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 Straight Line in Space exercise 27.5 question 3 sub question (iv) maths

Explain solution RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 3 sub question (iv) maths

Answers (1)

Answer: Given Lines are not interesting

Hint: Consider the given equation as (1) and (2)

Given: \frac{x-5}{4}=\frac{y-7}{-5}=\frac{z+3}{-5} \rightarrow(1) \text { and } \frac{x-8}{7}=\frac{y-7}{1}=\frac{z-5}{3} \rightarrow(2)

Solution: Since line (1) passes through the point (5,7,-3) and has direction ratios proportional to (4,-5,-5) its vector equation is \overrightarrow{\mathrm{\gamma}}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}

Here,

\begin{aligned} &\overrightarrow{a_{1}}=5 \hat{\imath}+7 \hat{\jmath}-3 \hat{k} \\\\ &\overrightarrow{b_{1}}=4 \hat{\imath}-5 \hat{\jmath}-5 \hat{k} \end{aligned}

Also line (2) passes through the point (8,7,5) and has direction ratios proportional to (7,1,3) its vector equation is \overrightarrow{\mathrm{\gamma }}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}

Here,

\begin{aligned} &\overrightarrow{a_{2}}=8 \hat{\imath}+7 \hat{\jmath}+5 \hat{k} \\\\ &\overrightarrow{b_{2}}=7 \hat{i}+1 \hat{\jmath}+3 \hat{k} \end{aligned}

Now,

\begin{aligned} &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=3 \hat{\imath}+8 \hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 4 & -5 & -5 \\ 7 & 1 & 3 \end{array}\right| \Rightarrow-10 \hat{\imath}-47 \hat{\jmath}+39 \hat{k} \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(3 \hat{\imath}+8 \hat{k}) \cdot(-10 \hat{\imath}-47 \hat{\jmath}+39 \hat{k}) \Rightarrow-30+312 \Rightarrow 282 \end{aligned}

We observe,   \left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) \neq 0

Thus, the given lines are not intersecting.

 

Posted by

infoexpert26

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

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question