RD Sharma Class 12 Exercise 28.2 The Plane Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 28.2 The Plane Solutions Maths - Download PDF Free Online

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

RD Sharma books are no less than a math-bible for CBSE students as they are comprehensive and detailed. They serve as the best material for exam preparation as they cover all concepts and provide a wide variety of information. In addition, many schools across the country refer to RD Sharma books to set up question papers. RD Sharma solutions This material is designed specifically to help students reduce their burden for preparation.

## The Plane Excercise: 28.2

The Plane exercise 28.2 question 1

The equation of plane whose intercepts on the co-ordinate axes are $2,-3,4 \; is\; 6x-4y+32=12$
Hint:
The equation of plane whose intercept on the co-ordinate axes are $p,q,r$ respectively given by
$\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1$
Given:
Intercepts on the co-ordinate axes are $2,-3,4.$
Explanation:
It is given that
Intercepts on the co-ordinate axes are
$2,-3,4.$
So, the equation of required plane is
\begin{aligned} &\frac{x}{2}+\frac{y}{-3}+\frac{z}{4}=1 \\\\ &\frac{6 x-4 y+3 z}{12}=1 \\\\ &\Rightarrow 6 x-4 y+3 z=12 \end{aligned}

### The Plane exercise 28.2 question 2(i)

Intercepts form of the given equation is $\frac{x}{3}+\frac{y}{4}+\frac{z}{-2}=1$ & its intercepts on co-ordinate axes are $3,-4$ & $-2$
Hint:
First divide the given equation by 12 then compare it with $\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1,$ where $p, q$ & $r$intercepts on co-ordinate axes.
Given:
Given equation is $4x+3y-6z-12=0$
Explanation:
It is given that,
equation of plane is $4x+3y-6z-12=0$ ….....(i)
Divide equation (i) by 12
\begin{aligned} \Rightarrow & \frac{4 x+3 y-6 z-12}{12}=0 \\ \end{aligned}
$\Rightarrow \frac{x}{3}+\frac{y}{4}-\frac{z}{2}=1 \\$
$\Rightarrow \frac{x}{3}+\frac{y}{4}+\frac{z}{(-2)}=1$
This is in the form of $\frac{x}{l}+\frac{y}{m}+\frac{z}{n}=1$
Here, $l=3, m=4, n=-2$

### The Plane exercise 28.2 question 2(ii)

Intercepts form of the given equation is $\frac{x}{3}+\frac{y}{4}+\frac{z}{-6}=1$ & its intercepts on co-ordinate axes are $3,2$ & $-6$
Hint:
First divide the given equation by 6 then compare it with $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1,$ where $a, b$ & $c$ intercepts on co-ordinate axes.
Given:
Equation of plane is $2x+3y-z=6$ ...........(i)
Explanation:
Divide equation (i) by 6
\begin{aligned} &\Rightarrow \frac{2 x+3 y-z}{6}=\frac{6}{6} \\\\ &\Rightarrow \frac{x}{3}+\frac{y}{2}-\frac{z}{6}=1 \end{aligned}
Compare above equation with $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Here, $a=3, b=2$ & $c=-6$
Thus, intercepts form is
$\frac{x}{3}+\frac{y}{2}+\frac{z}{(-6)}=1$
& intercepts on co-ordinate axes are $3,2$ & $-6$

The Plane exercise 28.2 question 2(iii)

Intercepts form of the given equation is $\frac{x}{\frac{5}{2}}+\frac{y}{-5}+\frac{z}{5}=1$ & its intercepts on co-ordinate axes are $\frac{5}{2},-5 \; \&\; 5$
Hint:
First divide the given equation by 5 then compare it with $\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1,$ where $p, q$ & $r$ intercepts on co-ordinate axes.
Given:
Equation of plane is $2x-y+z=5$ .......(i)
Explanation:
Divide equation (i) by 5
\begin{aligned} &\frac{2 x-y+z}{5}=\frac{5}{5} \\ &\Rightarrow \frac{x}{\frac{5}{2}}-\frac{y}{5}+\frac{z}{5}=1 \end{aligned}
Compare above equation with $\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1$
Here, $p=\frac{5}{2}, q=-5 \; \& \; r=5$
Hence, intercepts form is
$\frac{x}{\frac{5}{2}}+\frac{y}{-5}+\frac{z}{5}=1$
& intercepts on co-ordinate axes are $\frac{5}{2},-5\; \&\; 5$

