RD Sharma Class 12 Exercise 28.10 The Plane Solutions Maths

RD Sharma Class 12 Exercise 28.10 The Plane Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 11:52 AM IST

The majority of the class 12 students follow the RD Sharma solutions book for their reference purposes. Most students cannot visit tuitions or invite private tutors after school hours to help them with the homework. When the students solve mathematics sums, especially from the chapter, The Plane, they get confused and do not arrive at the solution. For such people, the RD Sharma Class 12th Exercise 28.10 guides them on the right path.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics

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

RD Sharma Class 12 Solutions Chapter28 The Plane - Other Exercise

The Plane Excercise: 28.10

The Plane Exercise 28.10 Question 1

Answer:

Distance is $\frac{25}{3\sqrt{14}}$
Hint: Use Formula $P=\frac{a x_{1}+b y_{1}+c_{z_{1}}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}$
Given: $-2x-y+3z-4=0$ and $6x-3y+9z+13=0$
Solution:Let $P\left ( x_{1},y_{1},z_{1} \right )$ be any point on $2x-y+3z-4=0$ Then
$2x_{1}-y_{1}+3z_{1}=4$ .......(1)
Let, P = Distance between $\left ( x_{1},y_{1},z_{1} \right )$ and the plane $6x-3y+9z+13=0.$
We know the distance of point $\left ( x_{1},y_{1},z_{1} \right )$ from the plane $ax+by+cz+d=0$ is given by
$\begin{aligned} &\mathrm{P}=\left|\frac{a x_{1}+b y_{1}+c_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right| \\ \end{aligned}$

Putting the values
$\begin{aligned} &\mathrm{P}=\left|\frac{6 x_{1}+(-3) y_{1}+9 z_{1}+13}{\sqrt{6^{2}+(-3)^{2}+9^{2}}}\right| \\ \end{aligned}$
$\begin{aligned} &\mathrm{P}=\left|\frac{3\left(2 x_{1}-y_{1}+3 z_{1}\right)+13}{\sqrt{36+9+81}}\right| \\ \end{aligned}$
$\begin{aligned} &\mathrm{P}=\left|\frac{3(4)+13}{3 \sqrt{14}}\right| \\ \end{aligned}$ (From equation (1))
$\begin{aligned} &\mathrm{P}=\frac{25}{3 \sqrt{14}} \end{aligned}$
Therefore, the distance between the parallel planes $2x-y+3z-4=0$ and $6x-3y+3z+13=0$ is $\frac{25}{3\sqrt{14}}$ units.

The Plane Exercise 28.10 Question 2

Answer: - Therefore, equation of the plane is $2x-3y+5z+11=0$ and distance of the plane is $\frac{4}{\sqrt{38}}$ units.
Hint: - Use formula $\mathrm{P}=\left|\frac{a x_{1}+b y_{1}+c_{z}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Given: - Point (3, 4, -1) and $2x-3y+5z+7=0$
Solution: - Since the plane is parallel to $2x-3y+5z+7=0$ it must be of the form:
$2x-3y+5z+\theta =0$ …… (i)
According to question:
The plane passes through (3, 4, - 1)
$2\left ( 3 \right )-3\left ( 4 \right )+5\left ( -1 \right )+\theta =0$
$\theta =1$
Putting the value of $\theta$ in eqn (i)
So, we get the equation of the plane is $2x-3y+5z+11=0$
Distance of the plane $2x-3y+5z+11=0$ from (3, 4, -1)
We know, the distance of point $\left ( x_{1},y_{1},z_{1} \right )$ from the plane ax + by + cz + d = 0 is given by $\begin{aligned} &\mathrm{P}=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right| \\ \end{aligned}$

Putting the values,
$\begin{aligned} &\mathrm{P}=\left|\frac{2(3)+(-3)(4)+5(-1)+7}{\sqrt{2^{2}+(-3)^{2}+5^{2}}}\right| \\ &\mathrm{P}=\left|\frac{-4}{\sqrt{38}}\right| \\ &\mathrm{P}=\frac{4}{\sqrt{38}} \end{aligned}$
Therefore, the distance of the plane $2x-3y+5z+7=0$ from (3, 4, - 1) is $\frac{4}{\sqrt{38}}$ units.

