RD Sharma Class 12 Exercise 28.13 The Plane Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 28.13 The Plane Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 28.13 The Plane Solutions Maths - Download PDF Free Online

Lovekush kumar sainiUpdated on 25 Jan 2022, 11:53 AM IST

The CBSE board institutions recommend that their students use the RD Sharma solution books to refer to the concepts when they encounter doubts at home. This helps the majority of the students who do not find time to visit tuitions after school hours and who cannot afford to go to tuitions. When topics like the plane in mathematics are a threat to most students, those who possess the RD Sharma Class 12th Exercise 28.13 solution book do not worry about any complexities.

This Story also Contains

RD Sharma Class 12 Solutions Chapter28 The Plane - Other Exercise
The Plane Excercise: 28.13
RD Sharma Chapter wise Solutions

Also Read - RD Sharma Solutions For Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter28 The Plane - Other Exercise

The Plane Excercise: 28.13

The Plane Exercise 28.13 Question 1

Answer: $\vec{r}.\left ( -\hat{i}+2\hat{j}-\hat{k} \right )=7$

Hint: use vector cross product to prove

Given:$\overline{r_{1}}=(2 \hat{j}-3 \hat{k})+\lambda(\hat{i}+2 \hat{j}+3 \hat{k}) \text { and } \vec{r}_{2}=(2 \hat{i}+6 \hat{j}+3 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+4 \hat{k})$

Solution:

It is given that

$\begin{aligned} &\overline{r_{1}}=(2 \hat{j}-3 \hat{k})+\lambda(\hat{i}+2 \hat{j}+3 \hat{k}) \\ \end{aligned}$

$\begin{aligned} &\vec{r}_{2}=(2 \hat{i}+6 \hat{j}+3 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+4 \hat{k}) \\ \end{aligned}$

Consider:$\begin{aligned} &\overline{r_{1}}=\overrightarrow{a_{1}}+\lambda \vec{b}_{1} \& \overrightarrow{r_{2}}=\overline{a_{2}}+\lambda \overline{b_{2}} \\ \end{aligned}$

$\begin{aligned} &\overrightarrow{a_{1}}=2 \hat{j}-3 \hat{k} \\ &\overrightarrow{b_{1}}=\hat{i}+2 \hat{j}+3 \hat{k} \\ &\overrightarrow{a_{2}}=2 \hat{i}+6 \hat{j}+3 \hat{k} \\ &\overline{b_{2}}=2 \hat{i}+3 \hat{j}+4 \hat{k} \\ \end{aligned}$

It can be written as

$\begin{aligned} &\overrightarrow{b_{1}} \times \overline{b_{2}}=\left(\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{array}\right) \\ \end{aligned}$

on further calculation

$\begin{aligned} &=\hat{i}(8-9)-\hat{j}(4-6)+\hat{k}(3-4) \\ &=-\hat{i}+2 \hat{j}-\hat{k} \\ \end{aligned}$

So we get

$\begin{aligned} \vec{a}_{1}\left(\vec{b}_{1} x \overline{b_{2}}\right)=(0 x(-1))+(2 x 2)+((-3) x(-1)) \\ =0+4+3=7 \end{aligned}$

$\begin{aligned} \vec{a}_{1}\left(\vec{b}_{1} \times \overline{b_{2}}\right)=7 \end{aligned}$ .....(i)

Similary :

$\begin{aligned} &\overline{a_{2}} \cdot\left(\vec{b}_{1} \times \overline{b_{2}}\right)=(2 \times(-1))+(6 \times 2)+((3) \times(-1)) \\ &=-2+12-3=7 \\ &\left.\overline{a_{2}} \cdot \vec{b}_{1} \times \overline{b_{2}}\right)=7 \\ \end{aligned}$....(ii)

Using both the equation we get

$\begin{aligned} &\left.\overrightarrow{a_{1}}\left(\vec{b}_{1} x \overline{b_{2}}\right)=\overline{a_{2}} \cdot \overrightarrow{b_{1}} x \overline{b_{2}}\right) \\ \end{aligned}$

So the lines $\vec{r_{1}}$ & $\vec{r_{2}}$ are coplanar

Hence the equation of plane containing $\begin{aligned} &\overline{r_{1}} \& \overrightarrow{r_{2}} \text { is } \vec{r} \cdot\left(\vec{b}_{1} \times \overline{b_{2}}\right)=\overrightarrow{a_{1}} \cdot\left(\vec{b}_{1} \times \overline{b_{2}}\right) \\ \end{aligned}$

$\begin{aligned} &\vec{r} \cdot(-\hat{i}+2 \hat{j}-\hat{k})=7 \\ \end{aligned}$

Hence, the given lines are coplanar and the equation of the plane determined by these lines is

