RD Sharma Class 12 Exercise 28.1 The Plane Solutions Maths

RD Sharma Class 12 Exercise 28.1 The Plane Solutions Maths - Download PDF Free Online

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

RD Sharma class 12th exercise 28.1 is a must-have NCERT solution for students in class 12. High school students in India have immense trust in RD Sharma Solutions. The RD Sharma class 12 solutions The Plane ex 28.1 will become an indispensable book for every student who takes their exam preparations seriously. The answers in the book can be incredibly helpful in checking students' performance and help them understand concepts better.

Latest: JEE Main: high scoring chapters | Past 10 year's papers

Don't Miss: Most scoring concepts for NEET | NEET papers with solutions

New: Aakash iACST Scholarship Test. Up to 90% Scholarship. Register Now

This Story also Contains

RD Sharma Class 12 Solutions Chapter28 The Plane - Other Exercise
The Plane Excercise: 28.1
The Plane exercise 28.1 question 3(i)
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter28 The Plane - Other Exercise

The Plane Excercise: 28.1

The Plane exercise 28.1 question 1(i)

Answer:-

The required equation of the plane is $7x+3y-z=17$
Hint:- Use equation of the plane $\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)$ and $\left(x_{3}, y_{3}, z_{3}\right)$
Given:- $(2,1,0), (3,-2,-2)$ and $(3,1,7)$
Solution:- We know that, the equation of the plane passing through three non-collinear points $\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)$ and $\left(x_{3}, y_{3}, z_{3}\right)$ is
$\left|\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{array}\right|=0$
$\begin{aligned} &\Rightarrow\left|\begin{array}{ccc} x-2 & y-1 & z-0 \\ 3-2 & -2-1 & -2-0 \\ 3-2 & 1-1 & 7-0 \end{array}\right|=0 \\\\ &\Rightarrow\left|\begin{array}{ccc} x-2 & y-1 & z \\ 1 & -3 & -2 \\ 1 & 0 & 7 \end{array}\right|=0 \end{aligned}$
$\begin{gathered} (x-2)(-21)-(y-1)(9)+z(3)=0 \\\\ \therefore 7 x+3 y-z=17 \end{gathered}$

The Plane exercise 28.1 question 1(ii)

Answer:-

The required equation of the plane is $2x-y-2z-2=0$
Hint:- Use equation of the plane $\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)$ and $\left(x_{3}, y_{3}, z_{3}\right)$
Given:- $(-5,0,-6), (-3,10,-9)$ and $(-2,6,-6)$
Solution:- We know that, the equation of the plane passing through three non-collinear points $\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)$ and $\left(x_{3}, y_{3}, z_{3}\right)$ is
$\left|\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{array}\right|=0$
Now, substitute the given value
$\begin{aligned} &\Rightarrow\left|\begin{array}{ccc} x+5 & y-0 & z+6 \\ -3+5 & 10-0 & -9+6 \\ -2+5 & 6-0 & -6+6 \end{array}\right|=0 \\\\ &\Rightarrow\left|\begin{array}{ccc} x+5 & y & z+6 \\ 2 & 10 & -3 \\ 3 & 6 & 0 \end{array}\right|=0 \end{aligned}$

$\begin{gathered} (x+5)(18)-y(9)+(z+6)(-18)=0 \\\\ 18 x+90-9 y-18 z-108=0 \\\\ \therefore 2 x-y-2 z-2=0 \end{gathered}$

The Plane exercise 28.1 question 1(iii)

Answer:-

The required equation of plane is $x-3y-6z+8=0$
Hint:- Use equation of the plane $\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)$ and $\left(x_{3}, y_{3}, z_{3}\right)$
Given:- $(1,1,1), (1,-1,2)$ and $(-2,-2,2)$
Solution:- We know that, the equation of the plane passing through given three points, $\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)$ and $\left(x_{3}, y_{3}, z_{3}\right)$ is
$\left|\begin{array}{lll} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{array}\right|=0$
Now, substitute the given values
$\begin{aligned} &\Rightarrow\left|\begin{array}{ccc} x-1 & y-1 & z-1 \\ 1-1 & -1-1 & 2-1 \\ -2-1 & -2-1 & 2-1 \end{array}\right|=0 \\\\ &\Rightarrow\left|\begin{array}{ccc} x-1 & y-1 & z-1 \\ 0 & -2 & 1 \\ -3 & -3 & 1 \end{array}\right|=0 \end{aligned}$

$\begin{gathered} (x-1)(-2+3)-(y-1)(0+3)+(z-1)(0-6)=0 \\\\ (x-1)(1)-(y-1) 3+(z-1)(-6)=0 \\\\ \therefore x-3 y-6 z+8=0 \end{gathered}$

