ii)

Answer:-

The required equation of the plane is $2x-y-2z-2=0$
Hint:- Use equation of the plane $(x_1,y_1,z_1),(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$
Given:- $(-5,0,-6),(-3,10,-9)$ and $(-2,6,-6)$
Solution:- We know that, the equation of the plane passing through three non-collinear points $(x_1,y_1,z_1),(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$ is
$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$
Now, substitute the given value
$\begin{array}{rl}& \Rightarrow \left|\begin{array}{ccc}x+5& y-0& z+6\\ -3+5& 10-0& -9+6\\ -2+5& 6-0& -6+6\end{array}\right|=0\\ \\ & \Rightarrow \left|\begin{array}{ccc}x+5& y& z+6\\ 2& 10& -3\\ 3& 6& 0\end{array}\right|=0\end{array}$

$\begin{array}{c}(x+5)(18)-y(9)+(z+6)(-18)=0\\ \\ 18x+90-9y-18z-108=0\\ \\ \therefore 2x-y-2z-2=0\end{array}$

The Plane exercise 28.1 question 1(iii)

Answer:-

The required equation of plane is $x-3y-6z+8=0$
Hint:- Use equation of the plane $(x_1,y_1,z_1),(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$
Given:-$(1,1,1),(1,-1,2)$ and $(-2,-2,2)$
Solution:- We know that, the equation of the plane passing through given three points, $(x_1,y_1,z_1),(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$ is
$\left|\begin{array}{lll}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$
Now, substitute the given values
$\begin{array}{rl}& \Rightarrow \left|\begin{array}{ccc}x-1& y-1& z-1\\ 1-1& -1-1& 2-1\\ -2-1& -2-1& 2-1\end{array}\right|=0\\ \\ & \Rightarrow \left|\begin{array}{ccc}x-1& y-1& z-1\\ 0& -2& 1\\ -3& -3& 1\end{array}\right|=0\end{array}$

$\begin{array}{c}(x-1)(-2+3)-(y-1)(0+3)+(z-1)(0-6)=0\\ \\ (x-1)(1)-(y-1)3+(z-1)(-6)=0\\ \\ \therefore x-3y-6z+8=0\end{array}$

The Plane exercise 28.1 question 1(iv)

Answer:-

The required equation of the plane is $x+y-z-1=0$
Hint:- Use equation of the plane $(x_1,y_1,z_1),(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$
Given:-$(2,3,4),(-3,5,1)$ and $(4,-1,2)$
Solution:- We know that, the equation of the plane passing through given three points, $(x_1,y_1,z_1),(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$ is
$\left|\begin{array}{lll}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$
Now, substitute the given value
$\begin{array}{rl}& \Rightarrow \left|\begin{array}{ccc}x-2& y-3& z-4\\ -3-2& 5-3& 1-4\\ 4-2& -1-3& 2-4\end{array}\right|=0\\ \\ & \Rightarrow \left|\begin{array}{ccc}x-2& y-3& z-4\\ -5& 2& -3\\ 2& -4& -2\end{array}\right|=0\end{array}$
$\begin{array}{c}(x-2)(-4-12)-(y-3)(10+6)+(z-4)(20-4)=0\\ \\ (x-2)(-16)-(y-3)(16)+(z-4)(16)=0\\ \\ -16x+32-16y+48+16z-64=0\end{array}$

$\begin{array}{c}-16x-16y+16z+16=0\\ \\ \therefore x+y-z-1=0\end{array}$

The Plane exercise 28.1 question 1(v)

Answer:-

The required equation of the plane is $4x-3y+2z-3=0$
Hint:- Use equation of the plane $(x_1,y_1,z_1),(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$
Given:- $(0,-1,0),(3,3,0)$ and $(1,1,1)$
Solution:- We know that, the equation of the plane passing through given three points, $(x_1,y_1,z_1),(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$ is
$\left|\begin{array}{lll}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$
Now, substitute the given value
$\begin{array}{rl}& \Rightarrow \left|\begin{array}{ccc}x-0& y+1& z-0\\ 3-0& 3+1& 0-0\\ 1-0& 1+1& 1-0\end{array}\right|=0\\ \\ & \Rightarrow \left|\begin{array}{lll}x& y+1& z\\ 3& 4& 0\\ 1& 2& 1\end{array}\right|=0\end{array}$
$\begin{array}{c}(x)(4-0)-(y+1)(3-0)+z(6-4)=0\\ \\ 4x-(y+1)(3)+z((2)=0\\ \\ 4x-3y-3+2z=0\\ \\ \therefore 4x-3y+2z-3=0\end{array}$

