A) Zero
B) \[\frac{\overrightarrow{\operatorname{IA}}+\overrightarrow{\operatorname{IB}}+\overrightarrow{\operatorname{IC}}}{3}\]
C) \[\overrightarrow{\operatorname{IA}}+\overrightarrow{\operatorname{IB}}+\overrightarrow{\operatorname{IC}}\]
D) none of these
Correct Answer: A
Solution :
If \[\overrightarrow{\left| BC \right|} =a;\overrightarrow{\left| CA \right|} = b;\overrightarrow{\left| AB \right|}=c then~\] |
\[\frac{a\overrightarrow{IA}+b\overrightarrow{IB}+c\overrightarrow{IC}}{a+b+c}\]is the position vector of I with respect to I & this is equal to zero. |
[\[\because \]. p.v. of incentre of a triangle is\[\frac{\operatorname{a}\vec{\alpha }+\operatorname{b}\vec{\beta }+\operatorname{c}\vec{\gamma }}{\operatorname{a}+b+c}\] |
Where \[\vec{\alpha },\vec{\beta }\operatorname{and}\vec{\gamma }\]are p.v. of vertices A, B and C of a triangle ABC respectively]. |
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