P and Q are the points of trisection of the l | vedclass

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(A) Let $\vec{a} = 2 \hat{i}-5 \hat{j}+3 \hat{k}$ and $\vec{b} = 4 \hat{i}+\hat{j}-6 \hat{k}$ be the position vectors of $A$ and $B$. Since $P$ and $Q

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$\frac{1}{15}(44 \hat{i}-33 \hat{j}-18 \hat{k})$

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(A) Let $\vec{a} = 2 \hat{i}-5 \hat{j}+3 \hat{k}$ and $\vec{b} = 4 \hat{i}+\hat{j}-6 \hat{k}$ be the position vectors of $A$ and $B$. Since $P$ and $Q$ trisect $AB$, the position vectors are $\vec{p} = \vec{a} + \frac{1}{3}(\vec{b}-\vec{a}) = \frac{2}{3}\vec{a} + \frac{1}{3}\vec{b}$ and $\vec{q} = \vec{a} + \frac{2}{3}(\vec{b}-\vec{a}) = \frac{1}{3}\vec{a} + \frac{2}{3}\vec{b}$. We need to find the position vector $\vec{r}$ of a point $R$ that divides $PQ$ in the ratio $2:3$. Using the section formula: $\vec{r} = \frac{3\vec{p} + 2\vec{q}}{2+3} = \frac{3(\frac{2}{3}\vec{a} + \frac{1}{3}\vec{b}) + 2(\frac{1}{3}\vec{a} + \frac{2}{3}\vec{b})}{5} = \frac{2\vec{a} + \vec{b} + \frac{2}{3}\vec{a} + \frac{4}{3}\vec{b}}{5} = \frac{\frac{8}{3}\vec{a} + \frac{7}{3}\vec{b}}{5} = \frac{8\vec{a} + 7\vec{b}}{15}$. Substituting the vectors: $\vec{r} = \frac{8(2 \hat{i}-5 \hat{j}+3 \hat{k}) + 7(4 \hat{i}+\hat{j}-6 \hat{k})}{15} = \frac{(16+28)\hat{i} + (-40+7)\hat{j} + (24-42)\hat{k}}{15} = \frac{44\hat{i}-33\hat{j}-18\hat{k}}{15}$.

$P$ and $Q$ are the points of trisection of the line segment $AB$. If $2 \hat{i}-5 \hat{j}+3 \hat{k}$ and $4 \hat{i}+\hat{j}-6 \hat{k}$ are the position vectors of $A$ and $B$ respectively, then the position vector of the point which divides $PQ$ in the ratio $2:3$ is

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