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Scorching goIITian

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18 Mar 2008 16:21:51 IST

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Distance between two parallel planes

None

Find the distance between the parallel planes r.(2i-3j+6k)=5 and r.(6i-9j+18k)+20=0

Ans. 5/3

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LAMPARD

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18 Mar 2008 16:26:33 IST

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(1,3,2) is a pt. on the 1st plane...to find dist. bet planes,we hav to find perp. dist of the pt. frm 2nd plane...calc. the perp. dist. of the pt. from the plane which comes out to be 5/3...

Akhil

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18 Mar 2008 16:29:35 IST

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@Lampard....sorry am very weak in 3d...
how did u get (1,3,2)?

LAMPARD

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18 Mar 2008 16:32:08 IST

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see,the eqn of the plane is 2x-3y+6z=5...so we jus need to find a pt. dat lies on the plane so that we can calc. its distance frm other plane...by trial n subst. method,i got (1,3,2)...

ANUPAM DEV

Scorching goIITian

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18 Mar 2008 16:34:46 IST

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@ Lampard
how u r going to write it in CBSE examination then?

LAMPARD

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18 Mar 2008 16:36:44 IST

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m not in cbse...in isc...so,i dunno bout cbse system...

ANUPAM DEV

Scorching goIITian

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18 Mar 2008 16:40:37 IST

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then how r u going to write in isc

LAMPARD

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18 Mar 2008 16:43:44 IST

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isc boards are over...n even if this ques was askd,i wud hav written it in de same way...sumthin is better dan nothin...n y wont de examiner giv ne marks...the method is a valid one...

ram kumar

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18 Mar 2008 17:32:20 IST

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

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

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.........

arpan1

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18 Mar 2008 17:49:09 IST

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or calculating dist. between 2 parallel planes u can use the following approach

find the distance between origin and the 2 planes..

add them

u get the answer...

i guess this is more easier to remember than the formula....although both are same..

pls rate me

joy francis

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18 Mar 2008 19:17:16 IST

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Eqn of the first plane can be written as

r.((2i-3j+6k) / 7) = 5/7...here 5/7 is the perpendicular distance of the plane from the orgin, since 2i-3j+6k/7 is a unit vector.

Similarly the second plane can be expressed as

r.((6i-9j+18k) / 21) = -20/21...

Hence the distance between the planes is

(5/7) - (-20/21)

= 5/3.......(ans)

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