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Question

If a,b,c are position vectors of the vertices of a triangle ABC, then length of the perpendicular drawn from C to AB is

A
a×b+b×c+c×aab
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B
a×b+b×c+c×abc
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C
a×b+b×c+c×a|ac|
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D
a×b+b×c+c×a
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Solution

The correct option is A a×b+b×c+c×aab



Solve:-
Let ABC is a given triangle
having position vector of vertices be a ,
b and c respectively
Area of triangle =12(AB)(CP)

=12(AB)(AC)sinA

AC×AB=|a×b+b×c+c×a|

|AB||AC|sinA=|a×b+b×c+c×a|

|ba||CP|=|a×b+b×c+c×a|

|CP|=|AC|sinA=|a×b+b×c+c×a||ba|

|CP|=|a×b+b×c+c×a||a¯b|

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