If $A(0,4,0), \quad B(0,0,3)$ and $C(0,4,3)$ are the vertices of $\Delta A B C$, then its incentre is

If $A(0,4,0), \quad B(0,0,3)$ and $C(0,4,3)$ are the vertices of $\Delta A B C$, then its incentre is
  1. (2,0,3)
  2. (3,0,2)
  3. (0,3,2)
  4. (0,2,3)

Solution

Let $\begin{array}{ll} \overline{\mathrm{a}}=4 \hat{\mathrm{j}} & \therefore|\overline{\mathrm{a}}|=\mathrm{a}=4 \\ \overline{\mathrm{b}}=3 \hat{\mathrm{k}} & \therefore|\overline{\mathrm{b}}|=\mathrm{b}=3 \\ \overline{\mathrm{c}}=4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} & \therefore|\overline{\mathrm{c}}|=\mathrm{c}=\sqrt{16+9}=5 \end{array}$ Let $H(\bar{h})$ be the incentre $\begin{array}{l} \text { Incentre is given by } \frac{a \bar{a}+b \bar{b}+c \bar{c}}{a+b+c}=\bar{h} \\ \Rightarrow \bar{h}=\frac{4(4 \hat{j})+3(3 \hat{k})+5(4 \hat{j}+3 \hat{k})}{4+3+5} \\ \quad=\frac{16 \hat{j}+9 \hat{k}+20 \hat{j}+15 \hat{k}}{12}=\frac{36 \hat{j}+24 \hat{k}}{12}=3 \hat{j}+2 \hat{k} \end{array}$ Thus co-ordinate of incentre is $(0,3,2)$

Asked in: MHT CET 2020 (14 Oct Shift 2)

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