RD Sharma Class 12 Ex 29.7 Solutions Chapter 29 The Plane

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Here we provide RD Sharma Class 12 Ex 29.7 Solutions Chapter 29 The Plane for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 29.7 Solutions Chapter 29 The Plane book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter29
Exercise29.7
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 29.7 Solutions Chapter 29 The Plane

Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed

Question 1. Find the vector equation of the following planes in scalar product form (\vec{r}.\vec{n}=d):

Solution:

(i) \vec{r}=\{2\hat{i}-\hat{k}\}+λ \hat{i}+µ\{\hat{i}-2\hat{j}-\hat{k}\}

Here, \vec{r}=\{2\hat{i}-\hat{k}\}+λ \hat{i}+µ\{\hat{i}-2\hat{j}-\hat{k}\}

We know that, \vec{r}=\vec{a}+λ\vec{b}+μ\vec{c}  represent a plane passing through a point having position vector \vec{a}  and parallel to vectors  \vec{b} and .\vec{c}

Here, \vec{a}=2\hat{i}-\hat{k},\ \vec{b}=\hat{i},\ \vec{c}=\hat{i}-2\hat{j}-\hat{k}

The given plane is perpendicular to a vector

\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&0&0\\1&-2&-1\end{vmatrix}\\ =\hat{i}(0-0)-\hat{j}(-1-0)+\hat{k}(-2-0)\\ =0\hat{i}+\hat{j}+2\hat{k}\\ \vec{n}=\hat{j}-2\hat{k}

We know that vector equation of plane in scalar product form is,

\vec{r}.\vec{n}=\vec{a}.\vec{n} —(Equation-1)

Put \vec{n}  and \vec{a}  in (Equation-1),

\vec{r}.(\hat{j}-2\hat{k})=(2\hat{i}-\hat{k})(\hat{j}-2\hat{k})\\ \vec{r}.(\hat{j}-2\hat{k})=(2)(0)+(0)(1)+(-1)(-2)\\ =0+0+2\\ \vec{r}.(\hat{j}-2\hat{k})=2

The equation is required form is,

\vec{r}.(\hat{j}-2\hat{k})=2

(ii) \vec{r}=(1+s-t)\hat{i}+(2-s)\hat{j}+(3-2s+2t)\hat{k}

Here,\vec{r}=(1+s-t)\hat{i}+(2-s)\hat{j}+(3-2s+2t)\hat{k}\\ \vec{r}=(1+2\hat{j}+3\hat{k})+s(\hat{i}-\hat{j}-2\hat{k})+t(-\hat{i}+2\hat{k})\\

We know that, \vec{r}=\vec{a}+λ\vec{b}+μ\vec{c}  represent a plane passing through a point having position vector \vec{a} and parallel to vectors \vec{b}  and \vec{c}

Here, \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}

The given plane is perpendicular to a vector

\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&-2\\-1&0&2\end{vmatrix}\\ =\hat{i}(-2-0)-\hat{j}(2-2)+\hat{k}(0-1)\\ \vec{n}=-2\hat{i}-\hat{k}

We know that, vector equation of a plane is scalar product is,

\vec{r}.\vec{n}=\vec{a}.\vec{n}  —(Equation-1)

Put value of \vec{a}  and \vec{n}  in (Equation-1)

\vec{r}.(-2\hat{i}-\hat{k})=(-2\hat{i}-k)(\hat{i}+2\hat{k}+3\hat{k})\\ \vec{r}.(-2\hat{i}-\hat{k})=(-2)(1)+(0)(2)+(-1)(3)\\ =-2+0-3\\ \vec{r}.(-2\hat{i}-\hat{k})=-5

Multiplying both the sides by (-1),

\vec{r}.(2\hat{i}+\hat{k})=5

The equation in the required form,

\vec{r}.(2\hat{i}+\hat{k})=5

(iii) \vec{r}=(\hat{i}+\hat{j})+λ(\hat{i}+2\hat{j}-\hat{k})+μ(-\hat{i}+\hat{j}-2\hat{k})

Given, equation of plane,

\vec{r}=(\hat{i}+\hat{j})+λ(\hat{i}+2\hat{j}-\hat{k})+μ(-\hat{i}+\hat{j}-2\hat{k})

We know that, \vec{r}=\vec{a}+λ\vec{b}+μ\vec{c}  is the equation of a plane passing through point \vec{a}  and parallel to \vec{b}  and \vec{c} .

