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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]4 Marks
Question
Using vector method, find the incenter of the triangle whose vertices are $A(0, 3, 0), B(0, 0, 4)$ and $C(0, 3, 4)$
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Answer

Let $\overline{ a }, \overline{ b }, \overline{ c }$ be the position vectors of points $A , B , C$ respectively of $\triangle ABC$ and $\overline{ h }$ be the position vector of its incentre $H$.
$ \therefore \overline{ h }=\frac{|\overline{ BC }| \overline{ a }+|\overline{ AC }| \overline{ b }+|\overline{ AB }| \overline{ c }}{|\overline{ BC }|+|\overline{ AC }|+|\overline{ AB }|}\ldots(i)$
$\therefore \overline{ a }=3 \hat{ j }, \overline{ b }=4 \widehat{ k }, \overline{ c }=3 \hat{ j }+4 \widehat{ k }$
$\therefore \overline{ BC }=\overline{ c }-\overline{ b }=(3 \hat{ j }+4 \widehat{ k })-4 \widehat{ k }=3 \hat{ j }$
$\overline{ AC }=\overline{ c }-\overline{ a }=(3 \hat{ j }+4 \widehat{ k })-3 \hat{ j }=4 \widehat{ k }$
$\overline{ AB }=\overline{ b }-\overline{ a }=4 \widehat{ k }-3 \hat{ j }$
$\therefore|\overline{ BC }|=\sqrt{9}=3$
$|\overline{ AC }|=\sqrt{16}=4 $
and
$ |\overline{ AB }|=\sqrt{16+9}=\sqrt{25}=5$
$\therefore \overline{ h }=\frac{3(3 \hat{ j })+4(4 \widehat{ k })+5(3 \hat{ j }+4 \widehat{ k })}{3+4+5}\ldots[From(i)]$
$\therefore \overline{ h }=\frac{9 \hat{ j }+16 \widehat{ k }+15 \hat{ j }+20 \widehat{ k }}{12}$
$=\frac{24 \hat{ j }+36 \widehat{ k }}{12}$
$=2 \hat{ j }+3 \widehat{ k } $
$\therefore$ Incentre of the triangle is $H (0,2,3)$.

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