$\begin{aligned} &\vec{r} \cdot(-\hat{i}+2 \hat{j}-\hat{k})=7 \end{aligned}$

The Plane Exercise 28.13 Question 2

Answer: Proud L.H.S =R.H.S
Hint: use vector cross product
Given:$\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1} \text { and } \frac{x}{1}=\frac{y-7}{-3}=\frac{8+7}{2}$
Solution :two lines
$\begin{aligned} &\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}} \text { and } \frac{x-x_{2}}{a_{2}} \\ &=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}} \end{aligned}$are coplanar if$\begin{aligned} \left(\begin{array}{ccc} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right) \\ \end{aligned}$
Here, $\begin{aligned} &x_{1}=-1, y_{1}=-1, z_{1}=-2 \\ \end{aligned}$
$\begin{aligned} &{x_{2}}=0 \quad y_{2}=7 \quad z_{1}=-7 \\ &a_{1}=-3 \quad b_{1}=2 \quad c_{1}=1 \\ &a_{2}=1 \quad b_{2}=-3 \quad c_{2}=2 \\ \end{aligned}$
$\begin{aligned} &\left(\begin{array}{ccc} 0-(-1) & 7-3 & -7-(-2) \\ -3 & 2 & 1 \\ 1 & -3 & 2 \end{array}\right) \\ &=\left(\begin{array}{ccc} 1 & 4 & -5 \\ -3 & 2 & 1 \\ 1 & -3 & 2 \end{array}\right) \\ &=1(7)-4(-7)-5(-7) \\ &=0 \end{aligned}$

The given lines are coplarer . equation of the plane containing the given line is

$\begin{aligned} &\left(\begin{array}{ccc} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right)=0 \\ &\left(\begin{array}{ccc} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ -3 & 2 & 1 \\ 1 & -3 & 2 \end{array}\right)=0 \\ &(x+1)(4+3)-(y-3)(-6+1)+(z+2)(9-0)=0 \\ &7 x+7 y+7 z=0 \\ &x+y+z=0 \end{aligned}$

The Plane Exercise 28.13 Question 3

Answer:$x+y+z=0$

Hint: use simultaneous equation to solve

Given:$\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point $\left ( 0,7,-7 \right )$

Solution: let the equation of the plane passing through $\left ( 0,7,-7 \right )$ be

$a\left ( 2-0 \right )+b\left ( y-7 \right )+c\left ( 2+7 \right )=0$ -----------(1)

The line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$passes through $\left ( -1,3-2 \right )$ and its direction ratios are proportional to $-3,2,1$

Since plane (1) contains this line, it must pass through the point $-3,2,1$

$\begin{aligned} &=a(-1-0)+b(3-7)+c(-2+7)=0 \\ &=-1-4 b+5 c=0 \\ &=a+4 b-5 c=0 \end{aligned}$......(2)

Since plane (1) contains this line, it must be parallel to the line

$=-3a+2b+c=0$............(3)

Solving (1)(2) and (3) we get

$\begin{aligned} &\left(\begin{array}{ccc} x-0 & y-7 & z+7 \\ 1 & 4 & -5 \\ -3 & 2 & 1 \end{array}\right)=0 \\ &=14 x+14(y-7)+14(z+7)=0 \\ &=14 x+14 y+14 z=0 \\ &=x+y+z=0 \end{aligned}$

The Plane Exercise 28.13 Question 4

Answer:$11x-y-3z-35=0$

Hint: use simultaneous equation to solve

Given:$\frac{x-y}{1}=\frac{y-3}{-4}=\frac{z-2}{5} \text { and } \frac{x-3}{1}=\frac{y+2}{-4}=\frac{2}{5}$

Solution: we know that equation of plane passing through $\left (x _{1}+y_{1}+ z_{1}\right )$ is given by

$a\left ( x-x_{1} \right )+b\left ( y-y_{1} \right )+c\left ( z-z_{1} \right )=0$....(i)

Since required plane contains lines $\frac{x-y}{1}=\frac{y-3}{-4}=\frac{z-2}{5} \text { and } \frac{x-3}{1}=\frac{y+2}{-4}=\frac{2}{5}$ , so we required plane passes through (4, 3, 2) and (3, -8, 0) so equation of required plane is

$a\left ( x-4_{1} \right )+b\left ( y-3\right )+c\left ( z-2 \right )=0$....(2)

Plane (2) also passes through (3, -3, 0) , so

$\begin{aligned} &a(3-4)+b(-2-3)+c(0-2) \\ &-a-5 b-2 z=0 \\ &a+5 b+2 c=0 \end{aligned}$....(3)

Now plane (2) is also parallel to the line with direction ratio (1, -4, 5) so

$\begin{aligned} &a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=(a)(1)+(6)(-4)+c(5)=0 \\ &a-4 c+5 c=0 \ldots-\ldots-(4) \\ \end{aligned}$

Solving equation (3) and (4) by cross multiplication

$\begin{aligned} &\frac{a}{(5)(5)-(-4)(2)}=\frac{b}{1(2)-(1)(5)}=\frac{c}{(z)(-4)-(1)(5)} \\ &\frac{a}{25+8}=\frac{b}{2-5}=\frac{c}{-5-5} \\ &\frac{a}{33}=\frac{b}{-3}=\frac{c}{-9} \\ \end{aligned}$

