RD Sharma Class 12 Exercise 28.7 The Plane Solutions Maths

RD Sharma Class 12 Exercise 28.7 The Plane Solutions Maths - Download PDF Free Online

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

Students of class 12 have a very vast syllabus when it comes to the subject of mathematics. Chapters like ‘The Plane’ are complex maths topics that make solving the problem challenging for students; that is why students are highly recommended to use the Class 12 RD Sharma chapter 28 exercise 28.7 solution to practice at home and improve their performance on the subject. The RD Sharma class 12th exercise 28.7 of The Plane of class 12 deals with various topics like,

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

Equation of plane
Cartesian form of the equation of the plane
Vector Equation of planes in scalar product form
Vector Equation of planes in nonparametric form

RD Sharma Class 12 Solutions Chapter28 The Plane - Other Exercise

The Plane Excercise: 28.7

The Plane exercise 28.7 question 1(i)

Answer:
Required equation is
$\vec{r}.(\hat{j}-2\hat{k})=2$
Hint:
Using scalar form
$\vec{r}.\vec{n}=\vec{a}.\vec{n}$
Given:
$\vec{r}=(2\hat{i}-k)+\lambda \hat{i}+\mu (\hat{i}-2\hat{j}-\hat{k})$
Solution:
We know that
$\vec{r}=\vec{a}+\lambda \vec{b}+\mu \vec{c}$
represent a plane passing through a point having position vector $\vec{a}$ and parallel to the vector $\vec{b}$ and $\vec{c}$
Clearly,
$\begin{aligned} &\vec{a}=2\hat{i}-\hat{k}\\ &\vec{b}=\hat{i}\\ &\vec{c}=\hat{i}-2\hat{j}-\hat{k} \end{aligned}$
Now the plane is perpendicular to $\vec{b}$ X $\vec{c}$
Hence
$\begin{aligned} &\vec{n}=\vec{b}\times \vec{c}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 1 &0 &0 \\ 1 &-2 &-1 \end{vmatrix}\\ &\Rightarrow \vec{n}=\hat{i}(0-0)-\hat{j}(-1-0)+\hat{k}(-2-0)\\ &\Rightarrow \vec{n}=\hat{j}-2\hat{k} \end{aligned}$
We know that vector equation of plane in scale product form is given by
$\vec{r}.\vec{n}=\vec{a}.\vec{n}$ (a)

Put $\begin{aligned} &\vec{a} \end{aligned}$ and $\begin{aligned} &\vec{n} \end{aligned}$ in equation (a) we get
$\begin{aligned} &\vec{r}.(\hat{j}-2\hat{k})=(2\hat{i}-\hat{k})(\hat{j}-2\hat{k})\\ &\Rightarrow \vec{r}.(\hat{j}-2\hat{k})=2(0)+(0)(1)+(-1)(-2)\\ &\Rightarrow \vec{r}.(\hat{j}-2\hat{k})=2 \end{aligned}$
Hence the require equation is
$\begin{aligned} &\vec{r}.(\hat{j}-2\hat{k})=2 \end{aligned}$

The Plane exercise 28.7 question 1(ii)

