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Cool goIITian

Joined:	26 Mar 2007
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26 Mar 2008 11:44:47 IST

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534

the toughest question of 3d geometry

None

if l,m,n are dc's of a line and -3,4,1 are three real no's find the range of values of -3l+4m+n?

can any one solve this

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varsha valli g.

Blazing goIITian

Joined: 31 Jan 2007

Posts: 515

26 Mar 2008 12:13:35 IST

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l,m,n are d.c's so range of l is between 0 and 1(range of cosx)
same is the range of m and n.
0 <= l <=1
multiply by -3,
0<= -3l <= -3_______________(1)
similarly for m,
0<= 4m <=4---------(2)
for n
0<= n <= 1-----(3)
(1)+(2)+(3),

0<= -3l+4m+n <= 2
so range is [0,2]

correct me if m wrong
!!!!!!!!!!!!!!!!!!!!!

nishant kumar

Cool goIITian

Joined: 26 Mar 2007

Posts: 81

26 Mar 2008 12:43:32 IST

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no i am sorry but the ans is

^[_

26,

26]

tanmoy sanyal

New kid on the Block

Joined: 24 Feb 2008

Posts: 22

6 Apr 2008 20:18:34 IST

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(-3,4,5) are the d.r's of some other line (say) ...so its d.c's are
(-3/p, 4/p, 5/p) where p= sqrt (26) so angle between these 2 lines is cos A = (1/p) (-3l+4m+n) now 0 deg<= A <= 180 deg ........that gives -1<= (1/p) (-3l+4m+n) <= 1.........so the reqd range is [-p,p] i.e [-sqrt 26, sqrt 26]

and u call that toughest ???

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