Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The equation of a plane Passing through the points (2,2,1) and (9,3,6) are perpendicular to the plane

Answers (1)

best_answer

Solution: The equation of a plane passing through $(2,2,1)$ is

$P:$ $a(x-2)+b(y-2)+c(z-1)=0$

It passes through $(9,3,6)$

$herefore$ $a(9-2)+b(3-2)+c(6-2)=7a+b+4c=0$ $...........(1)$

Since , plane $P$ , perpendicular to the plane $2x+6y+6z-1=0$

$Rightarrow$ $a+3b+3c=0$ $.........(2)$

From $(1)$ and $(2)$ we have ,

$a/(3-15)=b/(5-21)=c/(21-1)$ or $a/3=b/4=c/-5$

Hence equation of a plane , $3k(x-2)+4k(y-2)-5k(z-1)=0$

$Rightarrow$ $P_1:3x+4y-5z-9=0$

Posted by

Deependra Verma

View full answer

Similar Questions

Latest Question