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If the vertices $A, B, C$ of a triangle $A B C$ are $(1,2,3)$ $(-1,0,0),(0,1,2)$ respectively, then find $\angle A B C$.
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Let $\mathrm{O}$ be the origin then.
$\overrightarrow{\mathrm{OA}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{OB}}=-\hat{\mathrm{i}}$ and $\overrightarrow{\mathrm{OC}}=\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
$\therefore \quad \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{BA}}=\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OB}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
Now $\cos \angle \mathrm{ABC}=\frac{\overline{\mathrm{BC}} \cdot \overrightarrow{\mathrm{BA}}}{|\overrightarrow{\mathrm{BC}}||\overrightarrow{\mathrm{BA}}|}$
$=\frac{(\hat{i}+\hat{j}+2 \hat{k}) \cdot(2 \hat{i}+2 \hat{j}+3 \hat{k})}{|\hat{i}+\hat{j}+2 \hat{k}||2 \hat{i}+2 \hat{j}+3 \hat{k}|}=\frac{10}{\sqrt{102}}$
$\Rightarrow \quad \angle \mathrm{ABC}=\cos ^{-1}\left(\frac{10}{\sqrt{102}}\right)$
$\overrightarrow{\mathrm{OA}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{OB}}=-\hat{\mathrm{i}}$ and $\overrightarrow{\mathrm{OC}}=\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
$\therefore \quad \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{BA}}=\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OB}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
Now $\cos \angle \mathrm{ABC}=\frac{\overline{\mathrm{BC}} \cdot \overrightarrow{\mathrm{BA}}}{|\overrightarrow{\mathrm{BC}}||\overrightarrow{\mathrm{BA}}|}$
$=\frac{(\hat{i}+\hat{j}+2 \hat{k}) \cdot(2 \hat{i}+2 \hat{j}+3 \hat{k})}{|\hat{i}+\hat{j}+2 \hat{k}||2 \hat{i}+2 \hat{j}+3 \hat{k}|}=\frac{10}{\sqrt{102}}$
$\Rightarrow \quad \angle \mathrm{ABC}=\cos ^{-1}\left(\frac{10}{\sqrt{102}}\right)$
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