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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Jan 2008 23:07:54 IST
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1). if vector a, vector b ,vector c are the position vectors of the non- collinear points A,B,C resp. then show that a*b+b*c+c*a is perpendicular to the plane of ABC.(* is actually cross).,where a,b,c are vectors. 2). Let a ,b,c be 3 non zero vectors. if a.(b*c)=0 and b,c are not parallel then prove that a=  b+kc where  and k are some scalars. 3). if a,b,c represent the vectors BC,CA and AB of a triangle ABC then show that a*b=b*c=c*a. (* is actually cross product in vectors in all 3 sums)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Jan 2008 23:19:03 IST
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ans 4 3rd 1 a=OC-OB , b=OA-OC , c=OB-OA hence by cross product v wud get OA*OB+OB*OC+OC*OA
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Jan 2008 23:20:11 IST
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1) as a,b,c lie in same plane so a*b is perp. to the plane ABC b*c is perp. to the plane ABC, similarly for c*a...
as a*b, b*c, c*a all lie in plane perpendicular to tht of ABC, so a*b +b*c+c*a lies in plane perpendicular to tht of ABC
aliter
let a*b+b*c+c*a lie in any plane perp to ABC then its dot product with any of a,b,c shud be zero on taking dot product with a we get [aba] + [bca] + [caa] [aba]=[caa]=0 and since b,c,a are coplanar so [abc] =0 thus dot product is zero so the vector lies in plane perp to ABC
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The heights which great men reached and kept, Were not attained by sudden flight, They, whilst their companions slept, Were toiling upwards in the night.... |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Jan 2008 23:22:49 IST
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hey jv604, see i have a sol. to yr 3rd q. it is as follows :
see vector a =BC vctor, then vector b = AC vector and lastly vctor c=AB vctor.
now since these three vectors a,b n c for a , it is very obvious acc, to law of vector addition, that a+b+c=0.
Now thats a imp. relation.
now take cross product of a on both the sides,
a x (a+ b + c) = a x 0 a x b + a x c =0 axb=cxa.----(1)
now take cross product on both sides with b
b x (a+b+c)=b x 0 bxa+bxc=0 bxc=axb---(2)
hnce by 1 & 2, axb=bxc=cxa
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Jan 2008 23:22:58 IST
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4 2nd quesn lets cum from end its a=lambda b +kc hence
[ ( lb+kc)*b] . c
= k(c*b) . c =0 hence proved
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Jan 2008 23:23:57 IST
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2) a.(b*c)=[abc] =0
thus abc are coplanar vectors
and if three vectors lie in a plane then any vector can be expressed as a linear combination of other 2 vectors
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The heights which great men reached and kept, Were not attained by sudden flight, They, whilst their companions slept, Were toiling upwards in the night.... |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Jan 2008 23:27:51 IST
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hii
well u can ask only one question ... neways .. most of the answers are already there .. lemme tell something more about the 2nd question ... the vector (b*c) is not in the plane of b and c .. infact perpendicular to that plain ...
now, since the dot product of a with (b*c) is zero .. a must be in plain of b and c
and hence the answer
cheers
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Puneet Agrawal
IIT Delhi
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