The Plane Exercise 28 .10 Question 3

Answer: - Therefore, equation of the mid- parallel plane is $2x-2y+z+6=0$
Hint: - Use formula $\mathrm{P}=\left|\frac{a x_{1}+b y_{1}+c_{-1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Given:- $2x-2y+z+3=0$ and $2x-2y+z+9=0$
Let $\pi _{1}-2x-2y+z+3=0$ and $2x-2y+z+9=0$
And $\pi _{2}=2x-2y+z+9=0$
Solution: - Let the equation of the plane mid- parallel to these plane be
$\pi _{3}=2x-2y+z+\theta =0$
Now, let $P\left ( x_{1},y_{1},z_{1} \right )$ be any point on this plane.
$2x_{1}-2y_{1}+z_{1}+\theta =0$ ----- (1)
We know, the distance of point from the plane $ax +by +cz + d = 0$
$\begin{aligned} &\mathrm{P}=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|\\ \end{aligned}$
Distance of P from $\pi _{1}$ $\begin{aligned} &\mathrm{P}=\left|\frac{2 x_{1}-2 y_{1}+z_{1}+3}{\sqrt{2^{2}+(-2)^{2}+1^{2}}}\right|\\ &\mathrm{P}=\left|\frac{(-\theta)+3}{3}\right| \text { (using equation (1)) }\\ \end{aligned}$

Similarly, Distance of Q from $\pi _{2}$
$\begin{aligned} &\mathrm{Q}=\left|\frac{2 x_{1}-2 y_{1}+z_{1}+9}{\sqrt{2^{2}+(-2)^{2}+1^{2}}}\right|\\ &\mathrm{Q}=\left|\frac{(-\theta)+9}{3}\right| \text { (using equation (1)) }\\ \end{aligned}$

As $\pi _{3}$ is mid- parallel to $\pi _{1}$ and $\pi _{2}$
$\begin{aligned} &\mathrm{P}=\mathrm{Q}\\ \end{aligned}$
$\begin{aligned} &\left|\frac{(-\theta)+3}{3}\right|=\left|\frac{(-\theta)+9}{3}\right|\\ \end{aligned}$

Squaring both sides
$\begin{aligned} &\left(\frac{(-\theta)+3}{3}\right)^{2}=\left(\frac{(-\theta)+9}{3}\right)^{2}\\ \end{aligned}$
$\begin{aligned} &(3-\theta)^{2}=(9-\theta)^{2}\\ \end{aligned}$
$\begin{aligned} &9-6 \theta+\theta^{2}=81-18 \theta+\theta^{2}\\ &12 \theta=72 \end{aligned}$
$\theta =6$
Therefore, equation of the mid-parallel plane is $2x-2y+z+6=0$

The Plane Exercise 28.10 Question 4

Answer: - The required distance is $\frac{7}{\sqrt{56}}$ unit.
Hint: - Use formula $\mathrm{P}=\left|\frac{\vec{a} \cdot \vec{n}-d}{|\vec{n}|}\right|$
Given: - $r(\hat{i}+2 \hat{j}+3 \hat{k})+7=0 \text { and } \vec{r} \cdot(2 \hat{i}+4 \hat{j}+6 \hat{k})+7=0$
Solution: - Let $\bar{a}$ be the Position vector of any point on plane $\bar{r}.(\hat{i}+2 \hat{j}+3 \hat{k})+7=0 \: \: \:$
$\vec{a} \cdot( \hat{i}+2 \hat{j}+3 \hat{k})+7=0$ ----- (1)
We know the distance of $\vec{a}$ from the plane $\vec{r}.\vec{n}-d=0$ $\begin{aligned} &\mathrm{P}=\left|\frac{\vec{a} \cdot{n}-d}{|\vec{n}|}\right| \\ \end{aligned}$

