Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Class 12 RD Sharma Solutions- Chapter 29 The Plane – Exercise 29.7

  • Last Updated : 25 Jan, 2021

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


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!