If the position vectors of the points A and B | vedclass

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(B) Given position vectors are $\vec{OA} = 2 \hat{i} + 3 \hat{j} - \hat{k}$ and $\vec{OB} = \hat{i} - \hat{j} + 2 \hat{k}$.First,we find the vector $\

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$\frac{1}{\sqrt{26}}(-\hat{i}-4 \hat{j}+3 \hat{k})$

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(B) Given position vectors are $\vec{OA} = 2 \hat{i} + 3 \hat{j} - \hat{k}$ and $\vec{OB} = \hat{i} - \hat{j} + 2 \hat{k}$.First,we find the vector $\overrightarrow{AB} = \vec{OB} - \vec{OA} = (1-2)\hat{i} + (-1-3)\hat{j} + (2-(-1))\hat{k} = -\hat{i} - 4 \hat{j} + 3 \hat{k}$.Next,we find the vector $\overrightarrow{BA} = \vec{OA} - \vec{OB} = (2-1)\hat{i} + (3-(-1))\hat{j} + (-1-2)\hat{k} = \hat{i} + 4 \hat{j} - 3 \hat{k}$.The magnitude of $\overrightarrow{BA}$ is $|\overrightarrow{BA}| = \sqrt{1^2 + 4^2 + (-3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26}$.The unit vector along $\overrightarrow{BA}$ in the direction of $\overrightarrow{AB}$ is defined as $\frac{\overrightarrow{AB}}{|\overrightarrow{BA}|}$.Therefore,the required unit vector is $\frac{-\hat{i} - 4 \hat{j} + 3 \hat{k}}{\sqrt{26}}$.

If the position vectors of the points $A$ and $B$ are $2 \hat{i}+3 \hat{j}-\hat{k}$ and $\hat{i}-\hat{j}+2 \hat{k}$ respectively,then the unit vector along $\overrightarrow{BA}$ and in the direction of $\overrightarrow{AB}$ is

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