If $\mathrm{O}(0,0,0), \mathrm{A}(3,0,0), \mathrm{B}(0,4,0)$ form a triangle then the incentre of triangle…
If $\mathrm{O}(0,0,0), \mathrm{A}(3,0,0), \mathrm{B}(0,4,0)$ form a triangle then the incentre of triangle OAB is
- $(0,1,0)$
- $(0,1,1)$
- $(1,0,1)$
- $(1,1,0)$
Solution
$\vec{O}=0, \vec{a}=3 \hat{i}, \vec{b}=4 \hat{j}$
$\begin{aligned} & \vec{h}=\frac{|\overrightarrow{\mathrm{AB}}| \overrightarrow{\mathrm{O}}+|\overrightarrow{\mathrm{OB}}| \vec{a}+|\overrightarrow{\mathrm{OA}}| \vec{a}}{|\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{OB}}|+|\overrightarrow{\mathrm{OA}}|} \\ & \vec{h}=\frac{12 \hat{i}+12 \hat{j}}{5+4+3}=\hat{i}+\hat{j} \Rightarrow \text { Incentre }=(1,1,0)\end{aligned}$
Asked in: AP EAMCET 2024 (20 May Shift 2)
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