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Mathematics · 2024

JEE Main · 30 January 2024, Shift 1 · Q28

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…

If $\displaystyle \mathrm{d}_1$ is the shortest distance between the lines $\displaystyle x+1=2 y=-12 z, x=y+2=6 z-6$ and $\displaystyle \mathrm{d}_2$ is the shortest distance between the lines $\displaystyle \frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}, \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$, then the value of $\displaystyle \frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}$ is :
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.