Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let the plane contain the line <img alt="2 x+y-z-3=0=5 x-3 y+4 z+9" src="https://entrancecorner.oncodecogs.com/gif.latex?2%20

Let the plane P contain the line 2 x+y-z-3=0=5 x-3 y+4 z+9and be parallel to the line  \frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5} Then the distance of the point A(8,-1,-19) from the plane P measured parallel to the line \frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12} is equal to

Option: 1

26


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \text { Plane }=P_1=\lambda \cdot P_2=0 \\ & (2 x+y-z-3)+\lambda(5 x-3 y)+4 z+9)=0 \\ & (5 \lambda+2) x+(1-3 \lambda) y+(4 \lambda-1) z+9 \lambda-3=0 \\ & \overrightarrow{\mathrm{n}} \cdot \vec{b}=0 \text { where } \vec{b}(2,4,5) \\ & 2(5 \lambda+2)+4(1-3 \lambda)+5(4 \lambda-1)=0 \\ & \lambda=-\frac{1}{6} \\ & \text { Plane } 7 x+9 y-10 z-27=0 \end{aligned}

\text{ Fquation of line AB is }\\ \begin{aligned} & \frac{x-8}{-3}=\frac{y+1}{4}=\frac{x+19}{12}=\lambda \\ & \text { Let } B=(8-3 \lambda,-1+4 \lambda,-19+12 \lambda) \text { lies on plane } P \\ & \therefore 7(8-3 \lambda)+9(4 \lambda-1)-10(12 \lambda-19)=27 \\ & \lambda=2 \\ & \therefore \text { Point } B=(2,7,5) \\ & A B=\sqrt{6^2+8^2+24^2}=26 \end{aligned}

Posted by

avinash.dongre

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads
anam-khan

Bihar Board extends BSEB Super 50 free coaching application last date to March 26
anam-khan

JEE Main City Intimation Slip 2025 Live: NTA session 2 exam city slips soon at jeemain.nta.nic.in

Latest Question