The position vectors of A and B are (hat{i}+h | vedclass

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(D) Let the position vectors be $\vec{OA} = \hat{i}+\hat{j}+\hat{k}$ and $\vec{OB} = \frac{1}{3} \hat{j}+\frac{1}{3} \hat{k}$.Given that $B$ divides $

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$(\frac{-1}{2}, 0, 0)$

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(D) Let the position vectors be $\vec{OA} = \hat{i}+\hat{j}+\hat{k}$ and $\vec{OB} = \frac{1}{3} \hat{j}+\frac{1}{3} \hat{k}$.Given that $B$ divides $AC$ in the ratio $m:n = 2:1$.Using the section formula for internal division,the position vector of $B$ is given by:$\vec{OB} = \frac{m\vec{OC} + n\vec{OA}}{m+n}$Substituting the known values:$\frac{1}{3} \hat{j} + \frac{1}{3} \hat{k} = \frac{2\vec{OC} + 1(\hat{i}+\hat{j}+\hat{k})}{2+1}$$3(\frac{1}{3} \hat{j} + \frac{1}{3} \hat{k}) = 2\vec{OC} + \hat{i} + \hat{j} + \hat{k}$$\hat{j} + \hat{k} = 2\vec{OC} + \hat{i} + \hat{j} + \hat{k}$$2\vec{OC} = \hat{j} + \hat{k} - \hat{i} - \hat{j} - \hat{k}$$2\vec{OC} = -\hat{i}$$\vec{OC} = -\frac{1}{2} \hat{i} + 0 \hat{j} + 0 \hat{k}$Therefore,the position vector of $C$ is $(-\frac{1}{2}, 0, 0)$.

The position vectors of $A$ and $B$ are $(\hat{i}+\hat{j}+\hat{k})$ and $(\frac{1}{3} \hat{j}+\frac{1}{3} \hat{k})$. If $B$ divides the line segment $AC$ in the ratio $2:1$,then the position vector of $C$ is

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