The Plane exercise 28.1 question 2

Answer:-

The required equation of the plane is $5x-7y+11z+4=0$ and hence they are coplanar.
Hint:- Use equation of the plane $\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$
Given:-$(0,-1,-1),(4,5,1),(3,9,4)$ and $(-4,4,4)$
Solution:- Given that these four points are coplanar. So these four points lie on the same plane.
So, first let us take three points and find the equation of the plane passing through these four points and then let us substitute the fourth point in it. If is 0 then the points lies on the plane formed by these three points then they are coplanar. The equation of the plane passing through these three points is given
$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$
Now, let us take $(0,-1,-1),(4,5,1),(3,9,4)$ and find plane equation
$\begin{array}{rl}& \Rightarrow \left|\begin{array}{ccc}x-0& y+1& z+1\\ 4-0& 5+1& 1+1\\ 3-0& 9+1& 4+1\end{array}\right|=0\\ \\ & \Rightarrow \left|\begin{array}{ccc}x& y+1& z+1\\ 4& 6& 2\\ 3& 10& 5\end{array}\right|=0\end{array}$

$\begin{array}{c}x(30-20)-(y+1)(20-6)+(z+1)(40-18)=0\\ \\ 10x-14y+22z+8=0\\ \\ 5x-7y+11z+4=0\end{array}$
Now let us substitute fourth point $(-4,4,4)$ we get
$\begin{array}{c}5(-4)-7(4)+11(4)+4=0\\ \\ -20-28+44+4=0\\ \\ 48+48=0\\ \\ \therefore 0=0\end{array}$

∴$LHS=RHS$

So, as said above this fourth point satisfies. So, this point also lies on the same plane.
Hence they are coplanar.

The Plane exercise 28.1 question 3(i)

Answer:-

Hence they are coplanar.
Hint:- Use the equation of plane
$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$
Given:- $(0,-1,0),(2,1,-1),(1,1,1)$ and $(3,3,0)$
Solution:- Given that these four points are coplanar. So, these four points lie on the same plane.
So, first let us take three points and find the equation of plane passing through these four points and then let us substitute the fourth point in it. If it is 0 then the points lie on the plane formed by these three points then they are coplanar.
The equation of the plane passing through these three points is given
$\left|\begin{array}{lll}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$
Now, let us take $(0,-1,0),(2,1,-1),(1,1,1)$ and find plane equation.
$\begin{array}{rl}& \Rightarrow \left|\begin{array}{ccc}x-0& y+1& z+0\\ 2-0& 1+1& -1+0\\ 1-1& 1+1& 1-0\end{array}\right|=0\\ \\ & \Rightarrow \left|\begin{array}{ccc}x& y+1& z\\ 2& 2& -1\\ 1& 2& 1\end{array}\right|=0\end{array}$
$\begin{array}{c}x(2+2)-(y+1)(2+1)+z(4-2)=0\\ \\ 4x-3y-3+2z=0\\ \\ 4x-3y+2z-3=0\end{array}$
Now, let us substitute fourth point $(3,3,0)$ in plane equation $4x-3y+2z-3=0$
$4(3)-3(3)+2(0)-3=0$
$\begin{array}{c}12-9+0-3=0\\ \\ 12-12=0\\ \\ \therefore 0=0\end{array}$
$\therefore \mathrm{LHS}=\mathrm{RHS}$
So, this point also lies on the plane.
Hence they are coplanar.

The Plane exercise 28.1 question 3(ii)

Answer:-

Hence they are coplanar.
Hint:- Use equation of plane $\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$
Given:-$(0,4,3),(-1,-5,-3),(-2,-2,1)$ and $(1,1,-1)$
Solution:- Given that these four points are coplanar. So these four points lie on the same plane.
So, first let us take three points and find the equation of the plane passing through these four points and then let us substitute the fourth point in it. If it is 0 then the points lies on the plane formed by these three points then they are coplanar.
The equation of the plane passing through these three points is given by
$\left|\begin{array}{lll}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$
Now, let us take $(0,4,3),(-1,-5,-3),(-2,-2,1)$ and find plane equation
$\begin{array}{rl}& \Rightarrow \left|\begin{array}{ccc}x-0& y-4& z-3\\ -1-0& -5-4& -3-3\\ -2-0& -2-4& 1-3\end{array}\right|=0\\ \\ & \Rightarrow \left|\begin{array}{ccc}x& y-4& z-3\\ -1& -9& -6\\ -2& -6& -2\end{array}\right|=0\end{array}$
$\begin{array}{c}x(18-36)-(y-4)(2-12)+(z-3)(6-18)=0\\ \\ x(-18)-(y-4)(-10)+(z-3)(-12)=0\\ \\ -18x+10y-40-12z+36=0\\ \\ -18x+10y-12z-4=0\end{array}$
Now, let us substitute $(1,1,-1)$ in plane equation
$\begin{array}{c}-18x+10y-12z-4=0\\ \\ -18(1)+10(1)-12(-1)-4=0\\ \\ -18+10+12-4=0\end{array}$
$\begin{array}{c}-22+22=0\\ \\ \therefore 0=0\end{array}$
$\therefore \mathrm{LHS}=\mathrm{RHS}$
So, this point also lies on the plane.
Hence they are coplanar.

