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Question

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

A
a×b+b×c+c×a|ab|
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B
a×b+b×c+c×a|bc|
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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×a|ab|

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|ab|

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