RD Sharma Class 12 Exercise 28.4 The Plane Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 28.4 The Plane Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 11:54 AM IST

Class 12 students depend a lot on their teachers and tutors to clarify their doubts. And it is a known fact that more doubts arise only when the students solve the sums by themselves. When this takes place at their home while doing the homework, it is nearly impossible for them to contact their teachers at an instance. RD Sharma solutions To help the students with the mathematics chapter, the plane, the RD Sharma Class 12th Exercise 28.4, plays a vital role.

The Plane Excercise: 28.4

The Plane exercise 28.4 question 1

$\vec{r}.\hat{k}=3$
Hint:
You must know the rules of solving vector functions
Given:
Find the vector equation of a plane which is at a distance of 3 units from origin
and has k as the unit vector normal to it.
Solution:
We have
Normal vector,
$\vec{n}=\hat{k}$
Now,
$\vec{n}=\frac{\vec{n}}{\left | \vec{n} \right |}=\frac{\hat{k}}{\left | \hat{k} \right |}=\frac{\hat{k}}{1}=\hat{k}$
The equation of a plane in normal form is
$\vec{r}.\hat{n}=d \qquad \qquad \rightarrow (1)$
[d= distance of plane from origin]
By putting
$\vec{n}=\hat{k}$
d = 3 units in (1) relation
We get,
$\vec{r}.\hat{k}=3$
This is the required vector equation.

The Plane exercise 28.4 question 2

$\vec{r}.\left ( \frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k} \right )=5$
Hint:
You must know the rules of solving vector functions
Given:
Find the vector equation of a plane which is at a distance of 5 units from origin
and normal to the vector
$\hat{i}-2\hat{j}-2\hat{k}$
Solution:
We have
Normal vector,
$\vec{n}=\hat{i}-2\hat{j}-2\hat{k}$
Now,
\begin{aligned} &\hat{n}=\frac{\vec{n}}{\left | \vec{n} \right |}=\frac{\hat{i}-2\hat{j}-2\hat{k}}{\sqrt{(1)^{2}+(-2)^{2}+(-2)^{2}}}\\ &=\frac{\hat{i}-2\hat{j}-2\hat{k}}{\sqrt{1+4+4}}=\frac{\hat{i}-2\hat{j}-2\hat{k}}{\sqrt{9}}=\frac{\hat{i}-2\hat{j}-2\hat{k}}{3}\\ &\therefore \hat{n}=\frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k} \end{aligned}
The equation of a plane is in normal form is
\begin{aligned} &\vec{r}.\hat{n}=d \qquad \qquad \rightarrow (1) \end{aligned}
[d= distance of plane from origin]
By putting
\begin{aligned} &\hat{n}=\frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k} \end{aligned}
d = 5units in (1) relation
We get,
$\vec{r}.\left ( \frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k} \right )=5$
This is the required vector equation.

The Plane exercise 28.4 question 3

$2;\: \frac{2}{7},\frac{-3}{7},\frac{-6}{7}$
Hint:
You must know the rules of finding length of perpendicular from the origin from vector equation, also find direction cosines
Given:
Reduce the equation 2x - 3y - 6z = 14 to the normal form and hence,
find the length of perpendicular from the origin to the plane. Also find direction cosines
Solution:
We have
\begin{aligned} &2x-3y-6z=14 \qquad \qquad \qquad (1)\\ &=\sqrt{(2)^{2}+(-3)^{2}+(-6)^{2}}=\sqrt{4+9+36}\\ &=\sqrt{49}=7 \end{aligned}
Divide the equation (1) with 7
We get,
\begin{aligned} &\frac{2}{7}x-\frac{3}{7}y-\frac{6}{7}z=\frac{14}{7}\\ &\frac{2}{7}x-\frac{3}{7}y-\frac{6}{7}z=2 \qquad \qquad \qquad \rightarrow (2) \end{aligned}
Now compare equation (2) with cartesian equation.
The cartesian equation of normal form of a plane,
lx + my + nz = p
where l,m,n are direction cosines and P is the length of perpendicular.
By comparing we get,
Diection cosines :
\begin{aligned} &l=\frac{2}{7},m=-\frac{3}{7}, n=-\frac{6}{7} \end{aligned}
Length of perpendicular from the origin P = 2.

