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m putting this question on behalf of "Karthik-abhiram"... pls solve this for him
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The position vectors a,b,c and d {all in vectors} of four points A,B,C and D
on a plane are such that
(a-d).(b-c)=(b-d).(c-a)=0, {all in vectors}
then the pont D is
1]centroid of
ABC
ABC2] ortho centre of
ABC
ABC3]circumcentre of
ABC
ABC4]None of these.
plzz help ,and post a detailed solution plz and i want theoritical proof plzz as it would increase our mind to think better for other problems like these
Comments (7)
2 Apr 2007 12:56:15 IST
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Hey the answer is orthocentre option (2)
check that first of all, in all options u hav triangle ABC.....
so, keep it as fixed....
now, try for the positions of D...
the given equations imply that AD.BC=0 and BD.AC=0 (. means doy product)
so, AD and BC are perpendicular and so are BD and AC.....
for a given triangle ABC, this is possible only with the orthocentre.....
hope m clear...
thank u...
Goutham Harsha
5 Apr 2007 17:52:20 IST
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1 people liked this
Hello karthik_abiram
Most of the problems in the vector algebra are ended with only three major things
1. A line vector A (a) and B (b) are two vectors then their line vector AB is b-a
2. Dot product of two vectors a.b = |a||b| cos t
Since i and j and k are all one unit in length and they are all mutually perpendicular, we have
and i.j = j.i = i.k = k.i = j.k = k.j = 0.
3. Cross Product
If we let the angle between A and B be
, then the cross product of A and B can be expressed as
, then the cross product of A and B can be expressed as - A x B = A B sin(
)- Same way sin90=1 so you can calculate the different unit vaectors.



, for this and also for answering all the vectors.....









3]circumcentre of ABC.
Is it right?
If yes, I'll give the detailed solution.
DO RATE & REPLY....