Answer:
Required vector equation is
$\vec{r}.(-2\hat{i}-\hat{k})+5=0$
Hint:
Using scalar form
$\vec{r}.\vec{n}=\vec{a}.\vec{n}$
Given:
$\vec{r}=(1+s-t)\hat{i}+(2-s)\hat{j}+(3-2s+2t)\hat{k}$
Solution:
We know that the equation
$\vec{r}=\vec{a}+\lambda \vec{b}+\mu \vec{c}$
represent a plane passing through a point whose position vector $\vec{a}$
$\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+s(\hat{i}-\hat{j}-2\hat{k})+t(-\hat{i}+2\hat{k})$
Here
$\begin{aligned} &\vec{a}=\hat{i}+2\hat{j}+3\hat{k}\\ &\vec{b}=\hat{i}-\hat{j}-2\hat{k}\\ &\vec{c}=-\hat{i}+2\hat{k} \end{aligned}$
Normal vector
$\begin{aligned} &\vec{n}=\vec{b}\times \vec{c}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 1 &-1 &-2 \\ -1 &0 &2 \end{vmatrix}\\ &=(-2+0)\hat{i}-(2-2)\hat{j}+(0-1)\hat{k} \end{aligned}$
The vector equation of plane is in scalar product form in
$\begin{aligned} &\Rightarrow \vec{r}(-2\hat{i}-\hat{k})=(\hat{i}+2\hat{j}+3\hat{k}).(-2\hat{i}-\hat{k})\\ &\Rightarrow \vec{r}(-2\hat{i}-\hat{k})=-2+0-3=-5\\ &\therefore \vec{r}(-2\hat{i}-\hat{k})+5=0 \end{aligned}$
Is the required vector equation

The Plane exercise 28.7 question 1(iii)

Answer:
Required vector equation is
$\vec{r}.(-\hat{i}+\hat{j}+\hat{k})=0$
Hint:
Using scalar form
$\vec{n}=\vec{b}.\vec{n}$
Given:
$\vec{r}=(\hat{i}+\hat{j})+\lambda (\hat{i}+2\hat{j}-\hat{k})+\mu (-\hat{i}+2\hat{j}-2\hat{k})$
Solution:
We know that the equation
$\vec{r}=\vec{a}+\lambda \vec{b}+\mu \vec{c}$
represents a plane passing through a point whose vector is $\vec{a}$ and parallel to the vector $\vec{b}$ and $\vec{c}$
Here
$\begin{aligned} &\vec{a}=\hat{i}+\hat{j}+0\hat{k}\\ &\vec{b}=\hat{i}+2\hat{j}-\hat{k}\\ &\vec{c}=-\hat{i}+2\hat{j}-2\hat{k} \end{aligned}$
Normal vector
$\begin{aligned} &\vec{n}=\vec{b}\times \vec{c}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 1 &2 &-1 \\ -1 &1 &-2 \end{vmatrix}\\ &=-3\hat{i}+3\hat{j}+3\hat{k} \end{aligned}$
The vector equation of the plane in scalar product form is
$\begin{aligned} &\vec{r}.\vec{n}=\vec{a}.\vec{n} \end{aligned}$
$\begin{aligned} &\Rightarrow \vec{r}\: (-3\hat{i}+3\hat{j}+3\hat{k})= (\hat{i}+\hat{j}+0\hat{k}). (-3\hat{i}+3\hat{j}+3\hat{k})\\ &\Rightarrow \vec{r}\: (-3\hat{i}+3\hat{j}+3\hat{k})=-3+3\\ &\Rightarrow \vec{r}\: [3(-\hat{i}+\hat{j}+\hat{k})]=0\\ &\Rightarrow \vec{r}\: (-\hat{i}+\hat{j}+\hat{k})=0 \end{aligned}$

The Plane exercise 28.7 question 1(iv)