Multiplying by 3

$\begin{aligned} &\frac{a}{11}=\frac{b}{-1}=\frac{c}{-3}=x \\ &a=11 x, b=-x, c=3 x \\ \end{aligned}$

Put a, b, c in equation (2)

$\begin{aligned} &a(x-4)+b(y-3)+c(z-2)=0 \\ &(11 x)(2-4)+(-2)(y-3)+(-3 x)(z-2)=0 \\ &11 x_{2}-44 x-\lambda y+3 x z+6 x=0 \\ &11 x-x y-3 x_{2}-35 x=0 \end{aligned}$

Dividing by $x11x-y-3z-35=0$

So equation of required plane is $11x-y-3z-35=0$

The Plane Exercise 28.13 Question 5

Answer: equation of the plane $95x-17y+25z+53=0$ point of intersection is $\left ( 2,4,-3 \right )$
Hint: use simultaneous equation
Given:$\frac{x+y}{3}=\frac{y+6}{5}=\frac{z-1}{-2} \text { and } 3 x-2 y+z+3=0=2 x+3 y+2-6$

Solution: we have equation of the line is $\frac{x+y}{3}=\frac{y+6}{5}=\frac{z-1}{-2} =\lambda$

Point on the line is given by $\left ( 3x-4,5x-6,-2x+1 \right )$-------(1)

Another equation of line is $3x-2y+z+5=0$

$8z+2y+4=0$

Let a, b, c be the direction ratio of the lines it will be perpendicular to normal of $3x-2y+z+5=0$and $2x+3y+4z-4=0$ so using $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$

$\begin{aligned} &(3)(a)+(-2)(6)+(1)(c)=3 a-2 b+c=0 \\ \end{aligned}$............(2)

Agian $\begin{aligned} &(2)(a)+(3)(b)+(4)(c)=0 \\ \end{aligned}$

$\begin{aligned} &2 a+3 b+4 c=0 \ldots-\cdots-(3) \\ \end{aligned}$

Solving (2) and (3) by cross-multiplication

$\begin{aligned} &\frac{a}{(-2)(4)-(3)(1)}=\frac{b}{(2)(1)-(2)(4)}=\frac{c}{(3)(3)-(-2) 2)} \\ \end{aligned}$

$\begin{aligned} &\frac{a}{-8-3}=\frac{b}{2-12}=\frac{c}{9+4} \\ \end{aligned}$

$\begin{aligned} &\frac{a}{-11}=\frac{b}{-10}=\frac{c}{13} \\ \end{aligned}$

Directional ratios are proportional to

$\begin{aligned} &(-11,-10,13) \\ &\text { Let } z=0 \\ &3 x-2 y=-5 \ldots \cdots-(i) \\ &2 x+3 y=4 \ldots \ldots-(i i) \end{aligned}$

Solving (i) and (ii) by elimination method

$\begin{aligned} 6 x-4 y=-10 \\ \pm 6 x \pm 3 y=\pm 12 \\ \hline-13 y=-22 \\ y=\frac{22}{13} \end{aligned}$

Put y in equation (i)

$\begin{aligned} &3 x-2 y=-5 \\ &3 x-2 \frac{22}{13}=-5 \\ &3 x-\frac{44}{13}=-5 \\ &3 x=-5+\frac{44}{13} \\ &3 x=\frac{-21}{13} \\ &x=\frac{-7}{13} \end{aligned}$

So, the equation of the line (2) is symmetrical form

$\frac{x+\frac{7}{18}}{-11}=\frac{y-\frac{22}{13}}{-10}=\frac{2-0}{13}$

Put the general point of a line from equation (i)

$\begin{aligned} &\frac{3 x-4+\frac{7}{13}}{-11}=\frac{5 x-6-\frac{22}{13}}{-10}=\frac{-2 x+1}{13} \\ &\frac{39 x-52+7}{-11 x 13}=\frac{65 x-78-22}{-10 x 13}=\frac{-2 x+1}{13} \\ &\frac{39 x-45}{-11}=\frac{65 x-100}{-10}=\frac{-2 x+1}{1} \end{aligned}$

The equation of the plane is $45x-17y+25z+53$

There point of intersection is (2, 4, -3)

The Plane Exercise 28.13 Question 6

Answer: L.H.S = R.H.S
Hint: use vector set product.
Given: $\vec{r}=\left ( i+2j-k \right )=3$
Solution: we know that plane $\vec{r}.\vec{n}=d$ contains the line $\begin{aligned} &\pi=a+\lambda b+i j \\ &\vec{b} \cdot \vec{n}=0 \\ &\vec{a} \cdot \vec{n}=d \end{aligned}$
Given equation of plane;
$\bar{\pi }\left ( i+2j-k \right )=3$and equation of line $\bar{\pi }=i+j+\lambda \left ( 2i+j+4k \right )$
So,$\bar{n}=i+2j-k,\bar{a}=i+j$
$\begin{aligned} &\overline{3 \bar{b}}=2 i+j+4 k \\ &\vec{b} \cdot \vec{n}=(2 i+j 4 k)(i+2 j-k) \\ &=2+2-4 \\ &\vec{b} \cdot \vec{n}=0 \\ &\vec{a} \cdot \vec{n}=(i+j)(i+2 j-k) \\ &=1+2-0 \\ &=3 \\ &=\mathrm{d} \end{aligned}$