Putting the value of $\vec{a}$ and $\vec{n}$ .
$\begin{aligned} &\mathrm{P}=\left|\frac{\vec{a} \cdot(2 \hat{i}+4 \hat{j}+6 \hat{k})+7}{|(2 \hat{i}+4 \hat{j}+6 \hat{k})|}\right| \\ &\mathrm{P}=\left|\frac{2 \vec{a} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})+7}{\sqrt{2^{2}+4^{2}+6^{2}}}\right| \\ &\mathrm{P}=\left|\frac{2(-7)+7}{\sqrt{56}}\right| \\ &\mathrm{P}=\left|\frac{-7}{\sqrt{56}}\right| \\ &\mathrm{P}=\frac{7}{\sqrt{56}} \end{aligned}$
Therefore, the distance between the plane $\bar{r}.(\hat{i}+2 \hat{j}+3 \hat{k})+7=0 \: \: \:$ and $\vec{r} \cdot( 2\hat{i}+4 \hat{j}+6 \hat{k})+7=0$ is $\frac{7}{\sqrt{56}}$ units.

The syllabus of chapter 28, The Plane, has around fifteen exercises, ex 28.1 to 28.15. The tenth exercise, or ex 28.10, consists of finding the distance between two parallel planes, finding the plane's vector equation, and finding the distance between the planes. There are four questions asked in this exercise, and all are Level 1 questions. Whenever the students face an obstacle in solving the problem or are unclear about the steps and concept, the RD Sharma Class 12 Chapter 28 Exercise 28.10 reference material helps them a lot.

There are only four questions given in the textbook, which are not enough for the students to completely understand the concept. Therefore, numerous practice questions for this exercise are available in RD Sharma Class 12th Exercise 28.10 reference book. All the solutions are given according to the NCERT pattern, making the CBSE students grasp the methods easily. The Class 12 RD Sharma Chapter 28 Exercise 28.10 Solution book plays a major role in developing the confidence of the students to face the exams.

As the solutions are framed by a team of experts, the students need not worry about their accuracy. The RD Sharma Class 12 Solutions The Plane Ex 28.10 reference book can be used while the students are working on their homework, assignments, and even studying for their tests and examinations.

The accessibility of RD Sharma Class 12 Solutions Chapter 28 Ex 28.10 for free, has made many students make it their primary reference book. This free accessibility is provided by the career360 websites for the students to refer to the RD Sharma solution books. No one is asked to pay even a penny to download or refer to these books.

The questions for public examinations have also been taken from the RD Sharma Class 12th Exercise 28.10 practice questions section. Therefore, if a student prepares for their exam using the RD Sharma solution books, they are indirectly preparing for the questions which will be asked in the public exam. This escalated their score to a larger extent in the exams.

RD Sharma Chapter wise Solutions

Frequently Asked Questions (FAQs)

1. Why are the RD Sharma books recommended to the class 12 students?

The solutions provided in these books are given by various experts.
The practice questions help them understand the concepts in-depth.
Most of the students have benefited with the help

2. Which book contains the accurate solutions for chapter 28, Ex 20.10?

The right solutions for chapter 28, ex 28.10, can be found in the RD Sharma Class 12th Exercise 28.10 reference material.

3. Where can I find the RD Sharma solutions book online?

The RD Sharma books are present on the Career360 website, where they can be accessed for free.

4. Are there any possibilities of downloading the RD Sharma books to my device?

Yes, you can not only access the RD Sharma solution book for free at the Career360 website but the same can also be downloaded to your device for later reference purposes.

5. How many exercises does the class 12 mathematics chapter 28 contain?

There are fifteen exercises in the 28th chapter of class 12 mathematics. The solutions for the tenth exercise, ex 28.10, can be found in the RD Sharma Class 12th Exercise 28.10 book.