Answer:
Required vector equation is
$\vec{r}.(5\hat{i}+\hat{j}-6\hat{k})=4$
Hint:
The plane is perpendicular to $\vec{b}\times \vec{c}$ so find $\vec{n}= \vec{b}.\vec{n}$
Given:
$\vec{r}=(\hat{i}-\hat{j})+\lambda (\hat{i}+\hat{j}+\hat{k})+\mu (4\hat{i}-2\hat{j}+3\hat{k})$
Solution:
We know that the equation
$\vec{r}=\vec{a}+\lambda \vec{b}+\mu \vec{c}$
represents a plane passing through a point having position $\vec{a}$ parallel to the vectors $\vec{b}$ and $\vec{c}$
Clearly
$\begin{aligned} &\vec{a}=\hat{i}-\hat{j}\\ &\vec{b}=\hat{i}+\hat{j}+\hat{k}\\ &\vec{c}=4\hat{i}-2\hat{j}+3\hat{k} \end{aligned}$
Now the plane is perpendicular to $\vec{b}$ X $\vec{c}$
Hence
$\begin{aligned} &\vec{n}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ \hat{1} &\hat{1} &\hat{1} \\ \hat{4} &\hat{-2} &\hat{3} \end{vmatrix} \end{aligned}$
We know that vector equation of plane in scalar product form is given as
$\vec{r}. \vec{n}= \vec{a}.\vec{n} \qquad \qquad \dots (a)$
Put $\vec{a}$ and $\vec{n}$ in equation (a) we get
$\begin{aligned} &\vec{r}.(5\hat{i}+\hat{j}-6\hat{k})=(\hat{i}-\hat{j})(5\hat{i}+\hat{j}-6\hat{k})\\ &\Rightarrow \vec{r}.(5\hat{i}+\hat{j}-6\hat{k})=(1)(5)+(-1)(1)+(0)(-6)\\ &\Rightarrow \vec{r}.(5\hat{i}+\hat{j}-6\hat{k})=4 \end{aligned}$
Hence the required equation is
$\begin{aligned} &\vec{r}.(5\hat{i}+\hat{j}-6\hat{k})=4 \end{aligned}$

The Plane exercise 28.7 question 2(i)

Answer:
Required Cartesian equation is
$x-y+z=2$
Hint:
The plane is perpendicular to $\vec{n}$
Given:
$\vec{r}=(\hat{i}-\hat{j})+s(-\hat{i}+\hat{j}+2\hat{k})+t(\hat{i}+2\hat{j}+\hat{k})$
Solution:
$\vec{r}=(\hat{i}-\hat{j})+s(-\hat{i}+\hat{j}+2\hat{k})+t(\hat{i}+2\hat{j}+\hat{k})$
We know that the equation
$\vec{r}=\vec{a}+\lambda \vec{b}+\mu \vec{c}$
represents a plane passing through a point having position $\vec{a}$ parallel to the vectors $\vec{b}$ and $\vec{c}$
Here
$\begin{aligned} &\vec{a}=\hat{i}-\hat{j}\\ &\vec{b}=-\hat{i}+\hat{j}+2\hat{k}\\ &\vec{c}=\hat{i}+2\hat{j}+\hat{k} \end{aligned}$
The given plane is perpendicular to $\vec{b}$ X $\vec{c}$
$\begin{aligned} &\vec{b}\times \vec{c}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ -1 &1 &2 \\ 1 &2 &1 \end{vmatrix}\\ &=\hat{i}(1-4)-\hat{j}(-1-2)+\hat{k}(-2-1)\\ &=-3\hat{i}+3\hat{j}-3\hat{k} \end{aligned}$
Vector equation of the plane in scalar product form is
$\begin{aligned} &\vec{r}.\vec{n}=\vec{a}.\vec{n} \qquad \qquad \dots (1) \end{aligned}$
$\begin{aligned} &\vec{r}.(-3\hat{i}+3\hat{j}-3\hat{k})=(\hat{i}-\hat{j}).(-3\hat{i}+3\hat{j}-3\hat{k})\\ &\vec{r}.(-3\hat{i}+3\hat{j}-3\hat{k})=1(-3)+(-1)(3)+0(3)\\ &\vec{r}.(-3\hat{i}+3\hat{j}-3\hat{k})=-6\\ &\vec{r}.(-\hat{i}+\hat{j}-\hat{k})=-2\\ &\vec{r}.(\hat{i}-\hat{j}+\hat{k})=2 \end{aligned}$
Now put
$\begin{aligned} &\vec{r}=x\hat{i}+y\hat{j}+z\hat{k} \end{aligned}$
So, we have
$\begin{aligned} &(x\hat{i}+y\hat{j}+z\hat{k}).(-3\hat{i}+3\hat{j}-3\hat{k})=-6\\ &x(-3)+y(3)+2(-3)=-6\\ &-3x+3y-3z=-6 \end{aligned}$
Divide by -3, we get
$\begin{aligned} &x-y+z=2 \end{aligned}$
Hence the required equation is
$\begin{aligned} &x-y+z=2 \end{aligned}$

