RD Sharma Class 12 Exercise 28.12 The Plane Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 28.12 The Plane Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 11:53 AM IST

The class 12 students who are busy preparing for their public exams trust the solutions given in the RD Sharna books. Various concepts in different subjects are explained in a simple manner. It is the best source for students to clarify their doubts in mathematics too. For example, the solutions for the sums in chapter 28 are given in the RD Sharma Class 12th Exercise 28.12.

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

Mentioning chapter 28 consists of fifteen exercises. The ex 28.12, deals with concepts like finding the coordinates of the point, finding the distance of the point from the point of intersection of the line, direction cosines, finding the angle of the lines, finding the vector equations of the lines, and so on. This exercise consists of only the Level 1 part, where there are ten questions in total. Therefore, the RD Sharma Class 12 Chapter 28 Exercise 28.12 is the best reference material to look into if there are any doubts in this portion.

## The Plane Excercise: 28.12

The Plane Exercise 28 .12 Question 1(i)

Answer: Therefore, required coordinate is $\left ( 0,\frac{17}{3},\frac{-13}{2} \right )$

Hint: Use Formula $\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$

Given: (5,1,6) and (3,4,1)

Solution: Equation of line through (5,1,6)& (3,4,1) is

\begin{aligned} &\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6} \\ &\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=\lambda \end{aligned}
Another point on the line
$\left ( 5-2\lambda ,1+3\lambda ,6-5\lambda \right )$
Equation of YZ-plane is $x=0$
This point lies on $x=0$
$5-2\lambda =0\Rightarrow \lambda =\frac{5}{2}$
The required point

\begin{aligned} &=\left(5-\frac{10}{2}, 1+\frac{15}{2}, 6-\frac{25}{2}\right) \\ &=\left(0, \frac{17}{2}, \frac{-13}{2}\right) \end{aligned}

The Plane Exercise 28.12 Question 1(ii)

Answer: Therefore, required coordinate is $\left ( \frac{17}{3},0,\frac{23}{3} \right )$
Hint: Use Formula $\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$
Given:$\left ( 5,1,6 \right )$and $\left ( 3,4,1 \right )$
Solution:
Equation of line through $\left ( 5,1,6 \right )$& $\left ( 3,4,1 \right )$is
\begin{aligned} &\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6} \\ &\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=\lambda \end{aligned}
∴Another point on the line
$\left ( 5-2\lambda ,1+3\lambda ,6-5\lambda \right )$
Equation of ZX-plane is $y=0$
∴This point lies on$y=0$
$1+3\lambda =0\Rightarrow \lambda =\frac{-1}{3}$
The required point
\begin{aligned} &=\left(5+\frac{2}{3}, 1-\frac{3}{3}, 6+\frac{5}{3}\right) \\ &=\left(\frac{17}{3}, 0, \frac{23}{3}\right) \end{aligned}

The Plane Exercise 28.12 Question 2

Answer: Therefore, the required point $\left ( 1,-2,7 \right )$
Hint: Use formula $\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$
Given:$\left ( 3,-4,-5 \right )$ & $\left ( 2,-3,1 \right )$ and plane $2x+y+z=7$
Solution: Equation of line through $\left ( 3,-4,-5 \right )$ &$\left ( 2,-3,1 \right )$ is

$\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda$
$any\: point \: on \: the\: line (3-\lambda,-4+\lambda,-5+6 \lambda)$
$but \: this\: point \: lies \: on\: the\: plane\: 2 x+y+z=7 is$
$6-2 \lambda-4+\lambda-5+6 \lambda=7$
$\Rightarrow 5 \lambda=10 \Rightarrow \lambda=2$
$\therefore Required \: point \: is (1,-2,7)$

