RD Sharma Class 12 Exercise 28.9 The Plane Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 28.9 The Plane Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 11:55 AM IST

The RD Sharma solutions are the all in one problem solver for all the CBSE students who are preparing to appear for the board exams. The Class 12 RD Sharma chapter 28 exercise 28.9 solution is advised to go through while practicing the chapter ‘The Plane.’ It helps you to get a brief understanding of the concept while providing excellent examples that help you to understand better. The RD Sharma class 12th exercise 28.9 is the most recommended book by teachers for the students of CBSE as it matches the level of NCERT.

Also Read - RD Sharma Solutions For Class 9 to 12 Maths

## The Plane Exercise: 28.9

The Plane Exercise 28.9 Question 1

Answer : $\frac{47}{13}$
Hint : $\text { Distance }=\frac{\left|A x_{1}+B y_{1}+C z_{1}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}}$
Given : Point $(2 \hat{i}-\hat{j}-4 \hat{k})$ from the plane vector $\vec{r} \cdot(3 \hat{i}-4 \hat{j}+12 \hat{k})=9$
Solution :
For the given plane
Vector, $\vec{r} \cdot(3 \hat{i}-4 \hat{j}+12 \hat{k})=9$
Cartesian form is
$3 x-4 y+12 z=9$
We can write as,
$3 x-4 y+12 z-9=0$
The point given is $(2 \hat{i}-\hat{j}-4 \hat{k})$
Which can be written as $(2,-1,-4)$
We know that
\begin{aligned} &\text { Distance }=\frac{\left|A x_{1}+B y_{1}+C z_{1}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}} \\ &\text { Distance }=\frac{|(2 \times 3)+(-1 \times-4)+(-4 \times 12)+(-9)|}{\sqrt{3^{2}+(-4)^{2}+12^{2}}} \end{aligned}
So we get,
$=\frac{|6+4-48-9|}{\sqrt{9+16+144}}$
On further calculation
\begin{aligned} &=\frac{|-47|}{\sqrt{169}} \\ \end{aligned}
\begin{aligned} &=\frac{47}{13} \text { Units } \end{aligned}

The Plane Exercise 28.9 Question 2

Answer : The given points are equidistant from the plane
Hint : $\text { Distance }=\frac{\left|A x_{1}+B y_{1}+C z_{1}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}}$
Given :
Two points $(\hat{i}-\hat{j}+3 \hat{k})$ and $(3 \hat{i}+3 \hat{j}+3 \hat{k})$, plane $\vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9=0$
Solution :
Points given by the equation
$\vec{a}=\hat{i}-\hat{j}+3 \hat{k}, \vec{b}=3 \hat{i}+3 \hat{j}+3 \hat{k}$
Plane given by the equation
$\vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9=0$, where the normal is $\vec{n}=5 \hat{i}+2 \hat{j}-7 \hat{k}$
We know, the distance of $\vec{a}$ from the plane $\vec{r} \cdot \vec{n}-d=0$ Is given by
$p=\left|\frac{\vec{a} \cdot \vec{n}-d}{|\vec{n}|}\right|$
Distance of $\hat{i}-\hat{j}+3 \hat{k}$ from the plane
\begin{aligned} &=\left|\frac{(\hat{i}-\hat{j}+3 \hat{k}) \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9}{|5 \hat{i}+2 \hat{j}-7 \hat{k}|}\right| \\ &=\frac{9}{\sqrt{78}} \text { Units } \end{aligned}
And,
Distance of $3 \hat{i}+3 \hat{j}+3 \hat{k}$ from the plane
$=\left|\frac{(3 \hat{i}+3 \hat{j}+3 \hat{k}) \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9}{|5 \hat{i}+2 \hat{j}-7 \hat{k}|}\right|$
$= \frac{9}{\sqrt{78}}$ Units
The point $\hat{i}-\hat{j}+3 \hat{k} \; \text {and} \; 3 \hat{i}+3 \hat{j}+3 \hat{k}$ are equidistant from the plane
$\vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9=0$

The Plane Exercise 28.9 Question 3

Hint : $P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Given :
Point $A(2,3,-5)$
Plane $x+2 y-2 z-9=0$
Solution :
We know, the distance of point $\left(x_{1}, y_{1}, z_{1}\right)$ from the plane
$a x+b y+c z+d=0$ Is given by
$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Putting the necessary values
Distance of the plane from $A=\left|\frac{1 \times 2+2 \times 3+(-2) \times(-5)-9}{\sqrt{1^{2}+2^{2}+(-2)^{2}}}\right|=3 \text { Units }$
Hence the distance of the plane $(2,3,-5)$ from the plane $x+2 y-2 z-9=0$ is 3 units

