Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 6 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 6 Maths Textbook Solution.

Answers (1)

Answer – The answer of the given question is  \left ( -4,1,-3 \right )\frac{x-2}{-6}=\frac{y-4}{-3}=\frac{z-1}{-2}

(Hint – By using two point formula).

Given – Point (2,4,-1) and equation of the line \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}

Solution - Let PQ be the perpendicular drawn from P to given line whose endpoint/foot is Q point.

\begin{aligned} &* \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}=\gamma(\text { let }) \\ &* \quad \mathrm{x}=\gamma-5, y=4 \gamma-3, z=-9 \gamma+6 \\ \end{aligned}

:: Co-ordinates of  \begin{aligned} &\mathrm{Q}(\gamma-5,4 \gamma-3,-9 \gamma+6) \\ \end{aligned}

Hence,

Direction of PQ is

\begin{aligned} &=(\gamma-5-2),(4 \gamma-3-4),(-9 \gamma+6+1) \\ &=(\gamma+7),(4 \gamma-7),(-9 \gamma+7) \end{aligned}

And by comparing with given line equation, direction ratios of the given line are

( Hint – denominator terms of line equation).

= (1,4,9)

PQ  is perpendicular at the given line. Therefore, By “ Conditions of Perpendicularity”.

                                                                a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

\begin{aligned} &=1(\gamma-7)+(4)(4 \gamma-7)-9(-9 \gamma+7)=0 \\ &=\gamma-7+16 \gamma-28+81 \gamma-63=0 \\ &=98 \gamma-98=0 \\ &=\gamma=1 \\ \end{aligned}

:: Co-ordinates of Q.,

By putting the value of γ  in Q co-ordinate equation, we get

\begin{aligned} &=Q\{(1)-5,4(1)-3,-9(1)+6\} \\ &=Q \gamma\{-4,1,-3\} \\ \end{aligned}

Now,

So, equation of perpendicular PQ is

\begin{aligned} &=\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}} \\ &=\frac{x-2}{-4-2}=\frac{y-4}{1-4}=\frac{z+1}{-3+1} \\ &=\frac{x-2}{-6}=\frac{y-4}{-3}=\frac{z+1}{-2} . \end{aligned}

Posted by

infoexpert21

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