The Plane exercise 28.1 question 1(iv)

Answer:-

The required equation of the plane is $x+y-z-1=0$
Hint:- Use equation of the plane $\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)$ and $\left(x_{3}, y_{3}, z_{3}\right)$
Given:- $(2,3,4), (-3,5,1)$ and $(4,-1,2)$
Solution:- We know that, the equation of the plane passing through given three points, $\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)$ and $\left(x_{3}, y_{3}, z_{3}\right)$ is
$\left|\begin{array}{lll} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{array}\right|=0$
Now, substitute the given value
$\begin{aligned} &\Rightarrow\left|\begin{array}{ccc} x-2 & y-3 & z-4 \\ -3-2 & 5-3 & 1-4 \\ 4-2 & -1-3 & 2-4 \end{array}\right|=0 \\\\ &\Rightarrow\left|\begin{array}{ccc} x-2 & y-3 & z-4 \\ -5 & 2 & -3 \\ 2 & -4 & -2 \end{array}\right|=0 \end{aligned}$
$\begin{gathered} (x-2)(-4-12)-(y-3)(10+6)+(z-4)(20-4)=0 \\\\ (x-2)(-16)-(y-3)(16)+(z-4)(16)=0 \\\\ -16 x+32-16 y+48+16 z-64=0 \end{gathered}$

$\begin{gathered} -16 x-16 y+16 z+16=0 \\\\ \therefore x+y-z-1=0 \end{gathered}$

The Plane exercise 28.1 question 1(v)

Answer:-

The required equation of the plane is $4x-3y+2z-3=0$
Hint:- Use equation of the plane $\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)$ and $\left(x_{3}, y_{3}, z_{3}\right)$
Given:- $(0,-1,0), (3,3,0)$ and $(1,1,1)$
Solution:- We know that, the equation of the plane passing through given three points, $\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)$ and $\left(x_{3}, y_{3}, z_{3}\right)$ is
$\left|\begin{array}{lll} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{array}\right|=0$
Now, substitute the given value
$\begin{aligned} &\Rightarrow\left|\begin{array}{ccc} x-0 & y+1 & z-0 \\ 3-0 & 3+1 & 0-0 \\ 1-0 & 1+1 & 1-0 \end{array}\right|=0 \\\\ &\Rightarrow\left|\begin{array}{lll} x & y+1 & z \\ 3 & 4 & 0 \\ 1 & 2 & 1 \end{array}\right|=0 \end{aligned}$
$\begin{gathered} (x)(4-0)-(y+1)(3-0)+z(6-4)=0 \\\\ 4 x-(y+1)(3)+z((2)=0 \\\\ 4 x-3 y-3+2 z=0 \\\\ \therefore 4 x-3 y+2 z-3=0 \end{gathered}$

The Plane exercise 28.1 question 2

Answer:-

The required equation of the plane is $5x-7y+11z+4=0$ and hence they are coplanar.
Hint:- Use equation of the plane $\left|\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{array}\right|=0$
Given:- $(0,-1,-1), (4,5,1), (3,9,4)$ and $(-4,4,4)$
Solution:- Given that these four points are coplanar. So these four points lie on the same plane.
So, first let us take three points and find the equation of the plane passing through these four points and then let us substitute the fourth point in it. If is 0 then the points lies on the plane formed by these three points then they are coplanar. The equation of the plane passing through these three points is given
$\left|\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{array}\right|=0$
Now, let us take $(0,-1,-1), (4, 5,1), (3,9,4)$ and find plane equation
$\begin{aligned} &\Rightarrow\left|\begin{array}{ccc} x-0 & y+1 & z+1 \\ 4-0 & 5+1 & 1+1 \\ 3-0 & 9+1 & 4+1 \end{array}\right|=0 \\\\ &\Rightarrow\left|\begin{array}{ccc} x & y+1 & z+1 \\ 4 & 6 & 2 \\ 3 & 10 & 5 \end{array}\right|=0 \end{aligned}$

$\begin{gathered} x(30-20)-(y+1)(20-6)+(z+1)(40-18)=0 \\\\ 10 x-14 y+22 z+8=0 \\\\ 5 x-7 y+11 z+4=0 \end{gathered}$
Now let us substitute fourth point $(-4, 4, 4)$ we get
$\begin{gathered} 5(-4)-7(4)+11(4)+4=0 \\\\ -20-28+44+4=0 \\\\ 48+48=0 \\\\ \therefore 0=0 \end{gathered}$

∴ $LHS=RHS$

So, as said above this fourth point satisfies. So, this point also lies on the same plane.
Hence they are coplanar.