The Plane Exercise 28.12 Question 3

Answer:Therefore ,the distance is 13 Units .
Hint : Use Distance formula $\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-_y{1} \right )^{2}+\left ( z_{2}-z_{1} \right )^{2}}$
Given:
\begin{aligned} &\vec{r}=2 \hat{\imath}-\hat{\jmath}+2 \hat{k}+\lambda(3 \hat{\imath}+4 \hat{\jmath}+2 \hat{k}) \text { and } \\ &\vec{r} \cdot(\hat{\imath}-\hat{\jmath}+\hat{k})=5 \text { and point }(-1,-5,-10) \end{aligned}
Solution :
$General\; point \: on \: the \: line \: is$
$(2+3 \lambda,-1+4 \lambda, 2+2 \lambda)$
$As\: point \: lies \: on\: the\: plane$
$\therefore 2+3 \lambda+1-4 \lambda+2+2 \lambda=5$
$\Rightarrow \lambda=0$
$\therefore \: So\: the \: \: coordinate \: \: of \: \: point \: \: is (2,-1,2)$
$Distance =\sqrt{(2-(-1))^{2}+(-1-(-5))^{2}+(2-(-10))^{2}}$
$=\sqrt{(2+1)^{2}+(-1+5)^{2}+(2+10)^{2}}$
$=\sqrt{(3)^{2}+(4)^{2}+(12)^{2}}$
$=\sqrt{9+16+144}$
$=\sqrt{169}=13 unit$

The Plane Exercise 28.12 Question 4

Answer: Therefore, the distance is 13 units.
Hint: Use distance formula $\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}+\left ( z_{2}-z_{1} \right )^{2}}$
Given:
\begin{aligned} &\vec{r}=2 \hat{\imath}-4 \hat{\jmath}+2 \hat{k}+\lambda(3 \hat{\imath}+4 \hat{\jmath}+2 \hat{k}) \text { and } \\ &\vec{r} \cdot(\hat{\imath}-2 \hat{\jmath}+\hat{k})=0 \text { and point }(2,12,5) \end{aligned}
Solution:
\begin{aligned} &\text { General point on the line is }\\ &(2+3 \lambda,-4+4 \lambda, 2+2 \lambda)\\ &\text { The equation of the plane is } \vec{r} \cdot(\hat{\imath}-2 \hat{\jmath}+\hat{k})=0 \end{aligned}
The point of intersection of the line and the plane will be ,
Substituting general point of the line in the equation of plane and finding the perpendicular value of $\lambda$

$[(2+3 \lambda) \hat{\imath}+(-4+4 \lambda) \hat{\jmath}+(2+2 \lambda) \hat{k}](\hat{\imath}-2 \hat{\jmath}+\hat{k})=0$
$[(2+3 \lambda) 1+(-4+4 \lambda)(-2)+(2+2 \lambda) 1]=0$
$12-3 \lambda=0 \Rightarrow \lambda=4$$\therefore The \, point \: of \: intersection\: is (14,12,10)$
$Distance \: of \: this\: point \: from (2,12,5) is$
$=\sqrt{(14-2)^{2}+(12-12)^{2}+(10-5)^{2}}$
$=\sqrt{12^{2}+5^{2}}$
$=\sqrt{144+25}$

The Plane Exercise 28.12 Question 5

Answer:Therefore ,the distance is 13 units
Hint: Use distance formula $\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$
Given:
\begin{aligned} &\text { Point } P(-1,-5,-10) \\ &A(2,-1,2) \& B(5,3,4) \\ &\text { and } x-y+z=5 \end{aligned}
Solution:$Equation \: of \: line\: passing \: through (2,-1,2) \&(5,3,4) is$
$\frac{x-2}{5-2}=\frac{y+1}{3+1}=\frac{z-2}{4-2}$
$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=m$
$x=3 m+2, y=4 m-1, z=2 m+2$
$Now \: putting \: the \: value \: of \: \mathrm{x}, \mathrm{y} and \mathrm{z} in the \: equation \: of \: the \: palne \: x-y+z=5$
$(3 m+2)-(4 m-1)+(2 m+2)=5$
$m=0$
$So, the\: point \: of\: intersection \: of \: the \: line\: and\: the \: plane\: is (2,-1,2)$
$\therefore The\; distance \: of\: the \: point\; \mathrm{P}(-1,-5,-10)$
$=\sqrt{3^{2}+4^{2}+12^{2}}=13 \: unit$

