IIT JEE , AIEEE Forum - Mathematics , Vectors

Nithy

just give me hints...

sounds simple but not able to work out...

prove
( 1 + |a| ^2 ) * (1+|b|^2 ) = |1-a.b|^2 + | a+b+(a x b)|^2.

a , b vectors.

it was asked in our pre board too..........
pl.

raulrag009

Here's the soln

$As\\\\ {i\;have\;taken\;cross\;as\;x}\\\\ Ia.bI=IaIIbI\cos{\theta}\\\\ Squaring \;both\;sides\\\\ (Ia.bI)^{2}=IaI^{2}IbI^{2}-(axb)^{2}.........(1)\\\\ LHS\\\\ (1+IaI^{2})(1+IbI)^{2}\\\\ 1+IaI^{2}+IbI^{2}+IaI^{2}IbI^{2}\\\\ 1+IaI^{2}+IbI^{2}+(IaI.IbI)^{2}+(axb)^{2}\\\\ Add \;and\;subtract\;2Ia.bI\\\\ (1-Ia.bI)^{2}+IaI^{2}+IbI^{2}+2Ia.bI+(axb)^{2}\\\\ (1-Ia.bI)^{2}+Ia+b+(axb)I^{2}-2(a+b)(axb)\\ =RHS\\\\ As \;axb\;is \;a\;vector\;perpendicular\;to\;a\;and\;b\\\\ 2(a+b)(axb)=0$

Taara

5 th step???

Nithy

bcoz cos^x = 1 - sin^x na
and multiply inside u get (a x b)^2

& gr8 answer @raul.!!!!!!!

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