The Plane exercise 28.1 question 3(i)

Answer:-

Hence they are coplanar.
Hint:- Use the equation of plane
$\left|\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{array}\right|=0$
Given:- $(0,-1,0), (2,1,-1), (1,1,1)$ and $(3,3,0)$
Solution:- Given that these four points are coplanar. So, these four points lie on the same plane.
So, first let us take three points and find the equation of plane passing through these four points and then let us substitute the fourth point in it. If it is 0 then the points lie on the plane formed by these three points then they are coplanar.
The equation of the plane passing through these three points is given
$\left|\begin{array}{lll} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{array}\right|=0$
Now, let us take $(0,-1, 0), (2,1,-1), (1,1,1)$ and find plane equation.
$\begin{aligned} & \Rightarrow\left|\begin{array}{ccc} x-0 & y+1 & z+0 \\ 2-0 & 1+1 & -1+0 \\ 1-1 & 1+1 & 1-0 \end{array}\right|=0 \\\\ & \Rightarrow\left|\begin{array}{ccc} x & y+1 & z \\ 2 & 2 & -1 \\ 1 & 2 & 1 \end{array}\right|=0 \end{aligned}$
$\begin{gathered} x(2+2)-(y+1)(2+1)+z(4-2)=0 \\\\ 4 x-3 y-3+2 z=0 \\\\ 4 x-3 y+2 z-3=0 \end{gathered}$
Now, let us substitute fourth point $(3,3,0)$ in plane equation $4x-3y+2z-3=0$
$4(3)-3(3)+2(0)-3=0$
$\begin{gathered} 12-9+0-3=0 \\\\ 12-12=0 \\\\ \therefore 0=0 \end{gathered}$
$\therefore \mathrm{LHS}=\mathrm{RHS}$
So, this point also lies on the plane.
Hence they are coplanar.

The Plane exercise 28.1 question 3(ii)

Answer:-

Hence they are coplanar.
Hint:- Use equation of plane $\left|\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{array}\right|=0$
Given:- $(0,4,3), (-1,-5,-3), (-2,-2,1)$ and $(1,1,-1)$
Solution:- Given that these four points are coplanar. So these four points lie on the same plane.
So, first let us take three points and find the equation of the plane passing through these four points and then let us substitute the fourth point in it. If it is 0 then the points lies on the plane formed by these three points then they are coplanar.
The equation of the plane passing through these three points is given by
$\left|\begin{array}{lll} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{array}\right|=0$
Now, let us take $(0, 4,3), (-1,-5,-3), (-2,-2,1)$ and find plane equation
$\begin{aligned} &\Rightarrow\left|\begin{array}{ccc} x-0 & y-4 & z-3 \\ -1-0 & -5-4 & -3-3 \\ -2-0 & -2-4 & 1-3 \end{array}\right|=0 \\\\ &\Rightarrow\left|\begin{array}{ccc} x & y-4 & z-3 \\ -1 & -9 & -6 \\ -2 & -6 & -2 \end{array}\right|=0 \end{aligned}$
$\begin{gathered} x(18-36)-(y-4)(2-12)+(z-3)(6-18)=0 \\\\ x(-18)-(y-4)(-10)+(z-3)(-12)=0 \\\\ -18 x+10 y-40-12 z+36=0 \\\\ -18 x+10 y-12 z-4=0 \end{gathered}$
Now, let us substitute $(1, 1,-1)$ in plane equation
$\begin{gathered} -18 x+10 y-12 z-4=0 \\\\ -18(1)+10(1)-12(-1)-4=0 \\\\ -18+10+12-4=0 \end{gathered}$
$\begin{gathered} -22+22=0 \\\\ \therefore 0=0 \end{gathered}$
$\therefore \mathrm{LHS}=\mathrm{RHS}$
So, this point also lies on the plane.
Hence they are coplanar.