Here, \vec{a}=\hat{i}+\hat{j},\ \vec{b}=\hat{i}+2\hat{j}-\hat{k},\ \vec{c}=-\hat{i}+\hat{j}-2\hat{k}

The given plane is perpendicular to a vector

\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&-1\\-1&1&-2\end{vmatrix}\\ =\hat{i}(-4+1)-\hat{j}(-2-1)+\hat{k}(1+2)\\ -3\hat{i}+3\hat{j}+3\hat{k}

We know that, equation of plane in scalar product form is given by,  

\vec{r}.\vec{n}=\vec{a}.\vec{n}\\ \vec{r}.(-3\vec{i}+3\vec{j}+3\hat{k})=(\hat{i}+\hat{j})(-3\vec{i}+3\vec{j}+3\hat{k})\\ =(1)(-3)+(1)(3)+(0)(3)\\ =-3+3\\ \vec{r}.(-3\vec{i}+3\vec{j}+3\hat{k})=0

Dividing by 3, we get

\vec{r}.(-\vec{i}+\vec{j}+\hat{k})=0

Equation in required form is,

\vec{r}.(-\vec{i}+\vec{j}+\hat{k})=0

(iv) \vec{r}=\hat{i}-\hat{j}+λ(\hat{i}+\hat{j}+\hat{k})+μ(4\hat{i}-2\hat{j}-3\hat{k})

\vec{r}=\hat{i}-\hat{j}+λ(\hat{i}+\hat{j}+\hat{k})+μ(4\hat{i}-2\hat{j}-3\hat{k})

Plane is passing through (\hat{i}-\hat{j})  and parallel to b b(\hat{i}+\hat{j}+\hat{k})  and c(4\hat{i}-2\hat{j}+3\hat{k})

\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&1\\4&-2&3\end{vmatrix}\\ n=5i+j-6k\\ \vec{r}.n=(i-j)(5i+j-6k)=5-1=4\\ \vec{r}.(5i+j-6k)=4

Question 2. Find the cartesian form of the equation of the following planes:

Solution:

(i) \vec{r}=(\hat{i}-\hat{j})+s(-\hat{i}+\hat{j}+2\hat{k})+t(\hat{i}+2\hat{i}+\hat{k})\\

Here, given equation of plane is,

\vec{r}=(\hat{i}-\hat{j})+s(-\hat{i}+\hat{j}+2\hat{k})+t(\hat{i}+2\hat{i}+\hat{k})\\

We know that, \vec{r}=\vec{a}+λ\vec{b}+μ\vec{c}  represents the equation of a plane passing through a vector \vec{a}  and parallel to vector \vec{b}  and \vec{c}  . 

Here, \vec{a}=\hat{i}-\hat{j},\ \vec{b}=-\hat{i}+\hat{j}+2\hat{k},\ \vec{c}=\hat{i}+2\hat{j}+\hat{k}

Given plane is perpendicular to vector

\vec{n}=\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)\\ \vec{n}=-3\hat{i}+3\hat{j}-3\hat{k}

We know that, equation of plane in the scalar product form,

\vec{r}.\vec{n}=\vec{a}.\vec{n}  —Equation-1

Put the value of \vec{a}  and \vec{n}  in Equation-1,

\vec{r}.(\hat{i}-\hat{j})=(\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)\\ =-3-3+0\\ \vec{r}.(-3\hat{i}+3\hat{j}-3\hat{k})=-6

Put \vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\\ (x\hat{i}+y\hat{j}+z\hat{k})(-3\hat{i}+3\hat{j}-3\hat{k})=-6

(x)(-3) + (y)(3) + (z)(-3) = -6

-3x + 3y – 3z = -6

Dividing by (-3), we get

x – y + z = 2

Equation in required form is,

x – y + z = 2

(ii) \vec{r}=(1+s+t)\hat{i}+(2-s+t)\hat{j}+(3-2s+2t)\hat{k}

Given, equation of plane,

\vec{r}=(1+s+t)\hat{i}+(2-s+t)\hat{j}+(3-2s+2t)\hat{k}\\ =(\hat{i}+2\hat{j}+3\hat{k})+s(\hat{i}-\hat{j}-2\hat{k})+t(\hat{i}+\hat{j}+2\hat{k})

We know that, \vec{r}=\vec{a}+λ\vec{b}+μ\vec{c}   represents the equation of a plane passing through the vector \vec{a}  and parallel to vector \vec{b}  and \vec{c}

