RD Sharma Class 12 Exercise 28.14 The Plane Solutions Maths

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

Edited By Lovekush kumar saini | Updated on Jan 25, 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.

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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 $3x-y-2z+4=0$
$\begin{aligned} &D=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right| \\ &D=\left|\frac{3(1)-1(3)-2(-2)+4}{\sqrt{(3)^{2}+(-1)^{2}+(-2)^{2}}}\right| \end{aligned}$
$\begin{aligned} &D=\left|\frac{8}{\sqrt{14}}\right| \\ &D=\frac{8}{\sqrt{14}} \text { (answer) } \end{aligned}$

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Frequently Asked Question (FAQs)

1. 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.

2. 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.

3. 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.

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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.

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5. 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.