Lovekush kumar sainiUpdated on 25 Jan 2022, 11:54 AM IST

The RD Sharma solutions are a country-wide famous series of books for its extraordinary concepts that are so basic for students to understand. The Class 12 RD Sharma chapter 28 exercise 28.5 solution contains the questions for the chapter ‘The Plane’, which is also explained theoretically in a very simple language for any student to understand. The RD Sharma class 12th exercise 28.5 is the best study material to make your concept clear in this chapter.

RD Sharma Class 12 Solutions Chapter28 The Plane - Other Exercise

The Plane Excercise: 28.5

The Plane exercise 28.5 question 1

Answer:
The required vector equation of plane is
$\vec{r}.(3\hat{i}-4\hat{k})+1=0$
Hint:
Using
$\vec{r}.\vec{n}=\vec{a}.\vec{n}$
Given:
Vector equation of the plane passing through the points (1,1,1), (1,-1,1) and (-7,-3,-5)
Solution:
Let A (1,1,1), (1,-1,1) and (-7,-3,-5) be the coordinates.
The required plane passes through the point A (1, 1, 1) whose vector is
$\vec{a}=\hat{i}+\hat{j}+\hat{k}$
and is normal to the vector $\vec{n}$ given by
$\vec{n}=\vec{AB}\times \vec{AC}$ clearly.
$\begin{aligned} &\vec{AB}=\vec{OB}-\vec{OA}=(\hat{i}-\hat{j}+\hat{k})-(\hat{i}+\hat{j}+\hat{k})\\ &=0\hat{i}-2\hat{j}+0\hat{k}\\ &\vec{AC}=\vec{OC}-\vec{OA}=(-7\hat{i}-3\hat{j}-5\hat{k})-(\hat{i}+\hat{j}+\hat{k})\\ &=-8\hat{i}-4\hat{j}+6\hat{k}\\ &\vec{n}=\vec{AB}-\vec{AC}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 0 &-2 &0 \\ -8 &-4 &-6 \end{vmatrix}=12\hat{i}+0\hat{j}-16\hat{k} \end{aligned}$
The vector equation of required plane is: -
$\begin{aligned} &\vec{r}.(12\hat{i}+0\hat{j}-16\hat{k})=(\hat{i}+\hat{j}+\hat{k}).(12\hat{i}+0\hat{j}-16\hat{k})\\ &\vec{r}[4(3\hat{i}-4\hat{k})]=12+0-16\\ &\vec{r}[4(3\hat{i}-4\hat{k})]=-4\\ &\vec{r}[(3\hat{i}-4\hat{k})]=-1\\ &\vec{r}[(3\hat{i}-4\hat{k})]+1=0 \end{aligned}$

The Plane exercise 28.5 question 2

Answer:
The vector equation of required plane is
$\vec{r}.(2\hat{i}+3\hat{j}+4\hat{k})=7$
Hint:
Using formula
$\vec{r}.\vec{n}=\vec{a}.\vec{n}$
Given:
Vector equation of plane passing through the points P(2, 5, -3), Q(-2, -3, 5) and R(5, 3,-3)
Solution:
The required plane passes through the point (2, 5, - 3) whose position vector is
$\vec{a}=(2\hat{i}+5\hat{j}-3\hat{k})$
and is normal to the vector $\vec{n}$ given by
$\begin{aligned} &\vec{n}=\vec{PQ}\times \vec{PR}\\ &\vec{PQ}=\vec{OQ}- \vec{OP}=(-2\hat{i}-3\hat{j}+5\hat{k})-(2\hat{i}+5\hat{j}-3\hat{k})=-4\hat{i}-8\hat{j}+8\hat{k}\\ &\vec{PR}=\vec{OR}- \vec{OP}=(5\hat{i}+3\hat{j}-3\hat{k})-(2\hat{i}+5\hat{j}-3\hat{k})=3\hat{i}-2\hat{j}-0\hat{k}\\ &\vec{n}=\vec{PQ}\times \vec{PR}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ -4 &-8 &8 \\ 3 &-2 &0 \end{vmatrix}=16\hat{i}+24\hat{j}+32\hat{k} \end{aligned}$
The vector equation of the required plane is
$\begin{aligned} &\vec{r}.\vec{n}=\vec{a}.\vec{n}\\ &\vec{r}.(16\hat{i}+24\hat{j}+32\hat{k})=(2\hat{i}+5\hat{j}-3\hat{k}).(16\hat{i}+24\hat{j}+32\hat{k})\\ &\vec{r}.[8(2\hat{i}+3\hat{j}+4\hat{k})]=32+120-96\\ &\vec{r}.[8(2\hat{i}+3\hat{j}+4\hat{k})]=56\\ &\vec{r}.(2\hat{i}+3\hat{j}+4\hat{k})=7 \end{aligned}$

