If the position vectors of the vertices A, B | vedclass

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(B) The position vectors are $\vec{A} = \hat{i} + 2\hat{j} - 5\hat{k}$,$\vec{B} = -2\hat{i} + 2\hat{j} + \hat{k}$,and $\vec{C} = 2\hat{i} + \hat{j} -

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$\cos^{-1}\left(\frac{8}{\sqrt{105}}\right)$

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(B) The position vectors are $\vec{A} = \hat{i} + 2\hat{j} - 5\hat{k}$,$\vec{B} = -2\hat{i} + 2\hat{j} + \hat{k}$,and $\vec{C} = 2\hat{i} + \hat{j} - \hat{k}$.We need to find $\angle B$,which is the angle between vectors $\vec{BA}$ and $\vec{BC}$.$\vec{BA} = \vec{A} - \vec{B} = (1 - (-2))\hat{i} + (2 - 2)\hat{j} + (-5 - 1)\hat{k} = 3\hat{i} - 6\hat{k}$.$\vec{BC} = \vec{C} - \vec{B} = (2 - (-2))\hat{i} + (1 - 2)\hat{j} + (-1 - 1)\hat{k} = 4\hat{i} - \hat{j} - 2\hat{k}$.The cosine of the angle $\angle B$ is given by $\cos(\angle B) = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}| |\vec{BC}|}$.$\vec{BA} \cdot \vec{BC} = (3)(4) + (0)(-1) + (-6)(-2) = 12 + 0 + 12 = 24$.$|\vec{BA}| = \sqrt{3^2 + 0^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$.$|\vec{BC}| = \sqrt{4^2 + (-1)^2 + (-2)^2} = \sqrt{16 + 1 + 4} = \sqrt{21}$.$\cos(\angle B) = \frac{24}{3\sqrt{5} 	imes \sqrt{21}} = \frac{8}{\sqrt{5} 	imes \sqrt{21}} = \frac{8}{\sqrt{105}}$.Therefore,$\angle B = \cos^{-1}\left(\frac{8}{\sqrt{105}}ight)$.

If the position vectors of the vertices $A, B$ and $C$ of $\triangle ABC$ are $\hat{i}+2\hat{j}-5\hat{k}$,$-2\hat{i}+2\hat{j}+\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$ respectively,then $\angle B=$

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