If 2 hat{i}-hat{j}+hat{k} and hat{i}-3 hat{j} | vedclass

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(A) Let $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - 3\hat{j} - 5\hat{k}$ be the position vectors of $A$ and $B$. Since $C$ divid

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$5 \hat{i}-5 \hat{j}-3 \hat{k}$

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(A) Let $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - 3\hat{j} - 5\hat{k}$ be the position vectors of $A$ and $B$. Since $C$ divides $AB$ in the ratio $2:3$,the position vector of $C$ is given by $\vec{c} = \frac{2\vec{b} + 3\vec{a}}{2+3} = \frac{2(\hat{i} - 3\hat{j} - 5\hat{k}) + 3(2\hat{i} - \hat{j} + \hat{k})}{5} = \frac{(2+6)\hat{i} + (-6-3)\hat{j} + (-10+3)\hat{k}}{5} = \frac{8\hat{i} - 9\hat{j} - 7\hat{k}}{5}$. Thus,$5\vec{c} = 8\hat{i} - 9\hat{j} - 7\hat{k}$. Since $M$ is the mid-point of $AB$,the position vector of $M$ is $\vec{m} = \frac{\vec{a} + \vec{b}}{2} = \frac{(2+1)\hat{i} + (-1-3)\hat{j} + (1-5)\hat{k}}{2} = \frac{3\hat{i} - 4\hat{j} - 4\hat{k}}{2}$. Thus,$2\vec{m} = 3\hat{i} - 4\hat{j} - 4\hat{k}$. Finally,$5\vec{c} - 2\vec{m} = (8\hat{i} - 9\hat{j} - 7\hat{k}) - (3\hat{i} - 4\hat{j} - 4\hat{k}) = (8-3)\hat{i} + (-9+4)\hat{j} + (-7+4)\hat{k} = 5\hat{i} - 5\hat{j} - 3\hat{k}$.

If $2 \hat{i}-\hat{j}+\hat{k}$ and $\hat{i}-3 \hat{j}-5 \hat{k}$ are the position vectors of the points $A$ and $B$ respectively,$C$ divides $AB$ in the ratio $2:3$ and $M$ is the mid-point of $AB$,then $5(\text{position vector of } C) - 2(\text{position vector of } M) =$

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