data-orsrc="https://uploads-pracbee-ramanuj.s3.ap-south-1.amazonaws.com/pyq-question-images/batch-001/19ef8d670191.webp" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0px auto;max-height:40vh;vertical-align:baseline" alt="JEE Main 2023 (Online) 6th April Morning Shift Mathematics - 3D Geometry Question 110 English Explanation">

Equation of $O P$ is

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

Equation of edge parallel to $Z$-axis is

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

$\therefore$ Shortest distance

$$ =\frac{\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right|}{\sqrt{\begin{array}{r} \left(a_1 b_2-a_2 b_1\right)^2+\left(b_1 c_2-b_2 c_1\right)^2 +\left(c_1 a_2-c_2 a_1\right)^2 \end{array}}} $$

Here,

$$ \begin{aligned} & x_1=0, y_1=0, z_1=0 \\\\ & x_2=3, y_2=0, z_2=5 \\\\ & a_1=3, b_1=4, c_1=5 \\\\ & a_2=0, b_2=0, c_2=1 \end{aligned} $$

$$ \therefore\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right|=\left|\begin{array}{ccc} 3 & 0 & 5 \\ 3 & 4 & 5 \\ 0 & 0 & 1 \end{array}\right|=4(3)=12 $$

$\begin{aligned} & \text { and } \sqrt{\left(a_1 b_2-a_2 b_1\right)^2+\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2} \\\\ & =\sqrt{0+(4-0)^2+(0-3)^2} \\\\ & =\sqrt{16+9}=5 \\\\ & \therefore \text { Required shortest distance }=\frac{12}{5} \\\\ & \end{aligned}$

3d Geometry

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