Since $\vec{b}.\vec{n}=0$ and $\vec{a}.\vec{n}=d$ so the line is the given plane hence proved

The Plane Exercise 28.13 Question 7

Answer: $\frac{x}{2}=\frac{y}{0}=\frac{z}{-1}$

Hint: use simultaneous equation;
Given: $\frac{x+2}{3}=\frac{y}{-2}=\frac{z-7}{6} \text { and } \frac{x+6}{1}=\frac{x+5}{-3}=\frac{z-1}{-1}$
Solution:
Let $\begin{array}{r} \mathrm{L} 1=\frac{2+3}{3}=\frac{y}{-2}=\frac{z-7}{6} \end{array}$ and
$\begin{array}{r} { L2 }=\frac{2+6}{1}=\frac{y+5}{-3}=\frac{z-1}{2} \end{array}$
Equation of two lines .let the plane be $az+by+cz=0$ -------(i)
Given that the required plane through the intersection of the lines L1 and L2 hence the normal to the plane is perpendicular to the line L1 and L2
$\begin{array}{r} 3 x-3 y+6 z=0 \\ x-3 y+2 z=0 \\ \end{array}$

Using cross multiplication, we get
$\begin{array}{r} \frac{x}{-4+18}=\frac{y}{6-6}=\frac{z}{-9+8} \\ \end{array}$
$\begin{array}{r} \frac{x}{14}=\frac{y}{0}=\frac{z}{-1} \\ \frac{x}{2}=\frac{y}{0}=\frac{z}{-1} \end{array}$

The Plane Exercise 28.13 Question 8

Answer: $\vec{r}\left ( 5i+2j-3k \right )=17$
Hint: first find the coordinates
Given: (3,4,2) and (7,0,6)
Solution: let the equation of the plane be $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=1$ --------(1)
Plane is passing through (3,4,2) and (7,0,6)
$\begin{aligned} &\frac{3}{a}=\frac{4}{b}=\frac{2}{c}=1 \\ &\frac{7}{a}=\frac{0}{b}=\frac{6}{c}=1 \end{aligned}$

Required plane is perpendicular to

$\begin{aligned} &2 x-5 y-15=0 \\ &\frac{2}{a}+\frac{-5}{b}+\frac{0}{c}=1 \\ &2 b=5 a \\ &b=2.5 a \\ &\frac{3}{a}+\frac{4}{2.5 a}+\frac{2}{c}=1 \\ &\frac{7}{a}+\frac{6}{2}=1 \end{aligned}$

Solving the above equation

$\begin{aligned} &a=3.4=\frac{17}{5}, b=\frac{17}{2} \mathrm{and} \\ &c=\frac{-34}{6}=\frac{-17}{3} \end{aligned}$

Substituting the values (1)

$\begin{aligned} &\frac{x}{\frac{17}{5}}+\frac{y}{\frac{17}{2}}+\frac{z}{\frac{17}{-3}}=1 \\ &\frac{5 x}{17}+\frac{2 y}{17}-\frac{3 z}{17}=1 \\ &5 x+2 y-3 z=17 \end{aligned}$

$\begin{aligned} &(x i+y j+2 k) \cdot(5 i+2 j-3 k)=17 \\ &\vec{r}(5 i+2 j-3 k)=17 \end{aligned}$

Vector equation of the plane is $\begin{aligned} &\vec{r}(5 i+2 j-3 k)=17 \end{aligned}$

The line passes through B(1,3,-2) $5\left ( 1 \right )+2\left ( 3 \right )-3\left ( -2 \right )=17$

The point B lies on the plane the line; $\vec{r}=i+3 j-2 k+\lambda(i+j+k)$ lies on the plane$\begin{aligned} &\vec{r}(5 i+2 j-3 k)=17 \end{aligned}$

The Plane Exercise 28.13 Question 9(i)

