If the position vectors of A, B and C are res | vedclass

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(C) Let the position vectors be $\vec{a} = \overrightarrow{OA} = 2\hat{i}-\hat{j}+\hat{k}$, $\vec{b} = \overrightarrow{OB} = \hat{i}-3\hat{j}-5\hat{k}

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$\frac{35}{41}$

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(C) Let the position vectors be $\vec{a} = \overrightarrow{OA} = 2\hat{i}-\hat{j}+\hat{k}$, $\vec{b} = \overrightarrow{OB} = \hat{i}-3\hat{j}-5\hat{k}$, and $\vec{c} = \overrightarrow{OC} = 3\hat{i}-4\hat{j}-4\hat{k}$.To find $\cos A$, we consider the triangle $ABC$. The vectors forming the sides are $\vec{AB} = \vec{b} - \vec{a} = (1-2)\hat{i} + (-3+1)\hat{j} + (-5-1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k}$ and $\vec{AC} = \vec{c} - \vec{a} = (3-2)\hat{i} + (-4+1)\hat{j} + (-4-1)\hat{k} = \hat{i} - 3\hat{j} - 5\hat{k}$.The angle $A$ is the angle between vectors $\vec{AB}$ and $\vec{AC}$.$\cos A = \frac{\vec{AB} \cdot \vec{AC}}{|\vec{AB}| |\vec{AC}|}$.$\vec{AB} \cdot \vec{AC} = (-1)(1) + (-2)(-3) + (-6)(-5) = -1 + 6 + 30 = 35$.$|\vec{AB}| = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}$.$|\vec{AC}| = \sqrt{1^2 + (-3)^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$.$\cos A = \frac{35}{\sqrt{41} \sqrt{35}} = \sqrt{\frac{35}{41}}$.Therefore, $\cos^2 A = \frac{35}{41}$.

If the position vectors of $A, B$ and $C$ are respectively $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $3 \hat{i}-4 \hat{j}-4 \hat{k}$, then $\cos ^2 A$ is equal to

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