Three Dimensional Geometry Question 17
Question 17 - 2024 (30 Jan Shift 1)
If $d_{1}$ is the shortest distance between the lines $x+1=2 y=-12 z, x=y+2=6 z-6$ and $d_{2}$ is the shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}, \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$, then the value of $\frac{32 \sqrt{3} \mathrm{~d}{1}}{\mathrm{~d}{2}}$ is :
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Answer (16)
Solution
$\mathrm{L}{1}: \frac{\mathrm{x}+1}{1}=\frac{\mathrm{y}}{1 / 2}=\frac{\mathrm{z}}{-1 / 12}, \mathrm{~L}{2}: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}+2}{1}=\frac{\mathrm{z}-1}{\frac{1}{6}}$
$\mathrm{d}{1}=$ shortest distance between $\mathrm{L}{1} & \mathrm{~L}_{2}$
$=\left|\frac{\left(\vec{a}{2}-\vec{a}{1}\right) \cdot\left(\vec{b}{1} \times \vec{b}{2}\right)}{\text { matt }\left|\left(\vec{b}{1} \times \vec{b}{2}\right)\right| \text { ma }}\right|$
$d_{1}=2$
$\mathrm{L}{3}: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+8}{-7}=\frac{\mathrm{z}-4}{5}, \mathrm{~L}{4}: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-6}{-3}$
$\mathrm{d}{2}=$ shortest distance between $\mathrm{L}{3} & \mathrm{~L}_{4}$
$\mathrm{d}_{2}=\frac{12}{\sqrt{3}}$ Hence
$=\frac{32 \sqrt{3} \mathrm{~d}{1}}{\mathrm{~d}{2}}=\frac{32 \sqrt{3} \times 2}{\frac{12}{\sqrt{3}}}=16$