The Plane Exercise 28.9 Question 4

Answer : The answer of the given question is $x+2 y-2 z+4=0, x+2 y-2 z-8=0$
Hint :
$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Given : Plane: $x+2 y-2 z+8=0$, which are at a distance of 2 units from the point $(2,1,1)$
Solution :
The planes are parallel to $x+2 y-2 z+8=0$ they must be of the form
$x+2 y-2 z+\theta=0$
We know, the distance of point $\left(x_{1}, y_{1}, z_{1}\right)$ from the plane $a x+b y+c z+d=0$ given by
$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
According to the question, the distance of the plane from $(2,1,1)$ is 2 units
\begin{aligned} &\left|\frac{1 \times 2+2 \times 1+(-2) \times(1)+\theta}{\sqrt{1^{2}+2^{2}+(-2)^{2}}}\right|=2 \\ &\left|\frac{2+\theta}{3}\right|=2 \end{aligned}
\begin{aligned} &\frac{2+\theta}{3}=2 \text { Or } \frac{2+\theta}{3}=-2 \\ &\theta=4 \text { Or } \theta=-8 \end{aligned}
Hence the required planes are
\begin{aligned} &x+2 y-2 z+4=0 \\ &x+2 y-2 z-8=0 \end{aligned}

The Plane Exercise 28.9 Question 5

Answer: The answer of the given question is that the points are equidistant from the plane
Hint :
$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Given :
Points $A(1,1,1) \text { and } B(-3,0,1)$
Plane $p=3 x+4 y-12 z+13=0$
Solution :
We know that, the distance of points $\left(x_{1}, y_{1}, z_{1}\right)$ from the plane $p=a x+b y+c z+d=0$ Is given by
Distance of $(1,1,1)$ from the plane $=\left|\frac{3 \times 1+4 \times 1+(-12) \times 1+13}{\sqrt{3^{2}+4^{2}+(-12)^{2}}}\right|=\frac{8}{13} \text { Units }$
Distance of $(-3,0,1)$ from the plane $=\left|\frac{3 \times(-3)+4 \times 0+(-12) \times 1+13}{\sqrt{3^{2}+4^{2}+(-12)^{2}}}\right|=\frac{8}{13} \text { Units }$
The points $(1,1,1) \&(-3,0,1)$ are equidistant from the plane $3 x+4 y-12 z+13=0$

The Plane Exercise 28.9 Question 6

Answer : The answer of the given question are $x-2 y+2 z+2=0, x-2 y+2 z-4=0$
Hint :
$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Given :
$x-2 y+2 z-3=0 \text {, Point }(1,1,1)$
Solution :
Since the planes are parallel to $x-2 y+2 z-3=0$ they must be of the form $x-2 y+2 z+\theta=0$
We know, the distance of point $\left(x_{1}, y_{1}, z_{1}\right)$ from the plane $p: a x+b y+c z+d=0$ is given by
$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
According to the question, the distance of the plane from $(1,1,1)$ is 1 Unit
\begin{aligned} &\left|\frac{1 \times 1+(-2) \times 1+2 \times 1+\theta}{\sqrt{1^{2}+(-2)^{2}+2^{2}}}\right|=1 \\ &\left|\frac{1+\theta}{3}\right|=1 \end{aligned}
\begin{aligned} &\frac{1+\theta}{3}=1 \text { Or } \frac{1+\theta}{3}=-1 \\ &\theta=2 \text { Or }-4 \end{aligned}
The required planes are
$x-2 y+2 z+2=0 \& x-2 y+2 z-4=0$

The Plane Exercise 28.9 Question 7

Hint :
$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Given :
Points : $A(2,3,5)$
Plane : $z = 0$
Solution :
We know, the distance of point $\left(x_{1}, y_{1}, z_{1}\right)$ from the plane
$p: a x+b y+c z+d=0$ Is given by
$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Putting the values
$P=\left|\frac{0 \times 2+0 \times 3+1 \times 5+0}{\sqrt{0^{2}+0^{2}+1^{2}}}\right|=5 \text { Unit }$
The distance of the point $(2,3,5)$ from the xy-plane is 5 unit