The Plane exercise 28.7 question 2(ii)

Answer:
Required Cartesian equation is 2y - z = 1
Hint:
Using the equation of the plane
$\vec{r}.\vec{n}=d$
Given:
$\vec{r}=(1+s+t)\hat{i}+(2-s+t)\hat{j}+(3-2s+2t)\hat{k}$
Solution:
Since from the given equation
$\begin{aligned} &(\hat{i}+2\hat{j}+3\hat{k})+s(\hat{i}-\hat{j}-2\hat{k})+(\hat{i}+\hat{j}+2\hat{k})\\ &\vec{n}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 1 &-1 &-2 \\ 1 &1 &2 \end{vmatrix}\\ &=\hat{i}(-2+2)-\hat{j}(2+2)+\hat{k}(1+1)\\ &=4\hat{j}+2\hat{k} \end{aligned}$
We know that the vector equation of a plane is scalar product form is given as
$\begin{aligned} &\hat{r}.\hat{n}=\hat{a}.\hat{n} \qquad \qquad \dots(1) \end{aligned}$
Substituting the value of $\begin{aligned} &\vec{a} \end{aligned}$ and $\begin{aligned} &\vec{n} \end{aligned}$ in equation (1)
We get
$\begin{aligned} &\vec{r}.(-4\hat{j}+2\hat{k})=(\hat{i}+2\hat{j}+3\hat{k}).(-4\hat{j}+2\hat{k})\\ &=(1)(0)+(2)(-4)+(3)(2)\\ &\vec{r}.2(-2\hat{j}+\hat{k})=-2\\ &\vec{r}.(2\hat{j}-\hat{k})=1 \end{aligned}$
Now put
$\begin{aligned} &\vec{r}=x\hat{i}+y\hat{j}+z\hat{k} \end{aligned}$
So we have
$\begin{aligned} &(x\hat{i}+y\hat{j}+z\hat{k})(-4\hat{j}+2\hat{k})=-2\\ &(x)(0)+(y)(-4)+(z)(2)=-2\\ &-4y+2z=-2\\ &2y-z=1 \end{aligned}$
Hence the required equation of the plane is
$\begin{aligned} &2y-z=1 \end{aligned}$

The Plane exercise 28.7 question 3(i)

Answer:
The required equation is $\vec{r}.(2\hat{i}-5\hat{j}-\hat{k})+15=0$
Hint:
The plane is perpendicular to $\vec{b}$ X $\vec{c}$ so we find $\vec{n}=\vec{b}\times \vec{c}$
Given:
$\vec{r}=(\lambda -2\mu )\hat{i}+(3-\mu )\hat{j}+(2\lambda +\mu )\hat{k}$
Solution:
Here
$(3\hat{j})+\lambda (\hat{i}+2\hat{k})+\mu (-2\hat{i}-\hat{j}+\hat{k})$
We know that the equation
$\vec{r}=\vec{a}+\lambda \vec{b}+\mu \vec{c}$
represents a plane passing through a point having position $\vec{a}$ and parallel to the vectors $\vec{b}$ and $\vec{c}$
Clearly
$\begin{aligned} &\vec{a}=3\hat{j}\\ &\vec{b}=\hat{i}+2\hat{k}\\ &\vec{c}=-2\hat{i}-\hat{j}+\hat{k} \end{aligned}$
Now the plane is perpendicular to $\begin{aligned} &\vec{b}\times \vec{c} \end{aligned}$
Hence
$\begin{aligned} &\vec{n}=\vec{b}\times \vec{c}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 1 &0 &2 \\ -2 &-1 &1 \end{vmatrix}\\ &=(0+2)\hat{i}-(1+4)\hat{j}+(-1+0)\hat{k}\\ &=2\hat{i}-5\hat{j}-\hat{k} \end{aligned}$
We know that the vector equation of plane in scalar product form is given by
$\begin{aligned} &\vec{r}.\vec{n}=\vec{a}.\vec{n}\\ &\vec{r}.(2\hat{i}-5\hat{j}-\hat{k})=(3\hat{j}).(2\hat{i}-5\hat{j}-\hat{k})\\ &\vec{r}.(2\hat{i}-5\hat{j}-\hat{k})=-15 \end{aligned}$
Hence the required equation is
$\begin{aligned} &\vec{r}.(2\hat{i}-5\hat{j}-\hat{k})+15=0 \end{aligned}$