Here, \vec{a}=\hat{i}+2\hat{j}+3\hat{k}\\ \vec{b}=\hat{i}-\hat{j}-2\hat{k}\\ \vec{c}=\hat{i}+\hat{j}+2\hat{k}

The given plane is perpendicular to vector

\vec{n}=\vec{b}\times\vec{c}\\ =\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)\\ =0(\hat{i})-4\hat{j}+2\hat{k}\\ \vec{n}=-4\hat{j}+2\hat{k}

We know that, equation of plane in scalar product form is given by,

\vec{r}.\vec{n}=\vec{a}.\vec{n}  —Equation-1

Put, the value of \vec{a}  and \vec{n}  in equation-1

\vec{r}.(-4\hat{j}+2\hat{k})=(\hat{i}+2\hat{j}+3\hat{k})(-4\hat{j}+2\hat{k})\\ \vec{r}.(-4\hat{j}+2\hat{k})=(1)(0)+(2)(-4)+(3)(2)\\ =0-8+6\\ \vec{r}.(-4\hat{j}+2\hat{k})=-2

Put \vec{r}=x\hat{i}+y\hat{j}+z\hat{k}

(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

The equation in required form is,

2y – z = 1

Question 3. Find the vector equation of the following planes in non-parametric form:

Solution:

(i) \vec{r}=(λ-2μ)\hat{i}+(3-μ)\hat{j}+(2λ+μ)\hat{k}

Given, equation of plane is,

\vec{r}=(λ-2μ)\hat{i}+(3-μ)\hat{j}+(2λ+μ)\hat{k}\\ \vec{r}=(3\hat{j})+λ(\hat{i}+2\hat{k})+μ(-2\hat{i}-\hat{j}+\hat{k})

We know that, \vec{r}=\vec{a}+λ\vec{b}+μ\vec{c}  represents the equation of a plane passing through a point \vec{a}  and parallel to vector \vec{b}  and \vec{c}  .

Given,

\vec{a}=3\hat{j}\\ \vec{b}=\hat{i}+2\hat{k}\\ \vec{c}=-2\hat{i}-\hat{j}+\hat{k}

The given plane is perpendicular to

\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&0&2\\-2&-1&1\end{vmatrix}\\ =\hat{i}(0+2)-\hat{j}(1+4)+\hat{k}(1-0)\\ \vec{n}=2\hat{i}-5\hat{j}-\hat{k}

Vector equation of plane in non-parametric form is.

\vec{r}.\vec{n}=\vec{a}\vec{n}\\ \vec{r}.(2\hat{i}-5\hat{j}-\hat{k})=(3\hat{i})(2\hat{i}-5\hat{j}-\vec{k})

= (0)(2) + (3)(-5) + (0)(-1)

= 0 – 15 + 0

\vec{r}.(2\hat{i}-5\hat{j}-\hat{k})=-15\\ \vec{r}.(2\hat{i}-5\hat{j}-\hat{k})+15=0

The required form of equation is,

\vec{r}.(2\hat{i}-5\hat{j}-\hat{k})+15=0

(ii) \vec{r}=(2\hat{i}+2\hat{j}-\hat{k})+λ(\hat{i}+2\hat{j}+3\hat{k})+μ(5\hat{i}-2\hat{j}+7\hat{k})\\

Given, equation of plane is,

\vec{r}=(2\hat{i}+2\hat{j}-\hat{k})+λ(\hat{i}+2\hat{j}+3\hat{k})+μ(5\hat{i}-2\hat{j}+7\hat{k})\\

We know that, \vec{r}=\vec{a}+λ\vec{b}+μ\vec{c}  represents the equation of a plane passing through a vector \vec{a}  and parallel to vector \vec{b}  and \vec{c}  .

Here,

\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}

The given plane is perpendicular to vector

\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&3\\5&-2&7\end{vmatrix}\\ =\hat{i}(14+6)-\hat{j}(7-15)+\hat{k}(-2-10)\\ \vec{n}=20\hat{i}+8\hat{j}-12\hat{k}

We know that, equation of a plane in non-parametric form is given by,

\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})\\

= (2)(20) + (2)(8) – (-1)(-12)

=40 + 16 + 12

\vec{r}.(20\hat{i}+8\hat{j}-12\hat{k})=68

Dividing by 4,

\vec{r}.(5\hat{i}+2\hat{j}-3\hat{k})=17

Equation of plane in required form is,

\vec{r}.(5\hat{i}+2\hat{j}-3\hat{k})=17

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

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