The Plane exercise 28.4 question 4

$\vec{r}.\left ( -\frac{1}{3}\hat{i}+\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k} \right )=2\: ;\: 2$
Hint:
You must know the rules of solving vector functions
Given:
Reduce the equation
$\vec{r}.\left ( \hat{i}-2\hat{j}+2\hat{k} \right )+6=0$
to the normal form and hence,
find the length of perpendicular from the origin.
Solution:
We have
$\vec{r}.\left ( \hat{i}-2\hat{j}+2\hat{k} \right )+6=0$
$\vec{r}.\left ( \hat{i}-2\hat{j}+2\hat{k} \right )=-6$
The equation of a plane is in normal form is
$\vec{r}.\hat{n}=d$
[d= distance of plane from origin]
\begin{aligned} &\hat{n}=\hat{i}-2\hat{j}+2\hat{k}\\ &d=-6\\ &\left | \vec{n} \right |=\sqrt{(1)^{2}+(-2)^{2}+(2)^{2}}\\ &=\sqrt{1+4+4}=\sqrt{9}=3 \end{aligned}
For reducing the given equation to normal form
So, divide the given equation by
\begin{aligned} &\left | \vec{n} \right | \end{aligned}
We get
\begin{aligned} &\vec{r}.\left ( \frac{\hat{i}-2\hat{j}+2\hat{k}}{3} \right )=\frac{-6}{3}\\ &\vec{r}.\left ( \frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k} \right )=-2\\ &\vec{r}.\left ( -\frac{1}{3}\hat{i}+\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k} \right )=2 \qquad \qquad \rightarrow (1) \end{aligned}
The equation of a plane is in normal form is
\begin{aligned} &\vec{r}.\hat{n}=d \qquad \qquad \rightarrow (1) \end{aligned}
[d= distance of plane from origin]
By comparing equation (1) & (2)
we get d = 2 units
∴Length of perpendicular is d = 2 units

The Plane exercise 28.4 question 5

$\frac{-2}{7}x+\frac{3}{7}y-\frac{6}{7}z=2$
Hint:
You must know the rules of solving vector functions
Given:
Write the normal form of the equation of the plane 2x - 3y + 6z + 14 = 0
Solution:
We have
\begin{aligned} &2x-3y+6z+14=0\\ &2x-3y+6z=-14 \end{aligned}
Now,
\begin{aligned} &\sqrt{(2)^{2}+(-3)^{2}+(6)^{2}}\\ &=\sqrt{4+9+36}\\ &=\sqrt{49}\\ &=7 \end{aligned}
So, divide the given equation by 7
We get,
\begin{aligned} &\frac{2}{7}x-\frac{3}{7}y+\frac{6}{7}z=\frac{-14}{7}\\ &\frac{2}{7}x-\frac{3}{7}y+\frac{6}{7}z=-2 \end{aligned}
Balance the -ve sign,
\begin{aligned} &\frac{-2}{7}x+\frac{3}{7}y-\frac{6}{7}z=2 \end{aligned}
This is the normal form of the given equation of the plane.

The Plane exercise 28.4 question 6

$\frac{12}{13}x-\frac{3}{13}y+\frac{4}{13}z=5$
Hint:
You must know the rules of solving vector functions
Given:
The direction ratios of perpendicular from the origin to a plane are 12,-3,4 and the length of perpendicular is 5. Find the equation of plane.
Solution:
We have
Direction ratio are 12,-3,4 of vector $\vec{n}$
\begin{aligned} &\vec{n}=12\hat{i}-3\hat{j}+4\hat{k}\\ &\left | \vec{n} \right |=\sqrt{(12)^{2}+(-3)^2+(4)^2}\\ &=\sqrt{144+9+16}=\sqrt{169}=13 \end{aligned}
Now,
\begin{aligned} &\hat{n}=\frac{\vec{n}}{\left | \vec{n} \right |}=\frac{12\hat{i}-3\hat{j}+4\hat{k}}{13}\\ &\frac{12}{13}\hat{i}-\frac{3}{13}\hat{j}+\frac{4}{13}\hat{k} \end{aligned}
Also we have,
length of the perpendicular from the origin d=5
∴Equation of plane in normal form is
\begin{aligned} &\vec{r}.\hat{n}=d \end{aligned}
[d= distance of plane from origin]
\begin{aligned} &\vec{r}.\left ( \frac{12}{13}\hat{i}-\frac{3}{13}\hat{j}+\frac{4}{13}\hat{k} \right )=5 \end{aligned}
Normal form is
$\frac{12}{13}x-\frac{3}{13}y+\frac{4}{13}z=5$

