Vector Algebra - Section formula

Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$ respectively, in the ratio $2:1$ internally.

$\vec{r} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) + 1(\hat{i} + 2\hat{j} - \hat{k})}{2+1}$ $\vec{r} = \frac{-2\hat{i} + 2\hat{j} + 2\hat{k} + \hat{i} + 2\hat{j} - \hat{k}}{3}$ $\vec{r} = \frac{-\hat{i} + 4\hat{j} + \hat{k}}{3} = -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k}$

Substitute $m=2$, $n=1$, $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$, and $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ into the internal section formula $\vec{r} = \frac{m\vec{b} + n\vec{a}}{m+n}$ and simplify the components.

Find the position vector of the midpoint of the vector joining the points $P(2, 3, 4)$ and $Q(4, 1, -2)$.

Position vector of $P$ is $\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$. Position vector of $Q$ is $\vec{b} = 4\hat{i} + \hat{j} - 2\hat{k}$. $\vec{r} = \frac{(2\hat{i} + 3\hat{j} + 4\hat{k}) + (4\hat{i} + \hat{j} - 2\hat{k})}{2}$ $\vec{r} = \frac{6\hat{i} + 4\hat{j} + 2\hat{k}}{2} = 3\hat{i} + 2\hat{j} + \hat{k}$

To find the midpoint, add the position vectors of the two endpoints and divide the resulting vector by $2$.