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

Option: 1

\frac{12}{5 \sqrt{5}}


Option: 2

12 \sqrt{5}


Option: 3

\frac{12}{5}


Option: 4

\frac{12}{\sqrt{5}}


Answers (1)

best_answer

Equation of OP is \frac{x}{3}=\frac{y}{4}=\frac{z}{5}

\mathrm{a}_{1}=(0,0,0) \quad \mathrm{a}_{2}=(3,0,5)
\mathrm{b}_{1}=(3,4,5) \quad \mathrm{b}_{2}=(0,0,1)

Equation of edge parallel to \mathrm{z} axis

\frac{\mathrm{x}-3}{0}=\frac{\mathrm{y}-0}{0}=\frac{\mathrm{z}-5}{1}
S.D =\frac{\left(\overrightarrow{\mathrm{a}}_{2} \cdot \overrightarrow{\mathrm{a}}_{1}\right) \cdot\left(\overrightarrow{\mathrm{b}}_{1} \times \overrightarrow{\mathrm{b}}_{2}\right)}{\left|\overrightarrow{\mathrm{b}}_{1} \times \overrightarrow{\mathrm{b}}_{2}\right|}
\frac{\begin{vmatrix} 3& 0 &5 \\ 3&4 &5 \\ 0& 0& 1 \end{vmatrix}}{\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 3& 4 &5 \\ 0& 0& 1 \end{vmatrix}}= \frac{3\left ( 4 \right )}{\left | 4\hat{i}-3\hat{j} \right |}= \frac{12}{5}
 

Posted by

manish painkra

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