The Plane Exercise 28.9 Question 8

Answer : The answer of the given question is $\frac{9}{\sqrt{78}}$ unit
Hint :
$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Given :
Points : $A(3,3,3)$
Plane : $\vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9=0$, which in Cartesian form
Solution :
$5 x+2 y-7 z+9=0$ and point $(3,3,3)$
We know the distance of point $\left(x_{1}, y_{1}, z_{1}\right)$ from the plane
$p: a x+b y+c z+d=0$ Is given by
$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Putting the values
$P=\left|\frac{5 \times 3+2 \times 3+(-7) \times 3+9}{\sqrt{5^{2}+2^{2}+(-7)^{2}}}\right|=\frac{9}{\sqrt{78}} \text { Units }$
The distance of the point $(3,3,3)$ from the plane,
$\vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9=0 \text { is } \frac{9}{\sqrt{78}} \text { Units }$

The Plane Exercise 28.9 Question 9

Answer : The answer of the given question is $\lambda=4$
Hint :
$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Given : Points: $(1,1,1) \text { and } x-y+z+\lambda=0 \text { be } 5$
Solution :
The distance of the point $(1,1,1)$ from the origin
We know, distance of $\left(x_{1}, y_{1}, z_{1}\right)$ from the origin is $\sqrt{x_{1}^{2}+y_{1}^{2}+z_{1}^{2}}$
Putting values of $x_{1}, y_{1}, z_{1}=1$
Required distance $=\sqrt{3}$
Distance of the point $(1,1,1)$ from plane $x-y+z+\lambda=0$
We know, the distance of point $\left(x_{1}, y_{1}, z_{1}\right)$ from the plane $\pi=a x+b y+c z+d=0$
Is given by
$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Putting necessary values,
\begin{aligned} &P=\left|\frac{1 \times 1+(-1) \times 1+1 \times 1+\lambda}{\sqrt{1^{2}+(-1)^{2}+1^{2}}}\right| \\ &P=\left|\frac{1+\lambda}{\sqrt{3}}\right| \end{aligned}
According to the question
\begin{aligned} &\left|\frac{1+\lambda}{\sqrt{3}}\right| \times \sqrt{3}=5 \\ &|1+\lambda|=5 \\ &1+\lambda=5 \text { Or } 1+\lambda=-5 \\ &\lambda=4 \text { Or } \lambda=-6 \end{aligned}
$\lambda = 4$ [$\lambda \neq -6$ Because distance cannot be negative]

The Plane Exercise 28.9 Question 10

Answer : The answer of the given question is $37 x+20 y-21 z=61,67 x-20 y+99 z=121$
Hint :
$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Solution : Let $(x,y)$ be a point which is equidistant from the given plane. Then,
\begin{aligned} &\frac{|3 x-4 y+12 z-6|}{\sqrt{9+16+144}}=\frac{|4 x+3 z-7|}{\sqrt{16+9}} \\ &\frac{3 x-4 y+12 z-6}{13}=\pm \frac{4 x+3 z-7}{5} \end{aligned}
\begin{aligned} &15 x-20 y+60 z-30=52 x+39 z-91 ; 15 x-20 y+60 z-30=-52 x-39 z+91 \\ &37 x+20 y-21 z=61 ; 67 x-20 y+99 z=121 \end{aligned}
Therefore this is the required equation.

The Plane Exercise 28.9 Question 11

Answer : The answer of the given question is $\sqrt{29}$ Units
Hint :
$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Given :
3 Point $(7,2,4)$ and the plane determine by the point
$A(2,5,-3), B(-2,-3,5) \&(5,3,-3)$
Solution :
The equation of the plane passing through $\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right) \&\left(x_{3}, y_{3}, z_{3}\right)$ is given by the following equation
$\left|\begin{array}{lll} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{array}\right|=0$
According to questions
\begin{aligned} &\left(x_{1}, y_{1}, z_{1}\right)=(2,5,-3) \\ &\left(x_{2}, y_{2}, z_{2}\right)=(-2,-3,5) \\ &\left(x_{3}, y_{3}, z_{3}\right)=(5,3,-3) \end{aligned}
Putting these values
$\left|\begin{array}{ccc} x-2 & y-5 & z-(-3) \\ -4 & -8 & 8 \\ 3 & -2 & 0 \end{array}\right|=0$
\begin{aligned} &(x-2)(16)+(y-5)(24)+(z+3)(32)=0 \\ &2 x+3 y+4 z-7=0 \end{aligned}
Distance of $2 x+3 y+4 z-7=0$ from $(7,2,4)$
We know, the distance of point $\left(x_{1}, y_{1}, z_{1}\right)$ from the plane
$\pi: a x+b y+c z+d=0$ Is given by
\begin{aligned} &P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right| \\ &P=\left|\frac{2 \times 7+3 \times 2+4 \times 4-7}{\sqrt{2^{2}+3^{2}+4^{2}}}\right| \end{aligned}
$P= \sqrt{29}$ Units

