If the points P, Q and R have the position ve | vedclass

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(A) Let the position vectors of points $P, Q$ and $R$ be $\vec{p} = \hat{i}-2 \hat{j}+3 \hat{k}$,$\vec{q} = -2 \hat{i}+3 \hat{j}+2 \hat{k}$ and $\vec{

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collinear and $Q$ lies between $P$ and $R$.

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(A) Let the position vectors of points $P, Q$ and $R$ be $\vec{p} = \hat{i}-2 \hat{j}+3 \hat{k}$,$\vec{q} = -2 \hat{i}+3 \hat{j}+2 \hat{k}$ and $\vec{r} = -8 \hat{i}+13 \hat{j}$.First,we find the vectors $\vec{PQ}$ and $\vec{QR}$:$\vec{PQ} = \vec{q} - \vec{p} = (-2 \hat{i}+3 \hat{j}+2 \hat{k}) - (\hat{i}-2 \hat{j}+3 \hat{k}) = -3 \hat{i}+5 \hat{j}-\hat{k}$.$\vec{QR} = \vec{r} - \vec{q} = (-8 \hat{i}+13 \hat{j}) - (-2 \hat{i}+3 \hat{j}+2 \hat{k}) = -6 \hat{i}+10 \hat{j}-2 \hat{k}$.We observe that $\vec{QR} = 2(-3 \hat{i}+5 \hat{j}-\hat{k}) = 2 \vec{PQ}$.Since $\vec{QR}$ is a scalar multiple of $\vec{PQ}$,the vectors $\vec{PQ}$ and $\vec{QR}$ are parallel.Because they share a common point $Q$,the points $P, Q$ and $R$ are collinear.Since $\vec{QR} = 2 \vec{PQ}$,the point $Q$ lies between $P$ and $R$.

If the points $P, Q$ and $R$ have the position vectors $\hat{i}-2 \hat{j}+3 \hat{k}$,$-2 \hat{i}+3 \hat{j}+2 \hat{k}$ and $-8 \hat{i}+13 \hat{j}$ respectively,then these points are

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