The Plane Exercise 28.12 Question 6

Answer: Therefore, the distance is 7 units.
Hint : Use distance formula $\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}\left ( z_{2}-z_{1} \right )^{2}}$
Given:$P\left ( 3,4,4 \right ),A\left ( 3,-4,-5 \right )$ &$B\left ( 2,-3,1 \right )$
and $2x+y+z=7$
Solution:
$The \: direction \: ratios \: of\: line \: joining\: A(3,-4,-5) \& B(2,-3,1) are$
$[(2-3),(-3+4),(1+5)]=(-1,1,6)$
$Now\: \: the \: \: equation \: of \: line \: passing \: through \: (3,-4,-5) \& B(2,-3,1) and\: having DR's (-1,1,6) is$
$\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}$
\begin{aligned} &\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda(\text { say }) \\ &x=-\lambda+3, y=\lambda-4, z=6 \lambda-5 \end{aligned}
Thus the general point on the line is given by $\left ( 3-\lambda ,\lambda -4,6\lambda -5 \right )$
Since the intersection of the line and the plane intersects each other,the line lies on the plane

Hence,
$2 x+y+z=7$
$2(-\lambda+3)+(\lambda-4)+(6 \lambda-5)=7 \lambda=2$
$On \: substituting\: the \: value \: of\: \lambda,value of x, y \& z are$
$(3-2,2-4,6(2)-5) i.e (1,-2,7)$
$Distance\: \: between (3,4,4) and (1,-2,7)$
$=\sqrt{(3-1)^{2}+(4+2)^{2}+(4-7)^{2}}$
$=\sqrt{4+36+9}$
$=\sqrt{49}$
$= 7\: \: units$

The Plane Exercise 28.12 Question 7

Answer: Therefore, the distance is $10\sqrt{3}$units.

Hint:Use distance formula $\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}+\left ( z_{2}-z_{1} \right )^{2}}$
Given:
$\left ( 1,-5,9 \right )$ and $x-y+z=5$
$x=y=z$
Solution: The straight line is given by $x=y=z$
$\because \frac{x}{1}=\frac{y}{1}=\frac{z}{1}$
$Direction\: ratios \: are \: 1,1,1$
$Again\: the\: given\: point \: is (1,-5,9)$
$\therefore \frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}=t( say )$
$\Rightarrow x=t+1, y=t-5, z=t+9$
\begin{aligned} &\text { This point satisfies the plane equation }\\ &x-y+z=5\\ &t+1-t+5+t+9=5\\ &t+15=5\\ &t=-10 \end{aligned}
\begin{aligned} &x=-9, y=-15, z=-1\\ &\text { The distance between the coordinates }(-9,-15,-1) \text { and }(1,-5,9)\\ &=\sqrt{(1+9)^{2}+(-5+15)^{2}+(9+1)^{2}}\\ &=\sqrt{10^{2}+10^{2}+10^{2}}\\ &=10 \sqrt{3} \end{aligned}

The Plane Exercise 28.12 Question 8

Hint :Use distance formula $\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$
Given:$\frac{x-1}{3}=\frac{y+4}{7}=\frac{z+4}{2}$
Solution:
$If \: line \frac{x-1}{3}=\frac{y+4}{7}=\frac{z+4}{2} cuts \: the XY-plane$
$Then z=0$
$So, let \: coordinate\: of \: point \: be (x, y, 0)$
$\frac{x-1}{3}=\frac{y+4}{7}=\frac{z+4}{2}=k(s a y)$
$\Rightarrow x=3 k+1, y=7 k-4, z=2 k-4$
$Since\: z=0$
$\Rightarrow 2 k-4=0$
$\Rightarrow 2 k=4 \Rightarrow k=2$
$Now, x=7, y=10, z=0$

Therefore, required point (7,10,0)