## The Plane exercise 28.2 question 3

Equation of plane is $\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}=3$
Hint:
First using centroid, identify the values of $a, b$ & $c$ then using $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ , find equation of plane.
Given:
Plane meets axis in $A, B\; \& \; C.$
Explanation:
Assume, $A=(a, 0,0), B=(0, b, 0), \& \; C=(0,0, c)$
Here, centroid of triangle $AB$ is $\left ( \alpha ,\beta, \gamma \right )$
Now,
centroid $=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}, \frac{z_{1}+z_{2}+z_{3}}{3}\right)$
\begin{aligned} &\Rightarrow(\alpha, \beta, \gamma)=\left(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}\right) \\\\ &\Rightarrow \alpha=\frac{a}{3}, \quad \beta=\frac{b}{3}, \quad \gamma=\frac{c}{3} \\\\ &\Rightarrow a=3 \alpha, \quad b=3 \beta, \quad c=3 \gamma \end{aligned}
Put the values of $A, B\; \& \; C.$ in $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
$\Rightarrow \frac{x}{3 \alpha}+\frac{y}{3 \beta}+\frac{z}{3 \gamma}=3$
Here, equation of plane is
$\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}=3$

The Plane exercise 28.2 question 4

Equation of plane is $x+y+z=12$
Hint:
Use the formula, $\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1,$ where $p, q \; \&\; r$ intercepts on co-ordinate axes.
Given:
Intercept on co-ordinate axes are equal.
Explanation:
It is given that,
Intercepts on co-ordinate axes are equal.
We know if $p, q \; \&\; r$ are intercepts on co-ordinate axes then
$\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1$
According to question, plane makes equal intercept on the co-ordinate axes
so, $p=q=r$
As the equation of plane is $\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1$
$\frac{x}{p}+\frac{y}{p}+\frac{z}{p}=1 \quad\quad\quad\quad(p=q=r)$
$\Rightarrow x+y+z=p$ ...........(i)
We have, the plane is passing through the points $(2,4,6)$
Plane passes through the points $(2,4,6)$ so we substituting points in equation (i)
\begin{aligned} &2+4+6=P \\ &\Rightarrow p=12 \\ &\therefore x+y+z=12 \end{aligned}
Hence, the required equation of plane is $x+y+z=12$

The Plane exercise 28.2 question 5

Equation of plane is $6x-3y+2z=18$
Hint:
The equation of plane with intercepts $p,q \; \&\; r$ is $xp+yq+zr=1$
Given:
It is given that plane meets the co-ordinate axes at $A, B \; \& \; C$ with centroid of $\Delta ABC$ is $(1,-2,3).$
Explanation:

Centroid of triangle $ABC$ is $(1,-2,3).$
We know,
Centroid of triangle $=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}, \frac{z_{1}+z_{2}+z_{3}}{3}\right)$
\begin{aligned} &\Rightarrow(1,-2,3)=\left(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}\right) \\ &\Rightarrow(1,-2,3)=\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right) \\ &\Rightarrow \frac{a}{3}=1, \quad \frac{b}{3}=-2, \quad \frac{c}{3}=3 \\ &\Rightarrow a=3, \quad b=-6, \quad c=9 \end{aligned}\begin{aligned} &\Rightarrow(1,-2,3)=\left(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}\right) \\\\ &\Rightarrow(1,-2,3)=\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right) \\ \end{aligned}
$\Rightarrow \frac{a}{3}=1, \quad \frac{b}{3}=-2, \quad \frac{c}{3}=3 \\\\$
$\Rightarrow a=3, \quad b=-6, \quad c=9$
Put the values of $a,b \; \& \: c$ in
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, to get the equation of plane
\begin{aligned} &\Rightarrow \frac{x}{3}+\frac{y}{-6}+\frac{z}{9}=1 \\\\ &\frac{6 x-3 y+2 z}{18}=1 \end{aligned}
Thus, equation of plane is $6x-3y+2z=18$

RD Sharma class 12 solutions The Plane 28.2 is one of their top-selling books in mathematics. Chapter 28 of the NCERT is named The Plane. The ideas covered are The Plane and the concepts covered are Direction cosines and proportions, condition of Plane whose blocks on the organized tomahawks are given, Equation of Plane when the centroid is given.

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