24th_November_2008

AIEEE 2015 Coaching › Question Of the Day questions › 24th_November_2008

This one is a one liner..
A very objective paper question.

Pls use the Hide link.

Many of u might feel this not good enuf for a question of the day.. But then all ppl are not smart enuf either :)

find the smallest distance between teh points on (x-a)² + (y-b)²+(z-c)²=100

and the points on x²+y²+z² = 25

#Miscellaneous #Question Of the Day

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7 Answers

1

Rohan Ghosh ·2008-11-23 22:59:49

ans=
(âˆš(a²+b²+c²))-15

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62

Lokesh Verma ·2008-11-23 23:05:12

No rohan.. u are short of the full answer..

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Rohan Ghosh ·2008-11-23 23:08:38

i cant understand how

will i include the case when âˆša²+b²+c²<15
then it will be 0 of course
except the case when 10>5+âˆša²+b²+c²
then it will be
5- (âˆša²+b²+c²)

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62

Lokesh Verma ·2008-11-23 23:25:11

There is still some more to go rohan :)

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Rohan Ghosh ·2008-11-23 23:44:46

all possible cases =

let âˆša²+b²+c²=s in all cases

1 - s>=0 s<=5 then it will be 5-s
2 - s>=5 s<=15 then it will be 0
3 - s>=15 then it will be s-15

i have analyzed all possible values of s i cant understand what i am missing

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62

Lokesh Verma ·2008-11-23 23:47:11

oops sorry rohan... u are correct.. i overlooked the last case in ur previous answer :)

good work :)

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62

Lokesh Verma ·2008-11-23 23:52:45

haha.. rohan u edited ur post later :))

so i din relook at it.. u could have told me u did!

neways.. good work as always :)

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Your Answer

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