The Plane exercise 28.4 question 7

$\frac{1}{\sqrt{14}}\hat{i}+\frac{2}{\sqrt{14}}\hat{j}+\frac{3}{\sqrt{14}}\hat{k}$
Hint:
You must know the rules of solving vector functions
Given:
Find a unit normal vector to the plane x + 2y + 3z - 6 = 0
Solution:
We have
\begin{aligned} &x + 2y + 3z - 6 = 0\\ &x + 2y + 3z =6 \end{aligned}
Equation of plane in normal form is
\begin{aligned} &\vec{r}.\hat{n}=d\\ &\therefore \vec{r}.(\hat{i}+2\hat{j}+3\hat{k})=6\\ &\vec{r}.\hat{n}=d\\ &\text { So, }\vec{n}=\hat{i}+2\hat{j}+3\hat{k}\\ &\left | \vec{n} \right |=\sqrt{(1)^2+(2)^2+(3)^2}\\ &=\sqrt{1+4+9}\\ &=\sqrt{14} \end{aligned}
Unit vector to the plane ,
\begin{aligned} &\hat{n}=\frac{\vec{n}}{\left | \vec{n} \right |}\\ &=\frac{\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{14}}\\ &\hat{n}=\frac{1}{\sqrt{14}}\hat{i}+\frac{2}{\sqrt{14}}\hat{j}+\frac{3}{\sqrt{14}}\hat{k} \end{aligned}

The Plane exercise 28.4 question 8

x + y + z = 9
Hint:
You must know the rules of solving vector functions
Given:
Find the equation of the plane which is at a distance of $3\sqrt{3}$ units from the origin and the normal to which is equally inclined to the coordinate axes.
Solution:
Let α, β and γ be the angles made by n with x,y and z axis
It is given that,
\begin{aligned} &\alpha =\beta =\gamma \\ &cos\: \alpha =cos\: \beta =cos\: \gamma \\ &\Rightarrow l=m=n \end{aligned}
But
\begin{aligned} &l^2+m^2+n^2=1\\ &l^2+l^2+l^2=1\\ &3l^2=1\\ &l^2=\frac{1}{3}\\ &l=\frac{1}{\sqrt{3}}\\ &l=m=n=\frac{1}{\sqrt{3}} \end{aligned}
It is given that the length of perpendicular of the plane
from the origin = p = $3\sqrt{3}$
The normal form of plane is
\begin{aligned} &lx+my+nz=p\\ &\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z=3\sqrt{3}\\ &\frac{x+y+z}{\sqrt{3}}=3\sqrt{3}\\ &x+y+z=3\sqrt{3}(\sqrt{3})\\ &x+y+z=9 \end{aligned}

