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![[Post New]](/templates/default/images/icon_minipost_new.gif) 23 Apr 2008 20:57:45 IST
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area of triangle whose sides are 4i-j+k and 3i+j-k
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23 Apr 2008 21:05:49 IST
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want option ok
1)7root2
2)14root2
3)7root12/6
4)7/root2
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23 Apr 2008 21:19:47 IST
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is the answer 7root2
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23 Apr 2008 21:24:14 IST
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it is option d
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23 Apr 2008 21:25:09 IST
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how to solve tell me please
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23 Apr 2008 21:34:26 IST
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i guess u gotta find the angle between the two vectors.....using dot product by product of magnitude of the vectors....u get cosA....A being the angle b/w em....get sinA and then the area is got by usin the result..|a| |b| (sinA)...this shud b rite but i dunno if u get the rite ans....chk simplification.....
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23 Apr 2008 22:41:34 IST
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Ar = 1/2|(aXb)| aXb = 7j+7k.. |(aXb)| = 7root2 .: Ar = 7root2/2 = 7/root2
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23 Apr 2008 22:41:51 IST
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There is no better feeling in this world than being a winner! |
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24 Apr 2008 13:53:41 IST
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Area of triangle is simply 1/2(a cross b)
the ans ought to be d
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satyabharadwaja |
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