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∣∣∣|→a−→b|
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B
∣∣∣→a×→b+→b×→c+→c×→a∣∣∣|→b−→c|
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C
∣∣∣→a×→b+→b×→c+→c×→a∣∣∣|→a−→c|
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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∣∣∣|→a−→b|
Let ABC is a given triangle having position vector of vertices be →a,→band→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∣∣∣⇒|→b−→a||−−→CP|=∣∣∣→a×→b+→b×→c+→c×→a∣∣∣⇒|−−→CP|=|−−→AC|sinA=∣∣∣→a×→b+→b×→c+→c×→a∣∣∣|→b−→a|⇒|−−→CP|=∣∣∣→a×→b+→b×→c+→c×→a∣∣∣|→a−→b|