Answer: $-22x+19y+5z=31$
Hint: use vector cross product.
Given: $\begin{aligned} &\frac{x-1}{-3}=\frac{y-2}{-2 k}=\frac{z-3}{2} \text { and } \\ &\frac{x-1}{x}=\frac{y-2}{1}=\frac{z-3}{5} \end{aligned}$
Solution: the direction ratio of the line $\frac{x-1}{-3}=\frac{y-2}{-2 k}=\frac{z-3}{2} \text { is } \pi=(-3,-2 k, 2)$
The direction ratio of the line $\frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5} \text { is } \pi2=( k,1, 5)$
Since the line $\frac{x-1}{-3}=\frac{y-2}{-2 k}=\frac{z-3}{2}$ and $\frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5}$are perpendicular so
$\begin{aligned} &\pi_{1} \cdot \pi_{2}=0 \\ &(-3,-2 k, 2) \cdot(k, 1,5)=0 \\ &-3 k-2 k+10=0 \\ &-5 k=-10 \\ &k=2 \end{aligned}$
The equation of the line are $\frac{x-1}{-3}=\frac{y-2}{-2 k}=\frac{z-3}{2}$and $\frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5}$
The equation of the plane containing the line perpendicular lines $\frac{x-1}{-3}=\frac{y-2}{-2 k}=\frac{z-3}{2}$and $\frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5}$

The Plane Exercise 28.13 Question 9(ii)

Answer:

$\begin{aligned} &\left(\begin{array}{ccc} x & y & z \\ -3 & -4 & 2 \\ 2 & 1 & 5 \end{array}\right)=\mathrm{d} \\ &(-20-2) 2-y(-15-4)+2(-3+8)=d-22 z+19 y+5 z=d \end{aligned}$

The line $\frac{x-1}{-3}=\frac{y-2}{-2k}=\frac{z-3}{2}$ pass through the point (1,2,3) so putting $x=1,y=2,z=3$ is the equation we get

Therefore the equation of the plane containing the lines is $-22x+19y+5z=31$

The Plane Exercise 28.13 Question 10(i)

Answer:$\theta =\sin ^{-1}\left ( \frac{1}{\sqrt{3}\sqrt{29}} \right )$

Hint: use vector dot product

Since $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}$

Solution :any point on the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=k$ is of the form $\left ( 3k+2,4k-1,2k+2 \right )$

if the point $p\left ( 2k+2,4k-1,2k+2 \right )$ lies in the plane

$\begin{aligned} &x-y+z-5=0 \\ &(3 k+z)-(4 k-1)+(2 k+2)-5=0 \\ &3 k+2-4 k+1+2 k-5=0 \\ &k=0 \\ \end{aligned}$

Thus, the coordinates of the point of intersection of the line and the planes are

$\begin{aligned} &8(0)+2,4(0)-1,2(0)+2 \\ &p(2,-1,2) \\ \end{aligned}$

Let 0 be the angle between the line and the plane thus

$\begin{aligned} &\sin \theta=\frac{\mathrm{xl}+\mathrm{ym}+\mathrm{zn}}{\sqrt{\left(x^{2}+y^{2}+z^{2}\right)} \sqrt{l^{2}++n^{2}}} \end{aligned}$

Where l, m, and n are the direction ratio of the line and x, y and z are the direction ratios of the normal to the plane

The Plane Exercise 28.13 Question 10(ii)

Answer:

$\mathrm{L}=3, \mathrm{~m}=4, \mathrm{n}=2, \mathrm{a}=1, \mathrm{~b}=-1 and \; \mathrm{c}=1$
$\begin{aligned} &\sin \theta=\frac{1 \times 3+(-1) \times 4+1 \times 2}{\sqrt{\left(1^{2}+(-1)^{2}+1^{2}\right)} \sqrt{3^{2}+4^{2}+2^{2}}} \\ &\sin \theta=\frac{1}{\sqrt{3} \sqrt{29}} \\ &\theta=\sin ^{-1} \frac{1}{9.32} \end{aligned}$
Mention sin inverse 1/87

The Plane Exercise 28.13 Question 11(i)

Answer p(1,1,-2)
Hint: use vector cross product $i+j-2k,2i-j+k$ and $i+2j+k$
Solution: let A, B and C be three point with position vector
$i+j-2k,2i-j+k$ and$i+2j+k$
Thus $\begin{aligned} &\overrightarrow{A B}=\vec{b}-\vec{a}=(2 i-j+k)-(i+j-2 k) \\ \end{aligned}$
$\begin{aligned} &=i-2 j+3 k \\ &\overrightarrow{A C}=\vec{c}-\vec{a}=(i+2 j+k)-(i+j-2 k)=j+3 k \end{aligned}$
As we know that cross product of the line vectors gives a perpendicular vector so
$\begin{aligned} &\vec{m}=\overrightarrow{A B} x \overrightarrow{A C}=\left(\begin{array}{ccc} i & j & k \\ 1 & -2 & 3 \\ 0 & 1 & 3 \end{array}\right) \\ &\vec{m}=i(-6-3)-3 j+k=9 i-3 j+k \end{aligned}$
So the equation of the required plane is
$\begin{aligned} &(\vec{r}-\vec{a}) \vec{n}=0 \\ &(\vec{r}-\vec{n}) \cdot \vec{n}=(\vec{a}-\vec{n}) \\ &(\vec{r} \cdot(-9 i-3 j+k))=(i+j-2 k)-(-9 i-3 i+k) \\ &\vec{r}=(-9 i-3 j+k)=14 \end{aligned}$
Also we have to find the coordinates of the point of intersection of the plane and the line
$\vec{r}=3i-j-k+\lambda \left ( 2i-2j+k \right )$ any point on the line

