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 $\frac{32 \sqrt{3} d _1}{d _2}$ is :
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Answer (16)
Solution
$L _1: \frac{x+1}{1}=\frac{y}{1 / 2}=\frac{z}{-1 / 12}, L _2: \frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}}$
$d _1=$ shortest distance between $L _1 \& L _2$
$=\left|\frac{\left(\vec{a} _2-\vec{a} _1\right) \cdot\left(\vec{b} _1 \times \vec{b} _2\right)}{\text { matt }\left|\left(\vec{b} _1 \times \vec{b} _2\right)\right| \text { ma }}\right|$
$d _1=2$
$L _3: \frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}, L _4: \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$
$d _2=$ shortest distance between $L _3 \& L _4$
$d _2=\frac{12}{\sqrt{3}}$ Hence
$=\frac{32 \sqrt{3} d _1}{d _2}=\frac{32 \sqrt{3} \times 2}{\frac{12}{\sqrt{3}}}=16$
