One vertex of a rectangular parallelepiped is at the origin O and the lengths of its edges along and 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 -axis,

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Solution

Q.
One vertex of a rectangular parallelepiped 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 [2023]

1 $\frac{12}{5}$

2 $\frac{12}{5 \sqrt{5}}$

3 $\frac{12}{\sqrt{5}}$

4 $12 \sqrt{5}$

Ans.
(1)

$Equation of line O P, \frac{x - 0}{3 - 0} = \frac{y - 0}{4 - 0} = \frac{z - 0}{5 - 0}$
i.e., $\frac{x}{3} = \frac{y}{4} = \frac{z}{5}$

Now, equation of edge parallel to $z$ -axis passing through $(3, 0, 5)$ and having direction ratios $< 0, 0, 1 >$ is

$\frac{x - 3}{0} = \frac{y - 0}{0} = \frac{z - 5}{1}$

Here, $a_{1} = (0, 0, 0), a_{2} = (3, 0, 5); b_{1} = (3, 4, 5), b_{2} = (0, 0, 1)$

So, $\vec{a_{2}} - \vec{a_{1}} = 3 \hat{i} + 5 \hat{k}$

$\vec{b_{1}} \times \vec{b_{2}} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 0 & 0 & 1 \end{matrix} |$ $= \hat{i} (4 - 0) - \hat{j} (3 - 0) + \hat{k} (0 - 0) = 4 \hat{i} - 3 \hat{j}$

$∴$ $Required shortest distance =$ $| \frac{(3 \hat{i} + 5 \hat{k}) \cdot (4 \hat{i} - 3 \hat{j})}{| 4 \hat{i} - 3 \hat{j} |} |$

$= \frac{12}{\sqrt{16 + 9}} = \frac{12}{5} units .$

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