Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution RD Sharma class 12 Chapter 28 The Plane exercise 28.11 question 15

Explain solution RD Sharma class 12 Chapter 28 The Plane exercise 28.11 question 15

Answers (1)

Answer: Therefore, the line \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}  and plane \vec{r} \cdot \vec{n}=d are parallel. And line

\overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\lambda(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})  is parallel to \vec{r} \cdot(2 \hat{j}+\hat{k})=3 .. The distance between lines and plane is \frac{1}{\sqrt{5}}  units.

Hint: Use formula \mathrm{D}=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}-\mathrm{d}}{\overrightarrow{\mathrm{n}}}

Given:  \vec{r}=\hat{i}+\hat{j}+\lambda(3 \hat{i}-\hat{j}+2 \hat{k}) \text { and } \vec{r} \cdot(2 \hat{j}+\hat{k})=3

Solution: We know that line \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}  and plane \vec{r} \cdot \vec{n}=d  is parallel

if   \vec{b} \cdot \vec{n}=0                                            ……………… (1)

Given, equation of line \overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\mathrm{k}(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})   and equation of plane is \vec{r} \cdot(2 \hat{j}+\hat{k})=3

So, \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{n}}=2 \hat{\mathrm{j}}+\hat{\mathrm{k}}

Now,  \overrightarrow{\mathrm{b}} \cdot \vec{n}=(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})(2 \hat{\mathrm{j}}+\hat{\mathrm{k}})=-2+2=0

So, the line and plane are parallel.

We know that, the distance ‘D’ of plane is \vec{r} \cdot \vec{n}=d   from a point \vec{a}  is given by

\begin{aligned} &\mathrm{D}=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}-\mathrm{d}}{\overrightarrow{\mathrm{n}}} \\ & \end{aligned}

\overrightarrow{\mathrm{a}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}})

\begin{aligned} D &=\frac{(\hat{i}+\hat{j})(2 \hat{j}+\hat{k})-3}{\sqrt{2^{2}+1^{2}}} \\ & \end{aligned}

=\frac{2-3}{\sqrt{4+1}}=\frac{-1}{\sqrt{5}}

We take the mod value

            D=\frac{1}{\sqrt{5}}

 

Posted by

infoexpert27

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 result verification 2024 application out at cbse.gov.in; last date, fees
anam-khan

CBSE results 2024 later for 102 students over fake certificates; board issues show cause notice
anam-khan

What after CBSE Class 12 results 2024? From CUET UG to JEE Advanced, all you need to know

Latest Question