If the shortest distance between the lines $\frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$ and $\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}$ is $\frac{38}{3 \sqrt{5}} \mathrm{k}$, and $\int_0^{\mathrm{k}}\left[x^2\right] \mathrm{d} x=\alpha-\sqrt{\alpha}$, where $[x]$ denotes the greatest integer…

Question

If the shortest distance between the lines $\frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$ and $\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}$ is $\frac{38}{3 \sqrt{5}} \mathrm{k}$, and $\int_0^{\mathrm{k}}\left[x^2\right] \mathrm{d} x=\alpha-\sqrt{\alpha}$, where $[x]$ denotes the greatest integer function, then $6 \alpha^3$ is equal to________

Aicharya · Accepted Answer

$\begin{aligned} & \frac{38}{3 \sqrt{5}} \hat{\mathrm{k}}=\frac{(5 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-9 \hat{\mathrm{k}})}{\sqrt{5}} \cdot\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ 2 & 3 & 4 \ 1 & -3 & 2\end{array}ight| \ & \frac{38}{3 \sqrt{5}} \hat{\mathrm{k}}=\frac{19}{\sqrt{5}} \ & \mathrm{k}=\frac{19}{\sqrt{5}} \ & \mathrm{k}=\frac{3}{2} \ & \int_0^{3 / 2}\left[\mathrm{x}^2ight]=\int_0^1 0+\int_1^{\sqrt{2}} 1+\int_{\sqrt{2}}^{3 / 2} 2 \ & =\sqrt{2}-1+2\left(\frac{3}{2}-\sqrt{2}ight) \ & =2-\sqrt{2} \ & \alpha=2 \ & \Rightarrow 6 \alpha^3=48\end{aligned}$

If the shortest distance between the lines $\frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$ and…

Solution