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KVPY Sample Paper KVPY Stream-SX Model Paper-28

  • question_answer
    A point I is the centre of a inscribed in a triangle ABC, then the vector sum \[\overrightarrow{\left| \operatorname{BC} \right|}\,\,\overrightarrow{\operatorname{IA}}+\overrightarrow{\left| \operatorname{CA} \right|}\,\,\overrightarrow{\operatorname{IB}}+\overrightarrow{\left| \operatorname{AB} \right|}\,\,\overrightarrow{\operatorname{IC}}\]is

    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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