IIT JEE , AIEEE Forum - Mathematics , Vectors

ramkumar_november

$distance\;between\;two \;parallel\;planes\;is$

$\frac{d_2-d_1}{\sqrt{a^2+b^2+c^2}}$

in this problem we have two planes.....

2x-3y+6z-5=0 ...............(1)

6x-9y+18z+20=0

(or) 2x-3y+6z+(20/3) = 0

so we have $d_2=\frac{20}{3}\;\;and \;\;d_1=-5$

and $a=2\;\;\;b=-3\;\;\;c=6$

substituting the values in the formula we get

distance = $\frac{\frac{20}{3}+5}{\sqrt{2^2+3^2+6^2}}$

distance = 5\3

this formula is given in our text book .. so we can use it in board exam.........