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# The side of a parallelogram are $2 \hat i +4 \hat j -5 \hat k$ and $\hat i +2 \hat j +3 \hat k$, then the unit vector parallel to one of the diagonals is :

$\begin{array}{1 1} \frac{1}{7} (3 \hat i +6 \hat j -2 \hat k) \\\frac{1}{7} (-3 \hat i +6 \hat j -2 \hat k) \\ \frac{1}{7} (-3 \hat i +6 \hat j +2 \hat k) \\ \frac{1}{7} (3 \hat i +6 \hat j +2 \hat k) \end{array}$

Let $\overrightarrow{AB} = 2 \hat{i} + 4 \hat{j} -5 \hat{k}$
$\overrightarrow{BC} = \hat{i} + 2 \hat{j} + 3 \hat{k}$
$\therefore \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$
$\; \; \; \; \;= 3 \hat{i} + 6 \hat{j} - 2 \hat{k}$
$|\overrightarrow{AC} | = \sqrt{3^2 + 6^2+(-2)^2} = 7$
$\therefore$ unit vector parallel to $\overrightarrow{AC}$ is $\frac{1}{7} (3 \hat{i} + 6 \hat{j} -2 \hat{k})$