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

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.

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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