The Plane Exercise 28.12 Question 9

Answer: Therefore, required distance is $\vec{r}=-2 \hat{\imath}-4 \hat{\jmath}+7 \hat{k}+\mu(3 \hat{\imath}+3 \hat{\jmath}-3 \hat{k})$ & distance is $3\sqrt{3}$units
Hint: Use properties of vector
Given:
\begin{aligned} &\vec{r}=3 \hat{\imath}-2 \hat{\jmath}+6 \hat{k}+\lambda(2 \hat{\imath}-\hat{\jmath}+2 \hat{k}) \text { and }\\ &\vec{r} \cdot(\hat{\imath}-\hat{\jmath}+\hat{k})=6\\ \end{aligned}
Solution:
\begin{aligned} &\vec{r}=3 \hat{\imath}-2 \hat{\jmath}+6 \hat{k}+\lambda(2 \hat{\imath}-\hat{\jmath}+2 \hat{k})\\ &\vec{r}=(3+2 \lambda) \hat{\imath}+(-2-\lambda) \hat{\jmath}+(6+2 \lambda) \hat{k}\\ &\text { Plane is } \vec{r} \cdot(\hat{\imath}-\hat{\jmath}+\hat{k})=6\\ &\text { To find the point of intersection, }\\ &\therefore 3+2 \lambda+2 \lambda+6+2 \lambda=6\\ &5 \lambda=-5 \Rightarrow \lambda=-1\\ &\therefore \text { Point } \mathrm{Q}(1,-1,4)\\ &P(-2,-4,7)\\ &\text { Line } \mathrm{PQ} \text { is } \vec{r}=-2 \hat{\imath}-4 \hat{\jmath}+7 \hat{k}+\mu(3 \hat{\imath}+3 \hat{\jmath}-3 \hat{k})\\ &\text { Required distance of } \mathrm{PQ}=\sqrt{(1-(-2))^{2}+((-4)+1)^{2}+(7-4)^{2}}=\sqrt{3^{2}+3^{2}+3^{2}}=\sqrt{27} \text { units } \end{aligned}

It would not be sufficient for the student to be clear of the concept with only ten questions in it. Therefore, the RD Sharma Class 12th Exercise 28.12 solution book will enrich the students' knowledge as there are various additional questions for practice. As it follows the NCERT pattern, this gives another valid reason on how the CBSE students can adapt to it easily. Once the students work out all the sums given in this portion with concentration, they will be clear about this chapter.

No student needs to worry about the accuracy of the answers provided in this book. A team of experts in mathematics has committed their time and skill to create the solutions for this portion. Hence, Class 12 RD Sharma Chapter 28 Exercise 28.12 Solution is the ideal reference material for this chapter. The plane concept would be easy-peasy for the students to work out in a limited time duration after a good practice with the RD Sharma Class 12 Solutions the Plane Ex 28.12 book.

Another significant advantage of using the RD Sharma books is that no one is asked to pay even a single rupee to access this best solution book. All the RD Sharma books and the RD Sharma Class 12th Exercise 28.12 reference material are available for free on the Career360 website.

An abundance of previous batch students has already benefited by scoring good marks in their exams using the RD Sharma Class 12 Solutions Chapter 28 Ex 28.12 for their exam preparation. This book is recommended by most of the students to their fellow mates as they find it easy to adapt to it.

## RD Sharma Chapter wise Solutions

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1. What is the best solution guide to clarify the doubts on the mathematical concept of the plane?

The RD Sharma Class 12th Exercise 28.12 is the best solution material recommended for the students to clarify their doubts regarding the plane concepts in mathematics.

2. Are the RD Sharma books available only in hard copies?

You can find the softcopy PDF material of the RD Sharma solution books on the Career 360 website.

3. Do the RD Sharma solution books contain answers only for Level 1 questions?

The RD Sharma solution books contain answers for both the Level 1 and 2 sections given in the textbook.

4. What are the concepts present in class 12, mathematics chapter 28, ex 28.12 portions?

The concepts present in ex 28.12 are to find the coordinates of the point, the distance of the point, angle, and vector equations of the lines. You can find the solved sums in the RD Sharma Class 12th Exercise 28.12 solution book.

5. How much do the RD Sharma books cost?

You need not spend money buying solution books when you can access the RD Sharma books on the career360 websites.

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