158haa....i got that the sides are (a-b),(b-c) and (c-a)....but i couldn't get how ....area=(a-b)x(b-a)2......
Area of triangle ABC if position vector of A is a position vector of B is b and position vector of C is c
Area=(axb+bxc+cxa)2
Prove the above relation.
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6 Answers
1357sides of triangles are:
(a-b), (b-c), (c-a)
area=(1/2)*(a-b)X(a-c)
:)
- Indayan Bera M sry if Im wrong but, that area = (1/2) * (a-b) * (b-c) is only applicable when (a-b) is perpendicular to (a-c) because then u would b applying 1/2 *b*h ...where b perp. to h......
Apply the general heron's formula....where area = √((s1+s2+s3)/2)) .Then area = 0 O.O...its strange......u have to directly use the position vectors somehow...Upvote·0· Reply ·2012-06-15 22:35:30
- Anirban Mondal I THINK WHAT INDAYAN SAYS IS RIGHT.PLEASE SHOW ME ANOTHER PROCESS
49@Anik: To the last question I would ask: What is a position vector?
158
1(a-b) X (b-a) [cross product of two vectors has the magnitude of the area of the parallelogram considering (a-b) & (b-a) as two sides of it]...
area of the triangle is half of the area of the parallelogram!
hence the ans is right..