The perimeter of the triangle whose vertices have the position vectos $\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}, 5 \hat{\imath}+3 \hat{\jmath}-3 \hat{\mathrm{k}}$ and $2 \hat{\imath}+5 \hat{\jmath}+9 \hat{\mathrm{k}}$ is

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Aicharya · Accepted Answer

Let $\quad \overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{b}}=5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}, \overline{\mathrm{c}}=2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+9 \hat{\mathrm{k}}$ $	herefore \overline{\mathrm{AB}}=4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}} \quad \Rightarrow \mid \overline{\mathrm{AB}}=\sqrt{16+4+16}=6$ $\overline{\mathrm{BC}}=-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+12 \hat{\mathrm{k}} \Rightarrow|\overline{\mathrm{BC}}|=\sqrt{9+4+144}=\sqrt{157}$ $\overline{\mathrm{AC}}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+8 \hat{\mathrm{k}} \quad \Rightarrow \mid \overline{\mathrm{AC}}=\sqrt{1+16+64}=9$ Perimeter $=|\overline{\mathrm{AB}}|+|\overline{\mathrm{BC}}|+|\overline{\mathrm{AC}}|=15+\sqrt{157}$

The perimeter of the triangle whose vertices have the position vectos…

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