One vertex of a rectangular parallelopiped is at the origin O and the lengths of its edges…

Question

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 :

Accepted Answer

Equation of $OP$ is

$\begin{gathered} \frac{x-0}{3-0}=\frac{y-0}{4-0}=\frac{z-0}{5-0} \ \Rightarrow \frac{x}{3}=\frac{y}{4}=\frac{z}{5} \end{gathered}$

Equation of edge parallel to $Z$-axis is

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

$	herefore $ Shortest distance

$=\frac{\begin{vmatrix} {x}_{2}-{x}_{1} \quad  {y}_{2}-{y}_{1} \quad  {z}_{2}-{z}_{1} \ {a}_{1} \quad  {b}_{1} \quad  {c}_{1} \ {a}_{2} \quad  {b}_{2} \quad  {c}_{2} \end{vmatrix}}\sqrt{\begin{gathered} {({a}_{1}{b}_{2}-{a}_{2}{b}_{1})}^{2}+{({b}_{1}{c}_{2}-{b}_{2}{c}_{1})}^{2}+{({c}_{1}{a}_{2}-{c}_{2}{a}_{1})}^{2} \end{gathered}}$

Here,

$\begin{gathered} {x}_{1}=0,{y}_{1}=0,{z}_{1}=0 \ {x}_{2}=3,{y}_{2}=0,{z}_{2}=5 \ {a}_{1}=3,{b}_{1}=4,{c}_{1}=5 \ {a}_{2}=0,{b}_{2}=0,{c}_{2}=1 \end{gathered}$

$	herefore \begin{vmatrix} {x}_{2}-{x}_{1} \quad  {y}_{2}-{y}_{1} \quad  {z}_{2}-{z}_{1} \ {a}_{1} \quad  {b}_{1} \quad  {c}_{1} \ {a}_{2} \quad  {b}_{2} \quad  {c}_{2} \end{vmatrix}=\begin{vmatrix} 3 \quad  0 \quad  5 \ 3 \quad  4 \quad  5 \ 0 \quad  0 \quad  1 \end{vmatrix}=4(3)=12$

$\begin{gathered} \sqrt{{({a}_{1}{b}_{2}-{a}_{2}{b}_{1})}^{2}+{({b}_{1}{c}_{2}-{b}_{2}{c}_{1})}^{2}+{({c}_{1}{a}_{2}-{c}_{2}{a}_{1})}^{2}} \ =\sqrt{0+(4-0{)}^{2}+(0-3{)}^{2}} \ =\sqrt{16+9}=5 \ 	herefore =\frac{12}{5} \end{gathered}$

Solutions · 1