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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Apr 2008 21:26:04 IST
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A ray of light is incident on a plane mirror , along the direction given by vector 2i - 3j + 4k. Find the unit vector along the reflected ray.Take Normal to the mirror along the direction of vector B = 3i - 6j + 2k
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"Imagination is more important than knowledge."
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21 Apr 2008 16:51:22 IST
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angle of incidence = angle b/w -2i + 3j - 4k & 3i - 6j + 2k = cos -1 (-32/7  29) by dot product... let r be the unit vector along the reflected ray.. so angle of reflection = angle b/w r and 3i - 6j + 2k = cos-1(r.(3i - 6j + 2k) / 7) since angle of incidence = angle of ref -32/ 29 = r.(3i - 6j + 2k) ..ab daal daal ke dekh le options ..answer (a) aa jaayega  .. |
There is no better feeling in this world than being a winner! |
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21 Apr 2008 17:09:21 IST
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abbe correct hai dost :)
but i hate such bad values!! friggin' hell
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---------------------------------------------------------------
* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
* Your friendly neighborhood spideyunlimited |
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21 Apr 2008 17:53:05 IST
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abe yaar joy koi aisa method bata jisse options ki madad lene ki zarurat na pade isliye maine options post nahi kiye kyunki i want general soln assuming it's a subjective question
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"Imagination is more important than knowledge."
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23 Apr 2008 23:17:00 IST
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dude the formula is:
unit vector of reflected ray
= unit vector of incident ray - 2 unit vector of normal( into product of unit vector of incident ray and unit vector of normal)
thats it, put the values and get the anser...itz the shortest method...
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23 Apr 2008 23:47:12 IST
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Did this question come in any AIEEE paper?
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Will nip in at times to solve problems :)
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