The Plane Exercise 28.9 Question 12

Answer : The answer of the given question is $\frac{12}{\sqrt{29}}$ units
Hint :
$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Given : Plane makes intercept -6,3,4 respectively on the coordinate axes
Solution :
The equation of the plane which makes intercept a, b and c with the x, y and z axes respectively is
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Putting the values of a, b and c
Require equation of the plane
\begin{aligned} &\frac{x}{-6}+\frac{y}{3}+\frac{z}{4}=1 \\ &-2 x+4 y+3 z=12 \end{aligned}
We know, the distance of the point $\left(x_{1}, y_{1}, z_{1}\right)$ from the plane
$\pi: a x+b y+c z+d=0$ Is given by
$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Distance from the origin i.e. $(0,0,0):\left|\frac{d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Required distance
\begin{aligned} &=\left|\frac{-12}{\sqrt{(-2)^{2}+4^{2}+3^{2}}}\right| \\ &=\frac{12}{\sqrt{29}} \end{aligned}
The length of the perpendicular from the origin on the plane $=\frac{12}{\sqrt{29}} \text { Units }$

The Plane Exercise 28.9 Question 13

Answer : $2\sqrt{2}$
Hint :
$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Given : We have to find the distance of the point $(1,-2,4)$ from the plane passing through the point $(1,2,2)$
Solution : Let us consider the equation of plane passing through the point $(1,2,2)$ be
$a(x-1)+b(y-2)+c(z-2)=0$ (1)
The direction ratios of the normal to the plane are a, b and c
The equation of the plane are
\begin{aligned} &x-y+2 z=3 \\ &2 x-2 y+z+12=0 \end{aligned}
Since the plane passing through the point $(1,2,2)$ is perpendicular the given plane
Therefore,
\begin{aligned} &a-b+2 c=0\\ &2 a-2 b+c=0 \end{aligned}
Now, eliminating a, b and c from (1), (2) and (3)
We get
\begin{aligned} &\left|\begin{array}{ccc} x-1 & y-2 & z-2 \\ 1 & -1 & 2 \\ 2 & -2 & 1 \end{array}\right|=0 \\ &(x-1)(-1+4)-(y-2)(1-4)+(z-2)(-2+2)=0 \end{aligned}
\begin{aligned} &3(x-1)+3(y-2)+0=0 \\ &3 x-3+3 y-6=0 \\ &3 x+3 y-9=0 \\ &x+y-3=0 \end{aligned}
Now using distance formula
The distance of the point $(1,-2,4)$ from the plane $x+y-3=0$ is
\begin{aligned} &=\left|\frac{1-2-3}{\sqrt{1^{2}+1^{2}+0^{2}}}\right| \\ &=\left|\frac{-4}{\sqrt{2}}\right| \\ &=2 \sqrt{2} \text { Units } \end{aligned}

The RD Sharma class 12th exercise 28.9 covers the chapter 'The Plane.' There are about 19 questions in this exercise that are extremely basic and simple and covers all the essential concepts that are mentioned below-

• Intercept form of the equation

• Distance of the point from the plane

• Equation of plane passing through points

• Points are equidistant from the plane

• Equation of plane passing through three points

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• It has been found that most of the questions that are asked in the 12th board exams are similar to the ones that are provided in the RD Sharma solution, therefore having a thorough practice of the solution can help you score high in the exams.

• The exercises given in the RD Sharma class 12th exercise 28.9 are divided into two levels so that the students can distinguish easily between the low level and high level questions and practice accordingly.

• The RD Sharma solutions are the best study material for any students who have the ambition to score high marks in the board exams and therefore teachers of all the CBSE schools recommend making use of this solution to practice rigorously.

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## RD Sharma Chapter-wise Solutions

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