and thanx anurag !!
a great great great figure :)
that solves 75% of it :)
thanx :)
consider a triangle AOB in the x-y plane where A(1,0,0) , B(0,2,0) , O(0,0,0). the new position of O , when triangl is rotated about side AB by 90° can be ...
a) 4/5,3/5,2/√5
b)-3/5,√2/5,2/√5
c) 4/5,2/5,2/√5
d) 4/5,2/5,1/√5.
[got the ans C by an unconventional method (paper folding ..)] .. but plz someone solve 'nicely'..
Things that i think need to be kept in mind for this one..
1) the distances OA and OB are fixed. So the distance should be a constant.
OA=1 and OB=2
(1+9+20)/25
(64+2+20)/25
(1+4+20)/25
(1+4+5)/25
so what I am getting is C
sky could you confirm that D is the correct answer and not A, B or C?
got a doubt.
well won't A and b remain where they are ? triangle is rotated abt AB no ?
then we've got to find the prependicular dist of O from AB that's all , else . try with options . dist formula . that's what flashing nothign else. :P
yup srinath you are right.
but what if the point is shifted parallelly
I think even in finding the perpendicular distance, you will find points which are not the true rotation by 90 degree..
That is why i have taken the distance from the original A.
seems quite correct ..
but bhaiyya A and B will still remain fixed naa...
i m not moving A n B!
But I dont see a mistake in anurag's solution either :(
oh but i see a mistake in mine :P
bhaiya, am sry ... came oline after a long time...
thats no method actually..
its my method to get answers.. :)
it helps many tyms in the xam hall ... there wid a few approximations of .1~.2 we get nearly a correct ans..
bad method... to do maths.. but... :P
and thanx anurag !!
a great great great figure :)
that solves 75% of it :)
thanx :)