The Plane exercise 28.4 question 9

$x-y+3z-2=0\: ;\: \frac{2}{\sqrt{11}}$
Hint:
You must know the rules of solving vector functions
Given:
Find the equation of plane passing through (1,2,1) and perpendicular to the line joining the points (1,4,2) and (2,3,5). Find perpendicular distance.
Solution:
We have
The normal is passing through the points A(1,4,2) and B(2,3,5)
\begin{aligned} &\vec{n}=\vec{AB}=\vec{OB}-\vec{OA}\\ &=(2\hat{i}+3\hat{j}+5\hat{k})-(\hat{i}+4\hat{j}+2\hat{k})\\ &=(\hat{i}-\hat{j}+3\hat{k}) \end{aligned}
We know that the vector equation of the plane passing through a point (1,2,1)(\begin{aligned} &\vec{a} \end{aligned})
and normal to \begin{aligned} &\vec{n} \end{aligned} is
\begin{aligned} &\vec{r}.\vec{n}=\vec{a}.\vec{n} \end{aligned}
Putting
\begin{aligned} &\vec{a}=\hat{i}+2\hat{j}+\hat{k}\\ &\vec{n}=\hat{i}-\hat{j}+3\hat{k}\\ &\vec{r}.(\hat{i}-\hat{j}+3\hat{k})=(\hat{i}+2\hat{j}+\hat{k})(\hat{i}-\hat{j}+3\hat{k})\\ &\vec{r}.(\hat{i}-\hat{j}+3\hat{k})=(1-2+3)\\ &\vec{r}.(\hat{i}-\hat{j}+3\hat{k})=2 \qquad \qquad \qquad \rightarrow (1) \end{aligned}
To find the perpendicular distance of this plane from this origin,
we have
\begin{aligned} &\vec{n}=\hat{i}-\hat{j}+3\hat{k}\\ &\left | \vec{n} \right |=\sqrt{(1)^2+(-1)^2+(3)^2}\\ &=\sqrt{1+1+9}\\ &=\sqrt{11} \end{aligned}
Divide equation (1) by
\begin{aligned} &\sqrt{11} \end{aligned}
\begin{aligned} &\vec{r}.\left ( \frac{1}{\sqrt{11}}\hat{i}-\frac{1}{\sqrt{11}}\hat{j}+\frac{3}{\sqrt{11}}\hat{k} \right )=\frac{2}{\sqrt{11}} \qquad \qquad \rightarrow (2) \end{aligned}
The perpendicular distance of this plane from this origin
\begin{aligned} &=\frac{2}{\sqrt{11}} \end{aligned}
from equation (1)
\begin{aligned} &=\frac{1}{\sqrt{11}}x-\frac{1}{\sqrt{11}}y+\frac{3}{\sqrt{11}}z=\frac{2}{\sqrt{11}}\\ &x-y+3z=2\\ &x-y+3z-2=0 \end{aligned}

The Plane exercise 28.4 question 10

$\vec{r}\left ( \frac{2}{\sqrt{29}}\hat{i}-\frac{3}{\sqrt{29}}\hat{j}+\frac{4}{\sqrt{29}}\hat{k} \right )=\frac{6}{\sqrt{29}} \text { and }2x-3y+4z=6$
Hint:
You must know the rules of solving vector functions
Given:
find the vector equation of the plane which is at a distance of
$\frac{6}{\sqrt{29}}$
from the origin and it’s normal vector from the origin is
$2\hat{i}-3\hat{j}+4\hat{k}$
Solution:
We have
normal vector =
$\vec{n}=2\hat{i}-3\hat{j}+4\hat{k}$
Unit vector to the plane ,
\begin{aligned} &\hat{n}=\frac{\vec{n}}{\left | \vec{n} \right |}\\ &=\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{(2)^2+(-3)^2+(4)^2}}\\ &=\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{4+9+16}}\\ &=\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{29}}\\ &\hat{n}=\frac{2}{\sqrt{29}}\hat{i}-\frac{3}{\sqrt{29}}\hat{j}+\frac{4}{\sqrt{29}}\hat{k} \end{aligned}
Equation of plane in normal form is
\begin{aligned} &\vec{r}.\hat{n}=d\\ &\text { Put }\hat{n}=\frac{2}{\sqrt{29}}\hat{i}-\frac{3}{\sqrt{29}}\hat{j}+\frac{4}{\sqrt{29}}\hat{k}\text { and }d=\frac{6}{\sqrt{29}}\\ &\vec{r} \left ( \frac{2}{\sqrt{29}}\hat{i}-\frac{3}{\sqrt{29}}\hat{j}+\frac{4}{\sqrt{29}}\hat{k} \right ) =\frac{6}{\sqrt{29}}\\ \end{aligned}
This is the required vector equation.
For finding Cartesian form
Put
\begin{aligned} &\vec{r}=\left ( x\hat{i}+y\hat{j}+z\hat{k} \right )\\ &\left ( x\hat{i}+y\hat{j}+z\hat{k} \right ) \left ( \frac{2}{\sqrt{29}}\hat{i}-\frac{3}{\sqrt{29}}\hat{j}+\frac{4}{\sqrt{29}}\hat{k} \right ) =\frac{6}{\sqrt{29}}\\ &\frac{2x-3y+4z}{\sqrt{29}}=\frac{6}{\sqrt{29}}\\ &2x-3y+4z=\frac{6(\sqrt{29})}{\sqrt{29}}\\ &2x-3y+4z=6 \end{aligned}