The Plane exercise 28.7 question 3(ii)

Answer:
The required equation is $\vec{r}.(5\hat{i}+2\hat{j}-3\hat{k})=17$
Hint:
The plane is perpendicular to $\vec{b}\times \vec{c}$ so we find $\vec{n}=\vec{b}\times \hat{c}$
Given:
$\vec{r}=(2\hat{i}+2\hat{j}-\hat{k})+\lambda (\hat{i}+2\hat{j}+3\hat{k})+\mu (5\hat{i}-2\hat{j}+7\hat{k})$
Solution:
We know that $\vec{r}=\vec{a}+\lambda \vec{b}+\mu \vec{c}$ represents
A plane passing through a point having position vector $\vec{a}$ and parallel to the vectors $\vec{b}$ and $\vec{c}$
Clearly
$\begin{aligned} &\vec{a}=2\hat{i}+2\hat{j}-\hat{k}\\ &\vec{b}=\hat{i}+2\hat{j}+3\hat{k}\\ &\vec{c}=5\hat{i}-2\hat{j}+7\hat{k} \end{aligned}$
Hence
$\begin{aligned} &\vec{n}=\vec{b}\times \vec{c}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 1 &2 &3 \\ 5 &-2 &7 \end{vmatrix}\\ &=(14+6)\hat{i}-(7-15)\hat{j}+(-2-10)\hat{k}\\ &=20\hat{i}+8\hat{j}-12\hat{k} \end{aligned}$
We know that vector equation of a plane in scalar product is given as
$\begin{aligned} &\vec{r}.\vec{n}=\vec{a}.\vec{n}\\ &\vec{r}.(20\hat{i}+8\hat{j}-12\hat{k})=(2\hat{i}+2\hat{j}-\hat{k}).(20\hat{i}+8\hat{j}-12\hat{k})\\ &\vec{r}.(20\hat{i}+8\hat{j}-12\hat{k})=40+16+12\\ &\vec{r}.(20\hat{i}+8\hat{j}-12\hat{k})=68\\ &\vec{r}.(5\hat{i}+2\hat{j}-3\hat{k})=17 \end{aligned}$
Hence the require equation is
$\begin{aligned} &\vec{r}.(5\hat{i}+2\hat{j}-3\hat{k})=17 \end{aligned}$

Students can effectively start practicing RD Sharma class 12th exercise 28.7 of chapter 28 to get a hold on the concept to make it an asset while solving the questions. The RD Sharma class 12 solution of The Plane exercise 28.7 explains the equation of the plane among various logics and consists of a total of 8 questions.

Let's have a look at some of the benefits of using the RD Sharma class 12th exercise 28.7 mentioned below:-

The RD Sharma class 12 chapter 28 exercise 28.7 is used by teachers for conducting lectures and use an example to solve queries.
The questions assigned by the teachers for homework are also taken from the RD Sharma class 12th exercise 28.7 solution, therefore referring to the solution can help solve homework easily.
Most importantly, the questions are prepared by experts from across the country hence it is trusted by students and teachers as well that is why they use the questions from the solution to prepare question papers.
The RD Sharma class 12th exercise 28.7 is updated yearly to catch up with the syllabus of NCERT.
The RD Sharma solutions can be downloaded online from the Career360 website and it doesn't charge you a penny for getting the online PDFs from its website.

RD Sharma Chapter-wise Solutions