If the vertices A, B, C of a triangle ABC are | vedclass

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(A) The vertices of $	riangle ABC$ are $A(1,2,3), B(-1,0,0),$ and $C(0,1,2)$.$\angle ABC$ is the angle between the vectors $\overrightarrow{BA}$ and

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$\cos^{-1}\left(\frac{10}{\sqrt{102}}\right)$

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(A) The vertices of $	riangle ABC$ are $A(1,2,3), B(-1,0,0),$ and $C(0,1,2)$.$\angle ABC$ is the angle between the vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$.$\overrightarrow{BA} = (1 - (-1))\hat{i} + (2 - 0)\hat{j} + (3 - 0)\hat{k} = 2\hat{i} + 2\hat{j} + 3\hat{k}$.$\overrightarrow{BC} = (0 - (-1))\hat{i} + (1 - 0)\hat{j} + (2 - 0)\hat{k} = \hat{i} + \hat{j} + 2\hat{k}$.$\overrightarrow{BA} \cdot \overrightarrow{BC} = (2)(1) + (2)(1) + (3)(2) = 2 + 2 + 6 = 10$.$|\overrightarrow{BA}| = \sqrt{2^2 + 2^2 + 3^2} = \sqrt{4 + 4 + 9} = \sqrt{17}$.$|\overrightarrow{BC}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.Using the dot product formula: $\cos(\angle ABC) = \frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{|\overrightarrow{BA}| |\overrightarrow{BC}|} = \frac{10}{\sqrt{17} 	imes \sqrt{6}} = \frac{10}{\sqrt{102}}$.Therefore,$\angle ABC = \cos^{-1}\left(\frac{10}{\sqrt{102}}ight)$.

If the vertices $A, B, C$ of a triangle $ABC$ are $(1,2,3), (-1,0,0), (0,1,2)$ respectively,then find $\angle ABC$. $[\angle ABC \text{ is the angle between the vectors } \overrightarrow{BA} \text{ and } \overrightarrow{BC}]$.

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