If the shortest distance between the lines x+2/2 = y+3/2 = z-5/4 and x-3/1 = y-2/-3 = z+4/2 is

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 \(\int\limits_{0}^{\mathrm{k}}\left[\mathrm{x}^{2}\right] \mathrm{dx}=\alpha-\sqrt{\alpha}\), where \([\mathrm{x}]\) denotes the greatest integer function, then \(6 \alpha^{3}\) is equal to _______.

Answer 1

Correct answer: 48

\( \frac{38}{3 \sqrt{5}} \hat{k}=\frac{(5 \hat{i}+5 \hat{j}-9 \hat{k})}{\sqrt{5}} \cdot\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & -3 & 2\end{array}\right|\)

\(\frac{38}{3 \sqrt{5}} \hat{\mathrm{k}}=\frac{19}{\sqrt{5}}\)

\(\mathrm{k}=\frac{19}{\sqrt{5}}\)

\(\mathrm{k}=\frac{3}{2}\)

\(\int\limits_{0}^{3 / 2}\left[\mathrm{x}^{2}\right]=\int\limits_{0}^{1} 0+\int\limits_{1}^{\sqrt{2}} 1+\int\limits_{\sqrt{2}}^{3 / 2} 2\)

\(=\sqrt{2}-1+2\left(\frac{3}{2}-\sqrt{2}\right)\)

\(=2-\sqrt{2}\)

\(\alpha=2\)

\(\Rightarrow 6 \alpha^{3}=48\)