The Plane exercise 28.1 question 4

Answer:
Hence,$P$ externally divides the line segment $AB$ in the ratio $2:1.$
Hint:
Use determinant then solve the equation of plane.
Given:
The equation of the plane three points $L(2,2,1),M(3,0,1)\text{ and }\mathrm{N}(4,-1,0).$
Solution:
The equation of the plane passing through three points can be given by:
$L(2,2,1),M(3,0,1)\text{ and }\mathrm{N}(4,-1,0).$
$\left|\begin{array}{lll}x-2& y-2& z-1\\ x-3& y-0& z-1\\ x-4& y+1& z-0\end{array}\right|=0$
Operating $R_2\to R_2-R_1,\text{ and }{R}_3\to R_3-R_1$
$\left|\begin{array}{ccc}x-2& y-2& z-1\\ -1& 2& 0\\ -2& 3& 1\end{array}\right|=0$
Solving the above determinant, we get
$\begin{array}{rl}& (x-2)[2-0]-(y-2)[-1-0]+(z-1)[-3+4]=0\\ \\ & \Rightarrow (2x-4)+(y-2)+(z-1)=0\\ \\ & \Rightarrow 2x+y+z-7=0\end{array}$
Therefore, the equation of the plane is $2x+y+z-7=0$
Now, the equation of the line passing through two given points $A(3,-4,-5)$ and $B(2,-3,1)$ is
$\begin{array}{rl}& \frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}=\lambda \\ \\ & \Rightarrow \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda \end{array}$
$\begin{array}{rl}& \Rightarrow \frac{x-3}{-1}=\lambda ,\frac{y+4}{1}=\lambda ,\frac{z+5}{6}=\lambda \\ \\ & \Rightarrow x=(-\lambda +3),y=(\lambda -4),z=(6\lambda -5)\end{array}$
At the point of intersection, these points satisfy the equation of the plane
$2x+y+z-7=0$
Putting the value of $x,y$ and $z$ in the equation of the plane, we get the value of $\lambda$.
$\begin{array}{rl}& 2(-\lambda +3)+(\lambda +4)+(6\lambda -5)=0\\ \\ & \Rightarrow -2\lambda +6+\lambda -4+6\lambda -5-7=0\\ \\ & \Rightarrow 5\lambda =10\\ \\ & \Rightarrow \lambda =2\end{array}$
Thus, the point of intersection is $P(1,-2,7)$.
Now,
Let $P$ divides the line $AB$ in the ratio $m:n$
By the section formula, we have
Here,$A(3,-4,-5)\text{ and }\mathrm{B}(2,-3,1)\text{ and }P(1,-2,7)$
$\begin{array}{rl}& \Rightarrow \frac{2m+3n}{m+n}=1\\ \\ & \Rightarrow m+n=2m+3n\\ \\ & \Rightarrow m+2n=0\end{array}$
$\begin{array}{rl}& \Rightarrow m=-2n\\ \\ & \Rightarrow \frac{m}{n}=\frac{-2}{1}\end{array}$
Hence, $P$ externally divides the line segment $AB$ in the ratio $2:1$.

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RD Sharma Chapter wise Solutions

• Chapter 1 - Relations
• Chapter 2 - Functions
• Chapter 3 - Inverse Trigonometric Functions
• Chapter 4 - Algebra of Matrices
• Chapter 5 - Determinants
• Chapter 6 - Adjoint and Inverse of a Matrix
• Chapter 7 - Solution of Simultaneous Linear Equations
• Chapter 8 - Continuity
• Chapter 9 - Differentiability
• Chapter 10 - Differentiation
• Chapter 11 - Higher Order Derivatives
• Chapter 12 - Derivative as a Rate Measurer
• Chapter 13 - Differentials, Errors and Approximations
• Chapter 14 - Mean Value Theorems
• Chapter 15 - Tangents and Normals
• Chapter 16 - Increasing and Decreasing Functions
• Chapter 17 - Maxima and Minima
• Chapter 18 - Indefinite Integrals
• Chapter 19 - Definite Integrals
• Chapter 20 - Areas of Bounded Regions
• Chapter 21 - Differential Equations
• Chapter 22 - Algebra of Vectors
• Chapter 23 - Scalar Or Dot Product
• Chapter 24 - Vector or Cross Product
• Chapter 25 - Scalar Triple Product
• Chapter 26 - Direction Cosines and Direction Ratios
• Chapter 27 - Straight Line in Space
• Chapter 28 - The Plane
• Chapter 29 - Linear programming
• Chapter 30- Probability
• Chapter 31 - Mean and Variance of a Random Variable