The Plane Exercise 28.13 Question 11(ii)

Answer:

$\begin{aligned} &\vec{n}=3 i-j-k+\lambda(2 i-2 j+k) \end{aligned}$ is of the form $\begin{aligned} p(3+2 \lambda,-1-2 \lambda,-1+\lambda) \\ \end{aligned}$ lies in the plane
$\begin{aligned} &\vec{n} \cdot(-9 i-3 j+k) \\ &9(3+2 \lambda)-3(-1-2 \lambda)-(-1+2)=14 \\ &27+18 \lambda-3-6 \lambda+1-\lambda=14 \\ &11 \lambda=-11 \\ &\lambda=-1 \\ \end{aligned}$
Thus the required point of intersection is $\begin{aligned} &p(3+2 \lambda,-1-2 \lambda,-1+\lambda) \\ \end{aligned}$ put value$\lambda$ in the equation
$\begin{aligned} &p[3+2(-1),-1-2(-1),-1+(-1)] \\ &p(1,1,-2) \end{aligned}$

The Plane Exercise 28.13 Question 13

Answer:$x-y+z-1=0$
Hint: first find the value of coordinates, a, b and c
Given: (3, 2, 0) and contain the line $\frac{x-3}{1}=\frac{y-b}{5}=\frac{z-4}{7}$
Solution: given that a plane is passes through the point (3, 2, 0) so equation will be;
$\begin{aligned} &a(x-3)+b(y-2)+c(z-0)=0 \\ \end{aligned}$ -----(i)
$\begin{aligned} &a(x-3)+b(y-2)+c z=0 \end{aligned}$ --------------(ii)
Plane also contains the line
$\frac{x-3}{1}=\frac{y-b}{5}=\frac{z-4}{7}$
So it passes through the point (3, 2, 10)
$\begin{aligned} &a(3-3)+b(6-2)+c(4-0)=0 \\ &b+4 c=0 \\ \end{aligned}$ ------------------(2)
Also plane will be parallel to
$\begin{aligned} &a(1)+b(5)+c(4)=0 \\ &a+5 b+4 c=0------(3) \\ \end{aligned}$
Solving (2) and (3) by cross multiplication,
$\begin{aligned} &\frac{a}{16-20}=\frac{b}{4-0}=\frac{c}{0-4}=k \\ &\frac{a}{4}=\frac{b}{4}=-\frac{c}{4}=k \\ &a=-k, b=k, c=-k \\ \end{aligned}$
Put $a=-k,b=k,c=-k$in equation (ii) we get
$\begin{aligned} &(-k)(x-3)+(k)(y-2)+(-k)=0 \\ &-2+3+y-2-z=0 \\ &x-y+z-1=0 \end{aligned}$

The Plane Exercise 28.13 Question 14(i)

Answer: $x-2y+z=0$
Hint: use vector cross product
Given:$\frac{x+3}{-2}=\frac{y-1}{1}=\frac{z-5}{5} \text { and } \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$
Solution: we know that lines
$\frac{x+3}{-2}=\frac{y-1}{1}=\frac{z-5}{5} \text { and } \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$
Are coplanar if
$\begin{aligned} &\left(\begin{array}{ccc} \lambda-\lambda_{1} & y-y_{1} & z-z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}\right)=0 \\ &\text { here }: \lambda_{1}=-3, \lambda_{2}=-1, y_{1}=1, y_{2}=2, z_{1}=5, z_{2}=5 \\ &l_{1}=-3, l_{2}=-1, m_{1}=1, m_{2}=2, n_{1}=5, \ln _{2}=5 \\ &\left(\begin{array}{ccc} \lambda-\lambda_{1} & y-y_{1} & z-z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}\right) \\ &\left(\begin{array}{lll} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{array}\right) \\ &=2(5-10)-1(-15--5)+10(-6--1) \\ &=2(-5)-1(-10)=-10+10 \\ &=0 \end{aligned}$

So the given lines are coplanar , the equation of plane contains line is

$\begin{aligned} &\left(\begin{array}{ccc} \lambda+8 & y-1 & 3-5 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{array}\right)=0 \\ &(\lambda+3)(5-10)-(y-1)(-15-(-5))+(2-5)(-6-(-1)=0 \\ &-5 \lambda-15+10 y-10-5 z+25=0 \\ &-5 \lambda+10 y-5 z=0 \end{aligned}$