The Plane exercise 28.5 question 3

Answer:
The required vector equation of plane is:
$\vec{r}=(bc\hat{i}+ac\hat{j}+ab\hat{k})=abc$
Hint:
Using
$\vec{r}.\vec{n}=\vec{a}.\vec{n}$
Given:
Vector equation of the plane passing through the point A (a, 0, 0), B (0, b, 0) and
C (0, 0, c). Prove that
$\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}$
Solution:
The required plane passing through the point A (a, 0, 0) whose position vector is
$\vec{a}=(a\hat{i}+0\hat{j}+0\hat{k})$
and is normal to the vector given by
The vector equation of required plane is
$\begin{aligned} &\vec{r}.(bc\hat{i}+ac\hat{j}+ab\hat{k})=(a\hat{i}+0\hat{j}+0\hat{k}).(bc\hat{i}+ac\hat{j}+ab\hat{k})\\ &\vec{r}.(bc\hat{i}+ac\hat{j}+ab\hat{k})=abc+0+0\\ &\vec{r}.(bc\hat{i}+ac\hat{j}+ab\hat{k})=abc \qquad \qquad \dots(i)\\ &\left | \vec{n} \right |=\sqrt{(bc)^2+(ac)^2+(ab)^2}=\sqrt{b^2c^2+a^2c^2+a^2b^2} \end{aligned}$
For reducing (i) to normal form, we need to divide both sides of (i) by
$\begin{aligned} &\sqrt{b^2c^2+a^2c^2+a^2b^2} \end{aligned}$
Then, we get
$\begin{aligned} &\vec{r}.\left ( \frac{bc\hat{i}+ac\hat{j}+ab\hat{k}}{\sqrt{b^2c^2+a^2c^2+a^2b^2}} \right )=\frac{abc}{\sqrt{b^2c^2+a^2c^2+a^2b^2}} \end{aligned}$
Therefore, the normal form of plane (i) is
$\begin{aligned} &\frac{abc}{\sqrt{b^2c^2+a^2c^2+a^2b^2}} \end{aligned}$
So, the distance of plane (i) from the origin
$\begin{aligned} &P=\frac{abc}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}\\ &\frac{1}{P}=\frac{\sqrt{b^2c^2+a^2c^2+a^2b^2}}{abc}\\ &\frac{1}{P^2}=\frac{b^2c^2+a^2c^2+a^2b^2}{a^2b^2c^2}\\ &\frac{1}{P^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \end{aligned}$
Hence proved

The Plane exercise 28.5 question 4

Answer:
The required vector equation of plane is a(x - 1) + b(y - 1) + c (z + 1)=0,
Where 5a + 3b - 4c = 0
Hint:
Convert to the vector form by the given points.
Given:
Vector equation of plane passing through the points (1, 1, -1), (6, 4, -5) and (-4, -2, 3)
Solution:
Let A (1, 1, -1), B (6, 4, -5) and C (-4, -2, 3)
The required plane passes through the point A (1, 1, -1) whose vector is
$\vec{a}=\hat{i}+\hat{j}-\hat{k}$
and is normal to the vector $\vec{n}$ given by
$\begin{aligned} &\vec{n}=\vec{AB}\times \vec{AC}\\ &\vec{AB}=\vec{OB}\times \vec{OA}=(6\hat{i}+4\hat{j}-5\hat{k})-(\hat{i}+\hat{j}-\hat{k})=5\hat{i}+3\hat{j}-4\hat{k} \\ &\vec{AC}=\vec{OC}\times \vec{OA}=(-4\hat{i}-2\hat{j}+3\hat{k})-(\hat{i}+\hat{j}-\hat{k})=-5\hat{i}-3\hat{j}+4\hat{k} \\ &\vec{n}=\vec{AB}\times \vec{AC}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 5 &3 &-4 \\ -5 &-3 &4 \end{vmatrix}=0\hat{i}+0\hat{j}+0\hat{k}=\vec{0} \end{aligned}$
So, the given points are collinear.
Thus, there will be an infinite number of planes passing through these points.
Their equations passing through (1, 1, -1) are given by
a(x - 1) + b(y - 1) + c (z + 1)=0 .....(i)
Since this passes through B (6, 4, -5)
a(6 - 1) + b(4 - 1) + c (-5 + 1)=0
5a + 3b - 4c = 0 .....(ii)
From (i) and (ii) the equations of the infinite planes are
a(x - 1) + b(y - 1) + c (z + 1)=0,
Where 5a + 3b - 4c = 0

