If the position vectors of the points $A$ and $B$ are $2 \hat{i}+3 \hat{j}-\hat{k}$ and $\hat{i}-\hat{j}+2…
If the position vectors of the points $A$ and $B$ are $2 \hat{i}+3 \hat{j}-\hat{k}$ and $\hat{i}-\hat{j}+2 \hat{k}$ respectively, then the unit vector along $\overrightarrow{\mathrm{BA}}$ and in the direction of $\overrightarrow{\mathrm{AB}}$ is
$\overrightarrow{\mathrm{AB}}=-\hat{i}-4 \hat{j}+3 \hat{k}$ So, $\overrightarrow{\mathrm{BA}}=\hat{i}-4 \hat{j}+3 \hat{k}$
Unit vector along $\overrightarrow{\mathrm{BA}}$ and in direction $\overrightarrow{\mathrm{AB}}$ is $=\frac{\overrightarrow{\mathrm{AB}}}{|\overrightarrow{\mathrm{BA}}|}$
$=\frac{-\hat{i}-4 \hat{j}+3 \hat{k}}{\sqrt{1+16+9}}=\frac{-\hat{i}-4 \hat{j}+3 \hat{k}}{\sqrt{26}}$