The Plane exercise 28.1 question 4

Answer:
Hence, $P$ externally divides the line segment $AB$ in the ratio $2:1.$
Hint:
Use determinant then solve the equation of plane.
Given:
The equation of the plane three points $L(2,2,1), M(3,0,1) \text { and } \mathrm{N}(4,-1,0) .$
Solution:
The equation of the plane passing through three points can be given by:
$L(2,2,1), M(3,0,1) \text { and } \mathrm{N}(4,-1,0) .$
$\left|\begin{array}{lll} x-2 & y-2 & z-1 \\ x-3 & y-0 & z-1 \\ x-4 & y+1 & z-0 \end{array}\right|=0$
Operating $R_{2} \rightarrow R_{2}-R_{1}, \text { and } R_{3} \rightarrow R_{3}-R_{1}$
$\left|\begin{array}{ccc} x-2 & y-2 & z-1 \\ -1 & 2 & 0 \\ -2 & 3 & 1 \end{array}\right|=0$
Solving the above determinant, we get
$\begin{aligned} &(x-2)[2-0]-(y-2)[-1-0]+(z-1)[-3+4]=0 \\\\ &\Rightarrow(2 x-4)+(y-2)+(z-1)=0 \\\\ &\Rightarrow 2 x+y+z-7=0 \end{aligned}$
Therefore, the equation of the plane is $2 x+y+z-7=0$
Now, the equation of the line passing through two given points $A(3,-4,-5)$ and $B(2,-3,1)$ is
$\begin{aligned} &\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}=\lambda \\\\ &\Rightarrow \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{x-3}{-1}=\lambda, \frac{y+4}{1}=\lambda, \frac{z+5}{6}=\lambda \\\\ &\Rightarrow x=(-\lambda+3), y=(\lambda-4), z=(6 \lambda-5) \end{aligned}$
At the point of intersection, these points satisfy the equation of the plane
$2 x+y+z-7=0$
Putting the value of $x, y$ and $z$ in the equation of the plane, we get the value of $\lambda$ .
$\begin{aligned} &2(-\lambda+3)+(\lambda+4)+(6 \lambda-5)=0 \\\\ &\Rightarrow-2 \lambda+6+\lambda-4+6 \lambda-5-7=0 \\\\ &\Rightarrow 5 \lambda=10 \\\\ &\Rightarrow \lambda=2 \end{aligned}$
Thus, the point of intersection is $P(1,-2,7)$ .
Now,
Let $P$ divides the line $AB$ in the ratio $m:n$
By the section formula, we have
Here, $A(3,-4,-5) \text { and } \mathrm{B}(2,-3,1) \text { and } P(1,-2,7)$
$\begin{aligned} &\Rightarrow \frac{2 m+3 n}{m+n}=1 \\\\ &\Rightarrow m+n=2 m+3 n \\\\ &\Rightarrow m+2 n=0 \end{aligned}$
$\begin{aligned} &\Rightarrow m=-2 n \\\\ &\Rightarrow \frac{m}{n}=\frac{-2}{1} \end{aligned}$
Hence, $P$ externally divides the line segment $AB$ in the ratio $2:1$ .

RD Sharma class 12 solutions The Plane 28.1 is one book that will help students perfect their maths skills before board exams. The RD Sharma class 12th exercise 28.1 contains step-by-step solutions that are easy to understand and can also be understood by students who find maths difficult.

RD Sharma class 12th exercise 21.3 is a well known name among numerous NCERT solutions. There are certain good reasons why they are so popular. Here are a list of reasons which make RD Sharma class 12 chapter 28 exercise 28.1 such a great book:-

RD Sharma class 12th exercise 28.1 solutions contain answers that are created by subject experts with years of experience. Moreover, as they follow the CBSE syllabus, students can use them to practice and finish their homework as well.
Hundreds of students place their trust in RD Sharma class 12th exercise 28.1 every year and have benefitted from the answers. Students who have followed the book diligently throughout the school year have even found common questions in their board exams.
RD Sharma class 12 solutions chapter 28 ex 28.1 can work as a guide for students as they contain solved answers. This helps students to explore different ways of solving a problem and choose the best one for them.
The RD Sharma class 12th exercise 21.3 is updated to the latest version of the book. Students can save a lot of time by using this material as it has the best answers that are accessible through the website.
It enables a convenient and efficient way of preparation as the entire material is available on Career360’s website and is free of cost. Students can use any device with an internet connection to access this material.

RD Sharma Chapter wise Solutions

Frequently Asked Question (FAQs)

1. How can I download the RD Sharma class 12th exercise 28.1 pdf?

Students can easily download the RD Sharma class 12th exercise 28.1 pdf from the Career360 website. It can be accessed and downloaded for free.

2. How can students use RD Sharma class 12 solutions chapter 28 ex 28.1 for practice?

Students can test themselves at home by responding to the RD Sharma class 12 solutions chapter 28 ex 28.1 books and then contrasting the appropriate responses with really taking a look at their scores.

3. Can I use class RD Sharma class 12 chapter 28 exercise 28.1 solution for exam preparations?

Students can utilize the RD Sharma class 12 chapter 28 exercise 28.1 for their exam arrangements. The prospectus followed by these books covers exams like school tests, sheets, and JEE mains.

4. What are the concepts covered in class 12 RD Sharma chapter 28 exercise 28.1 solution?

The class 12 RD Sharma chapter 28 exercise 28.1 solution contains answers to the first exercise of the chapter. The questions in this exercise are on the concepts of the equation of planes.

5. Who makes the appropriate responses in the RD Sharma solutions?

Specialists and talented experts are liable for making the appropriate responses in the RD Sharma solutions. This is because they have a great deal of involvement with the instruction area, and their answers are exact and top caliber.