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Blazing goIITian

Joined:	18 Feb 2007
Post:	1042

20 Apr 2008 21:26:04 IST

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457

Aieee Ques Plz Solve

None

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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joy francis

Blazing goIITian

Joined: 19 Feb 2007

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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

Gaurav |spideyunlimited| Ragtah

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Joined: 16 Dec 2006

Posts: 3373

21 Apr 2008 17:09:21 IST

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abbe correct hai dost :)

but i hate such bad values!! friggin' hell

Priyesh

Blazing goIITian

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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

svj 29

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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...

Karthik M

Blazing goIITian

Joined: 1 May 2007

Posts: 2830

23 Apr 2008 23:47:12 IST

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Did this question come in any AIEEE paper?

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