Divided by -5

$x-2y+z=0$

The Plane Exercise 28.13 Question 15

Answer: $I^{2}+m^{2}=2$
Hint: use simultaneous equation to solve
Given: $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$
Solution: we know that the lines
$\begin{aligned} &\frac{x-x_{1}}{l_{1}}=\frac{y-y_{1}}{m_{1}}=\frac{z-z_{1}}{n_{1}} \text { lies in plane } a_{1}+b y+c z+d=0 \text { then }\\ &a_{1} \lambda_{1}+b y_{1}+c z_{1}+d=0 \text { and } a l+b m+c n=0 \text { here, }\\ &\lambda_{1}=3, y_{1}=-2, z_{1}=-4 \text { and } l=2, m=-1, n=3 \end{aligned}$
and
$\begin{aligned} &2 l-m-3=0 \\ &3 l-2 m=5 \\ &3 l-2 m=5------(1) \\ &2 l-m=3------(2) \end{aligned}$

Multiply equation (1) by 2 and equation (2) by 3 and then subtract we get

$m=-1$

$l=1$

$l^{2}+m^{2}=2$

The Plane Exercise 28.13 Question 16(i)

Answer: $x=\pm \sqrt{2}$
Hint: use vector cross product
Given: $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{\lambda ^{2}}$and $\frac{x-3}{1}=\frac{y-2}{\lambda ^{2}}=\frac{z-1}{2}$
Solution: we know that the lines
$\begin{aligned} &\frac{x-x_{1}}{l_{1}}=\frac{y-y_{1}}{m_{1}}=\frac{z-z_{1}}{n_{1}}\\ &\text { and }\\ &\frac{x-x_{2}}{l_{2}}=\frac{y-y_{2}}{m_{2}}=\frac{z-z_{2}}{n_{2}} \text { are } \end{aligned}$
Coplanar if
$\left(\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}\right)=0$
Here:
$\begin{aligned} &x_{1}=1 \cdot x_{2}=3 . y_{1}=2 \cdot y_{2} \\ &=2.21,=3.2=1 \\ &l_{1}=1 . l_{2}=m_{1}=2 m_{2} \\ &=\lambda^{2} n_{1}=\lambda^{2} n_{2}=2 \\ \end{aligned}$
$\begin{aligned} &\left(\begin{array}{ccc} \lambda-\lambda_{1} & y-y_{1} & z-z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}\right)=0 \\ &\left(\begin{array}{ccc} 2 & 0 & -2 \\ 2 & 2 & \lambda^{2} \\ 1 & \lambda^{2} & 2 \end{array}\right)=0 \end{aligned}$

The Plane Exercise 28.13 Question 16(ii)

Answer:

$2\left(4-\lambda^{2}\right)-0\left(2-\lambda^{2}\right)-2\left(\lambda^{2}-2\right)=0$
$8-2 \lambda^{2}-2 \lambda^{2}+4=0$
$\lambda^{2}+\lambda^{2}-6=0$
Let $\lambda^{2}=t$ then
$(t+3)(t-2)=0$
Put value of t
$\left(\lambda^{2}+3\right)\left(\lambda^{2}-2\right)=0$is neglected because direction cosine cannot be imaginary
$\lambda^{2}=-3$
$Therefore, \lambda=\pm \sqrt{2}$

The Plane Exercise 28.13 Question 17(i)

Answer: $a=1,4,5$
Hint: use vector cross product
Given: $x=5,\frac{y}{3-2}=\frac{z}{-2}$ and $x=\alpha ;\frac{y}{-1}=\frac{z}{2-\alpha }$
Solution: we know that the lines
$\begin{aligned} &\frac{x-x_{1}}{l_{1}}=\frac{y-y_{1}}{m_{1}}=\frac{z-z_{1}}{n_{1}}\\ &\text { and }\\ \end{aligned}$are coplar of
$\begin{aligned} &\frac{x-x_{2}}{l_{2}}=\frac{y-y_{2}}{m_{2}}=\frac{z-z_{2}}{n_{2}} \end{aligned}$
$\begin{aligned} &\left(\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}\right)=0\\ \end{aligned}$
$\begin{aligned} &\text { here }\\ &x_{1}=5, x_{2}=2, y_{1}=0, y_{2}\\ &=0, z_{1}=0, z_{2}=0\\ &l_{1}=1, l_{2}=1, m_{1}=3-a\\ &m_{2}=-1, n_{1}=-2, n_{2}=2-\alpha \end{aligned}$
$\begin{gathered} \left(\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}\right)=0 \\ \left(\begin{array}{ccc} \alpha-5 & 0 & 0 \\ 1 & 3-\alpha & -2 \\ 1 & -1 & 2-\alpha \end{array}\right)=0 \end{gathered}$

The Plane Exercise 28.13 Question 17

Answer:

$\begin{aligned} &(a-5)[(3-a)(2-a)-2]=0 \\ &(a-5)(6-3 a-2 a+a 2-2)=0 \\ &a^{3}-10 a^{2}+29 a-20=0 \\ &0(a-1)(a-4)(a-5)=0 \\ &a=1,4,5 \end{aligned}$

The Plane Exercise 28.13 Question 18(i)

