What do you mean by plane?

A plane is a two-dimensional flat surface with infinite dimensions but no thickness. Such that if any two points on it are taken, the line segment connecting them lies entirely on the surface.

Is it enough to study from RD Sharma Class 12 solutions from an exam point of view?

RD Sharma Solutions for Class 12 are compiled in a step-by-step manner in simple language for students' ease of comprehension. Referring to these solutions while solving can help students improve their conceptual knowledge.

Which website is the best for studying RD Sharma Solutions for Class 12 Maths?

You can find Rd Sharma Class 12th Exercise 28.14 on the Career360 website, where you can get step-by-step answers to all of the questions in the RD Sharma textbook. As a result, students in Class 12 should learn all of the concepts covered in the syllabus in order to understand the important topics.

What is the primary goal of the RD Sharma textbook for Class 12?

1. The primary goal of RD Sharma Textbooks is to provide a necessary perspective of each chapter, which in turn helps students to understand every thought clearly. 2. RD Sharma Textbooks present all of the textbook's content in simple language. 3. The primary goal of RD Sharma Solutions is to inculcate confidence in students before they sit for exams.

How do Rd Sharma Class 12th Exercise 28.14 Solutions help with CBSE Board Exams?

Rd Sharma Class 12th Exercise 28.14 Solutions for Class 12 provides a large number of questions to practice on a regular basis. This made it easier to complete all of the questions on time in exams. The short answer and multiple-choice questions keep them busy solving problems throughout the academic year.

RD Sharma Class 12 Exercise 28.14 The Plane Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 28.14 The Plane Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 28.14 The Plane Solutions Maths - Download PDF Free Online

Updated on 25 Jan 2022, 11:53 AM IST

For a long time, RD Sharma books have set parameters for board exams. This book is essential in terms of exams because teachers often assign question papers from it. The Rd Sharma Class 12th Exercise 28.14 solutions that you will find here are a team of subject experts who constantly strive for student success. This particular exercise has three questions and few examples to clarify the concept in a better way. The chapter 'The Plane' discusses 2D surfaces having infinite dimensions with no thickness such that when two lines join them, both of them lie on the surface entirely and find the shortest distance between the lines.

The Plane Excercise: 28.14

The Plane Exercise 28.14 Question 1

Answer : $\frac{65}{\sqrt{122}}$
Hint : Distance between shortest lines
$D=\left|\frac{\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b_{2}}\right|}\right|$
Given : Find the shortest distance between lines $\frac{x-2}{-1}=\frac{y-5}{2}=\frac{z-0}{3}$ and $\frac{x-0}{2}=\frac{y+1}{-1}=\frac{z-1}{2}$
Solution : $\frac{x-2}{-1}=\frac{y-5}{2}=\frac{z-0}{3}$
$\vec{r}=(2 \hat{\imath}+5 \hat{\jmath})+\lambda(-\hat{\imath}+2 \hat{\jmath}+3 \hat{k}) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\vec{r}=\vec{a}+\lambda \vec{b}]$
$\begin{aligned} &\vec{a}_{1}=2 \hat{i}+5 \hat{j} \\ &\vec{b}_{1}=-\hat{i}+2 \hat{j}+3 \hat{k} \end{aligned}$
$\frac{x-0}{2}=\frac{y+1}{-1}=\frac{z-1}{2}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\right]$
$\begin{aligned} &\vec{r}=(-\hat{\jmath}+\hat{k})+\mu(2 \hat{\imath}-\hat{\jmath}+2 \hat{k}) \\ &\vec{a}_{2}=-\hat{j}+\hat{k}, \vec{b}_{2}=2 \hat{i}-\hat{j}+2 \hat{k} \\ &D=\left|\frac{\left.\overrightarrow{(a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right| \end{aligned}$
$\vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 3 \\ 2 & -1 & 2 \end{array}\right|$
$\begin{aligned} &=\hat{i}(4+3)-\hat{j}(-2-6)+\hat{k}(1-4)=7 \hat{i}+8 \hat{j}-3 \hat{k} \\ &\left|\vec{b}_{1} \times \vec{b}_{2}\right|=\sqrt{7^{2}+8^{2}+(-3)^{2}}=\sqrt{49+64+9}=\sqrt{122} \\ &D=\left|\frac{(-2 \hat{\imath}-6 \hat{\jmath}+\hat{k}) \cdot(7 \hat{\imath}+8 \hat{\jmath}-3 \hat{k})}{\sqrt{122}}\right| \\ &=\left|\frac{-14-48-3}{\sqrt{122}}\right|=\frac{|-65|}{\sqrt{122}} \\ &D=\frac{65}{\sqrt{122}} \end{aligned}$

