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Vectors

Cool goIITian

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20 Mar 2008 15:17:32 IST

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todays board ques

None

todays board ques find the distance of the pt -2,3,-4 from the line x+2/3=2y+3/4=3z+4/5 measured parallel to the plane4x+ 12y-3z+1=0

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Scorching goIITian

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20 Mar 2008 15:19:01 IST

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edited

arpan1

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20 Mar 2008 15:31:58 IST

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see this is easy..

eqn of line parallel to given line can b found out....

then find the co-ordinates of general point

that point also satisfies the equation of the plane

from here u get the co-ordinate of that point on the plane

then use distance formula...

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ram kumar

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20 Mar 2008 16:31:41 IST

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answer is 17/2 units.......

dheer Bhatia

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20 Mar 2008 16:35:23 IST

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lambda = 1/13 or 1/9??????? in dat plane ques

set 1

BILLA

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20 Mar 2008 16:39:36 IST

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answer is 17/2!!! a easy one only!!!! whats the answer for that sin-1 question in set2???that 28th question???

pratik Agarwal

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20 Mar 2008 16:45:17 IST

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machine waale sawaal ka ans pooch raha to uska ans hai 6,0

ram kumar

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20 Mar 2008 16:46:27 IST

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answer for that sin inverse question in set 2 is

$\frac{a(\pi-2)}{2}$

BILLA

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20 Mar 2008 17:06:31 IST

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yeah rite even i got the same answer which school do you study in???

ram kumar

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20 Mar 2008 17:10:47 IST

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see your nudgebook....

BILLA

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20 Mar 2008 21:44:02 IST

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hey whats the answer for our graph question in set 2???

sreeraman nagasubramaniyan

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21 Mar 2008 14:12:41 IST

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$[tex] \\ \\ \mbox{Given plane equation is } \ 4x+12y-3z+1 = 0 \\ \\ \mbox{Equation of a parallel plane would be } \ 4x+12y-3z+d = 0 \\ \\ \mbox{Since this plane passes through } \ (-2,3,-4) \ \mbox{We get the equation of the parallel plane as} \\ \\ 4x+12y-3z-40 = 0 \\ \\ \mbox{Given plane equation is } \ \frac{x+2}{3} = \frac{y+\frac{3}{2}}{2} = \frac{z+\frac{4}{3}}{\frac{5}{3}} = \lambda \\ \\ \mbox{So any pt on this line is} \ (3\lambda - 2,2\lambda - \frac{3}{2} , \frac{5}{3}\lambda - \frac{4}{3}) \\ \\ \mbox{Substitute this in plane equation} \ 4(3\lambda - 2) + 12(2\lambda - \frac{3}{2}) - 3(\frac{5}{3}\lambda - \frac{4}{3}) - 40 = 0 \\ \\ \lambda = 2 \\ \\ \mbox{Thus the point where the line meets the plane is } \\ \\ (4,\frac{5}{2},2) \\ \\ \mbox{Now all the remains is to find distance between } \ (-2,3,-4) \ \mbox{and} \ (4,\frac{5}{2},2) \\ \\ \mbox{which is nothing but } \ \frac{17}{2} \mbox{or} \ 8.5 \ \mbox{units}$

puneet

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26 Mar 2008 15:07:23 IST

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hii

well .. sboosy has explained it very well .. i cannot do better .. well done dear ..
nice and clearn answer ..

cheers

p.s. using tex. .. gud habbit .. well done

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