Answer: $y+z+1=0$
Hint: use vector cross product
Given: $\frac{x-1}{2}=\frac{y+1}{1}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$
Solution: we know that the lines $\frac{x-1}{2}=\frac{y+1}{1}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar if
$\left(\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}\right)=0$
here
$\begin{aligned} &x_{1}=1, x_{2}=-1, y_{1}=-1, y_{2}=-1, z_{1}=0, z_{2}=0\\ &I_{1}=2, l_{2}=5, m_{1}=k, m_{2}=2, n_{1}=2, n_{2}=k\\ &\left(\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}\right)=0\\ &\left(\begin{array}{ccc} -2 & 0 & 0 \\ 1 & k & 2 \\ 5 & 2 & k \end{array}\right)=0\\ &-2\left(k^{2}-4\right)=0\\ &k=\pm 2 \end{aligned}$

The Plane Exercise 28.13 Question 18(ii)

Answer:

The equation of plane contain lines is

$\left(\begin{array}{ccc} x-1 & y+1 & 2 \\ 2 & 2 & 2 \\ 5 & 2 & n_{2} \end{array}\right)=0$

When k=2

$\begin{aligned} &(x-1)(4-4)-(y+1)(4-10)+(2)(4-10)=0 \\ &6 y-6 z+6=0 \\ &y+z+1=0 \end{aligned}$

The equation of plane contain line

$\left(\begin{array}{ccc} x-1 & y+1 & 2 \\ 2 & 2 & 2 \\ 5 & 2 & -2 \end{array}\right)=0$

When k=-2

$\begin{aligned} &(x-1)(4-4)-(y+1)(-4-10)+(z)(4+10)=0 \\ &14 y+14 z+14=0 \\ &y+z+1=0 \end{aligned}$

The Plane Exercise 28.13 Question 19(i)

Answer:$\sqrt{3}$ units

Hint: use vector cross products

Given: $\vec{r}=\left ( i+j \right )+\lambda \left ( i+2j-k \right )$

Solution: equation of given line is

$\vec{r}=\left ( i+j \right )+\lambda \left ( i+2j-k \right )$ and $\vec{a}=\vec{r}+\vec{\lambda b}$...................(i)

where

$\vec{a}=i+j$

$\vec{a}=i+2j-k$

Again: $\vec{r}=\left ( i+j \right )+\lambda \left ( i+2j-k \right )$

$\begin{aligned} &\overline{a^{n}}=(i+j)+u(-i+j-2 k) \\ &=\overline{a^{n}}+\bar{u}^{\bar{b}} \\ \end{aligned}$

Where$\begin{aligned} &\overline{a^{1}}=(i+j) \\ &\vec{b}=-i+j-2 k \end{aligned}$

The vector equation of the plane containing the line (i) and (ii) is given by

$\begin{aligned} &\vec{b} x \bar{b}^{\overline{1}}=\left(\begin{array}{ccc} i & j & k \\ 1 & 2 & -1 \\ -1 & 1 & -2 \end{array}\right) \\ &i(-4+1)-j(-2-1)+k(1+2)=3 i+3 j+3 k \end{aligned}$

The Plane Exercise 28.13 Question 19(ii)

Answer:

$$$ \begin{aligned} &r 1(-3 i+3 j+3 k)=(i+j)(-3 i+3 j+3 k) \\ &r 1(-3 i+3 j+3 k)=-3+3=0 \\ &r 1=-3 x+3 y+3 z=0 \\ &\text { le. } x-y-z=0 \end{aligned}$
$Distance \; from \: (2,1,-4) \: is$
$\left(\frac{2-1-4}{\sqrt{1^{2}+1^{2}+1^{2}}}\right)=\frac{3}{\sqrt{3}}=\sqrt{3} \text {.units }$

The portions in the class 12 mathematics chapter 28 include fifteen exercises. The thirteenth exercise or the ex 28.13 consists of nineteen questions given in the maths textbook which deals with the concepts like to show the lines are coplanar, find the vector equation of the plane, and find the coordinates of the given points. All these concepts have various questions under level 1; there are no level 2 questions in this exercise. To clarify the doubts in this concept, the RD Sharma Class 12 Chapter 28 Exercise 28.13 will lend a helping hand.

The intensity of the questions gets more profound in exercise 13, where all the previously learned concepts are included. This makes the students struggle to solve the sums if they are unclear about the previous exercises. In such cases, they can refer to the RD Sharma Class 12th Exercise 28.13 solution book and the previous exercises to understand the concepts in-depth. All these books are based on the NCERT pattern, which explains why CBSE students must prefer these books.

Whenever students encounter a doubt, they can immediately refer to the Class 12 RD Sharma Chapter 28 Exercise 28.13 Solution material to clear those doubts. Many staff members and experts have diligently checked the accuracy of every solution given by them in the RD Sharma solution books. Any student who finds the concept of the plane challenging can use these reference materials to develop their knowledge in this topic.

Many previous batch students suggest the RD Sharma Class 12 Solutions the Plane Ex 28.13 for their juniors to practice the sums in the 28th chapter. The career 360 website provides free access to the RD Sharma Class 12th Exercise 28.13 solution material. The students can also download it in PDF format for later reference.

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