One vertex of a rectangular parallelopiped is at the origin O and the lengths of its edges along x, y and z axes are 3, 4 and 5 units respectively. Let P be the vertex (3, 4, 5). Then the shortest distance between the diagonal OP and an edge parallel to z axis, not passing through O or P is

The rectangular parallelepiped has vertices at the origin O(0,0,0) and edges along the axes with lengths 3, 4, 5. The vertex P is at (3,4,5).

The diagonal OP has direction ratios (3,4,5) and passes through O(0,0,0).

Edges parallel to the z-axis not passing through O or P are at vertices (3,0,z), (0,4,z), and (3,4,z). But (3,4,z) passes through P at z=5, so the relevant edges not through O or P are at (3,0,z) and (0,4,z).

For edge at (3,0,z): The line is $$\vec{r_1} = (3,0,0) + s(0,0,1)$$.

The line OP is $$\vec{r_2} = t(3,4,5)$$.

The shortest distance formula between two skew lines is:

$$ d = \frac{|(\vec{a_2}-\vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} $$

Here $$\vec{a_1} = (0,0,0)$$, $$\vec{a_2} = (3,0,0)$$, $$\vec{d_1} = (3,4,5)$$, $$\vec{d_2} = (0,0,1)$$.

$$ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 0 & 0 & 1 \end{vmatrix} = (4)\hat{i} - (3)\hat{j} + (0)\hat{k} = (4, -3, 0) $$

$$|\vec{d_1} \times \vec{d_2}| = \sqrt{16+9} = 5$$

$$(\vec{a_2}-\vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2}) = (3,0,0) \cdot (4,-3,0) = 12$$

$$ d = \frac{|12|}{5} = \frac{12}{5} $$

The shortest distance is $$\dfrac{12}{5}$$.