The Plane Exercise 28.14 Question 2

Answer : $2\sqrt{29}$
Hint: Formula of distance between shortest lines
$D=\left|\frac{\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b_{2}}\right|}\right|$
Given: Find the shortest distance between lines
$\frac{x+1}{7}=\frac{y+1}{6}=\frac{z+1}{1} \text { and } \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$
Solution : $\frac{x+1}{7}=\frac{y+1}{6}=\frac{z+1}{1}$
$\vec{r}=(-\hat{\imath}-\hat{\jmath}+\hat{k})+\lambda(7 \hat{\imath}-6 \hat{\jmath}+\hat{k}) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\vec{r}=\vec{a}+\lambda \vec{b}]$
$\vec{a}=-\hat{i}-\hat{j}-\hat{k}, \vec{b}=7 \hat{i}-6 \hat{j}+\hat{k}$
Now $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{7} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\right]$
$\begin{aligned} &\vec{r}=(3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k})+\lambda(\hat{\imath}-2 \hat{\jmath}+\hat{k}) \\ &\vec{a}_{2}=3 \hat{i}+5 \hat{j}+7 \hat{k}, \quad \vec{b}_{2}=\hat{i}-2 \hat{j}+\hat{k} \\ &\therefore\left(\vec{a}_{2}-\vec{a}_{1}\right)=(3 \hat{i}+5 \hat{j}+7 \hat{k})-(-\hat{i}-\hat{j}-\hat{k}) \\ &=4 \hat{i}+6 \hat{j}+8 \hat{k} \end{aligned}$
$\therefore \vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{array}\right|$
$=\hat{\imath}(-6+2)-\hat{\jmath}(7-1)+\hat{k}((-14)+6))$
$\begin{aligned} &\vec{b}_{1} \times \vec{b}_{2}=-4 \hat{i}-6 \hat{j}-8 \hat{k} \\ &D=\left|\frac{(4 \hat{i}+6 \hat{j}+8 \hat{k}) \cdot(-4 \hat{i}-6 \hat{j}-8 \hat{k})}{\sqrt{16+36+64}}\right| \\ &D=\left|\frac{-16-36-64}{\sqrt{122}}\right| \end{aligned}$
$\begin{aligned} &=\frac{-116}{\sqrt{116}}=\sqrt{116}=2 \sqrt{29} \\ &d=2 \sqrt{29} \text { (answer) } \end{aligned}$

The Plane Exercise 28.14 Question 3

Answer: $\frac{8}{\sqrt{14}}$
$\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$
Given – Find the shortest distance between the lines $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1}$ and $3 x-y-2 z+4=0=2 x+y+z+1$
$\frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{x-x_{1}}{\alpha}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\right]$
$x_{1}=1, y_{1}=3, z_{1}=-2$
length of perpendicular from $(1,3,-2)$ to the plane $3x-y-2z+4=0$
$\begin{aligned} &D=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right| \\ &D=\left|\frac{3(1)-1(3)-2(-2)+4}{\sqrt{(3)^{2}+(-1)^{2}+(-2)^{2}}}\right| \end{aligned}$
$\begin{aligned} &D=\left|\frac{8}{\sqrt{14}}\right| \\ &D=\frac{8}{\sqrt{14}} \text { (answer) } \end{aligned}$

To attain an outstanding academic score in Class 12 Maths, students should wrap up all of the RD Sharma Class 12 Maths Textbook exercises from Rd Sharma Class 12th Chapter 28 Exercise 28.14. A team of Math experts created RD Sharma Solutions for Class 12 to assist students in performing well in their exams. The benefits of these solutions are:

1. Guaranteed good grades:

Career360's solutions are created by a team of experts who collaborate to provide perfect answers in the simplest way possible. Students who regularly practise Rd Sharma Class 12th Exercise 28.14 have a good chance of passing their board exams.

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Students can easily review RD Sharma Class 12 Solutions exercise 28.14 if they have any questions because it contains all of the solutions needed to get good grades. Using the solved examples, a student can quickly refer to the problem they are having difficulty with.

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Teachers do not cover all of Class 12 RD Sharma Chapter 28 Exercise 28.14 Solutions problems due to time constraints. As a result, this material is easily accessible for demonstrating capability.

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RD Sharma Class 12 Solutions Chapter 28 ex 28.14 are expert-created solutions that are straightforward and simple to grasp. As a result, students who are unfamiliar with the chapter can use this material to gain a thorough understanding from the start.

5. NCERT-related questions

The NCERT is usually used to answer questions in exams. Teachers assign question papers based on NCERT textbooks, so learning NCERT concepts and answering questions is required.

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