The Plane exercise 28.4 question 11

$\frac{6}{\sqrt{29}}$
Hint:
You must know the rules of solving vector functions
Given:
Find the distance of the plane 2x - 3y + 4z - 6 = 0 from the origin
Solution:
We have
\begin{aligned} &2x-3y+4z-6=0\\ &2x-3y+4=6 \qquad \qquad \dots (1)\\ &\vec{r}(2\hat{i}-3\hat{j}+4\hat{k})=6 \end{aligned}
Now,
\begin{aligned} &\left | \vec{n} \right |=\sqrt{(2)^2+(-3)^2+(4)^2}\\ &=\sqrt{4+9+16}\\ &=\sqrt{29} \end{aligned}
Divide equation (1) by \begin{aligned} &\sqrt{29} \end{aligned}
We get,
\begin{aligned} &\frac{2}{\sqrt{29}}x-\frac{3}{\sqrt{29}}y+\frac{4}{\sqrt{29}}z=\frac{6}{\sqrt{29}} \end{aligned}
Which is the normal of the plane
Therefore the length of perpendicular from the origin to the plane is
\begin{aligned} &\frac{6}{\sqrt{29}} \end{aligned}

The syllabus given in chapter 28 of mathematics is challenging for the students to solve without proper guidance. This chapter consists of 15 exercises, ex 28.1 to ex 28.15. The fourth exercise, ex 28.4, consists of 25 questions in the concepts like Direction cosines and ratios, vector equation of a place, equation of a Plane, the perpendicular distance from the origin, and equation of line under planes condition, and so on. The RD Sharma Class 12 Chapter 28 Exercise 28.4 helps the students in understanding the concepts with clarity. Both the Level 1 and level 2 solutions can be found in the same book.

Various mathematical experts provide the sums present in the RD Sharma Class 12th Exercise 28.4. The order of the questions is present in the same way as given in the textbook. There are frequent updates made according to the recent changes in the NCERT books. Therefore, it is easy for the students to follow it without any confusion. The plenty of practice questions available in the RD Sharma books develops the ability of the students to work out the sums quickly during the exams. This eventually saves a lot of time that they can spend on checking their answers.

Many staff members tend to pick the questions from the RD Sharma books when they conduct tests and exams. This gives another valid point of why the Class 12 RD Sharma Chapter 28 Exercise 28.4 Solution book must be used by the students to prepare for their exams. The verified solutions present in RD Sharma Class 12th Exercise 28.4 book give the guidance towards the proper method in which a sum must be solved in a concept.

The best website to download the RD Sharma Class 12 Solutions The Plane Ex 28.4 is the Career 360 site, where all these books are available for free of cost. Furthermore, no one is asked to pay any monetary charge, and there are no restrictions on who can access the books. Therefore, anyone can visit the Career 360 website and download the RD Sharma Class 12 Solutions Chapter 28 Ex 28.4 for free.

RD Sharma Chapter wise Solutions

1. Which mathematics guide can I use to solve the sums regarding the plane concept in mathematics?

The RD Sharma Class 12th Exercise 28.4 helps the students solving their doubts regarding the concept of the plane in mathematics.

2. Are the RD Sharma books costly to be bought?

You need not think of buying the hard copy of the RD Sharma books when they are available on the Career 360 website for free of cost.

3. Can only students access the RD Sharma books on the Career 360 website?

The RD Sharma books on the Career 360 website are open to all and have no restrictions on accessing the reference materials.

4. How many questions are asked in the class 12 mathematics book in chapter 28, ex 28.4?

There are 25 questions in total asked in chapter 28, ex 28.4, in the class 12 mathematics book. In addition, the students can use the RD Sharma Class 12th Exercise 28.4 reference book to refer to the solved sums.

5. Do the RD Sharma books follow the NCERT pattern in framing the questions and the syllabus?

Yes, the RD Sharma books follow the NCERT pattern in framing the questions and solved sums accordingly.

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