Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please Solve R.D.Sharma class 12 Chapter 28 The Plane Exercise 28.12 Question 2 Maths textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 28 The Plane Exercise 28.12 Question 2 Maths textbook Solution.

Answers (1)

Answer: Therefore, the required point \left ( 1,-2,7 \right )

Hint: Use formula \frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}

Given:\left ( 3,-4,-5 \right ) & \left ( 2,-3,1 \right ) and plane 2x+y+z=7

Solution: Equation of line through \left ( 3,-4,-5 \right ) &\left ( 2,-3,1 \right ) is 


\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda

any\: point \: on \: the\: line (3-\lambda,-4+\lambda,-5+6 \lambda)

but \: this\: point \: lies \: on\: the\: plane\: 2 x+y+z=7 is     

6-2 \lambda-4+\lambda-5+6 \lambda=7

\Rightarrow 5 \lambda=10 \Rightarrow \lambda=2

\therefore Required \: point \: is (1,-2,7)



 

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

Latest Question