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

# 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.

## The Plane Excercise: 28.1

The Plane exercise 28.1 question 1(i)

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)

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)

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)

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)

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

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)

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)

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

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 $2:1$.

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