The Plane exercise 28.5 question 5

Answer:
The required vector equation of plane is 9x + 2y - 7z - 15 = 0
Hint:
Using
$\frac{a}{c_2b_1-b_1c_2}=\frac{b}{a_2c_1-a_1c_2}=\frac{c}{b_2a_1-b_1a_2}$
Given:
Evaluation of the plane passing through the points
$3\hat{i}+4\hat{j}+2\hat{k}, \; 2\hat{i}-2\hat{j}-\hat{k},\; 7\hat{i}+6\hat{k}$
Solution:
Co-ordinates of the given plane are A (3, 4, 2), B (2, -2, -1) and C (7, 0, 6)
General equation of plane is given by-
$\begin{aligned} &a(x-3)+b(y-4)+c(z-2)=0 \qquad \qquad \dots (i)\\ &a(2-3)+b(-2-4)+c(-1-2)=0\\ &a(-1)+b(-6)+c(-3)=0\\ &a+6b+3c=0 \qquad \qquad \dots (ii) \end{aligned}$
Again putting point C in general equation
$\begin{aligned} &a(7-3)+b(0-4)+c(6-2)=0\\ &4a-4b+4c=0\\ &a-b+c=0 \qquad \qquad \dots (ii) \end{aligned}$
Let’s cross multiply;
$\begin{aligned} &\frac{a}{6+3}=\frac{b}{3-1}=\frac{c}{-1-6}\\ &\frac{a}{9}=\frac{b}{2}=\frac{c}{-7}=\lambda \\ &\therefore a=9\lambda ,\; b=2\lambda ,\: c=-7\lambda \end{aligned}$
Substituting a, b and c by their values in equation (i)$\begin{aligned} &9\lambda (x-3)+2\lambda (y-4)+(-7\lambda )(z-2)=0 \end{aligned}$

Divide by λ

$\begin{aligned} &\therefore 9x+2y-7z-21=0 \end{aligned}$

Also See:

RD Sharma Solutions Class 12 mathematics chapter 28 exercise 28.1
RD Sharma Solutions Class 12 mathematics chapter 28 exercise 28.2
RD Sharma Solutions Class 12 mathematics chapter 28 exercise 28.3
RD Sharma Solutions Class 12 mathematics chapter 28 exercise 28.4
RD Sharma Solutions Class 12 mathematics chapter 28 exercise 28.6
RD Sharma Solutions Class 12 mathematics chapter 28 exercise 28.7
RD Sharma Solutions Class 12 mathematics chapter 28 exercise 28.8
RD Sharma Solutions Class 12 mathematics chapter 28 exercise 28.9
RD Sharma Solutions Class 12 mathematics chapter 28 exercise 28.10
RD Sharma Solutions Class 12 mathematics chapter 28 exercise 28.11
RD Sharma Solutions Class 12 mathematics chapter 28 exercise 28.12
RD Sharma Solutions Class 12 mathematics chapter 28 exercise 28.13
RD Sharma Solutions Class 12 mathematics chapter 28 exercise 28.14
RD Sharma Solutions Class 12 mathematics chapter 28 exercise 28.15
RD Sharma Solutions Class 12 mathematics chapter 28 exercise MCQ
RD Sharma Solutions Class 12 mathematics chapter 28 exercise FBQ
RD Sharma Solutions Class 12 mathematics chapter 28 exercise VSA

The RD Sharma class 12th exercise 28.5 deals with the chapter The plane, which is a flat, 2-dimensional surface with infinite dimensions but zero thickness. The RD Sharma class 12 solution of The plane exercise 28.5 consists of a total of 5 questions that cover up all the essential concepts of the chapter like the equation of a plane, equation of the plane passing through points, equation of line under planes condition, equation of the plane passing through three points and Vector equation of plane.

Listed below are few reasons that explains the benefits of the RD Sharma class 12 solution chapter 28 exercise 28.5 :-

The RD Sharma class 12th exercise 28.5 are best solutions to refer to when you are solving your NCERT questions as it contains all the possible solutions.
The solutions are updated according to the NCERT, so if you face any problem while problem solving the solutions are present for your help.
The RD Sharma class 12 chapter 28 exercise 28.5 is trusted by thousands of students from across the country, as any student who has preferred the RD Sharma solutions over other books have definitely seen positive results.
If you face any trouble while solving the exercise questions you can always refer to the solved questions given in the RD Sharma class 12th exercise 28.5 for understanding the concept.
The RD Sharma solutions are available for every student for free of cost on the Career360 website.
You just have to make one click on the Career360 website and in one second the online PDF will be downloaded on your device.

RD Sharma Chapter-wise Solutions