Question

If the shortest distance between the lines \[ \frac{x+2}{2} = \frac{y+3}{3} = \frac{z-5}{4} \quad \text{and} \quad \frac{x-3}{1} = \frac{y-2}{-3} = \frac{z+4}{2}\] is \(\frac{38}{3\sqrt{5}} k\) and \[\int_{0}^{k} \lfloor x^2 \rfloor dx = \alpha - \sqrt{\alpha}, \]where \([x]\) denotes the greatest integer function, then \( 6\alpha^3 \) is equal to _____

Answer 1

Correct Answer: 48

Given lines: 
\[ \frac{x + 2}{2} = \frac{y + 3}{3} = \frac{z - 5}{4} \quad \text{and} \quad \frac{x - 3}{1} = \frac{y - 2}{-3} = \frac{z + 4}{2}. \] 
Direction vectors: 
The direction vector of the first line is: \[ \vec{d}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}. \]
The direction vector of the second line is: \[ \vec{d}_2 = \hat{i} - 3\hat{j} + 2\hat{k}. \] 
Shortest Distance Formula: 
The shortest distance between skew lines with direction vectors \(\vec{d}_1\) and \(\vec{d}_2\) is given by:
\[ \text{Distance} = \frac{|\vec{d}_1 \times \vec{d}_2 \times \vec{PQ}|}{|\vec{d}_1 \times \vec{d}_2|}, \]
where \(\vec{PQ}\) is the vector between points on the lines. 

Cross Product: 
Calculating \(\vec{d}_1 \times \vec{d}_2\):
\[ \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & -3 & 2 \end{vmatrix}. \] 
Simplifying: \[ \vec{d}_1 \times \vec{d}_2 = \hat{i}(3 \times 2 - 4 \times (-3)) - \hat{j}(2 \times 2 - 4 \times 1) + \hat{k}(2 \times (-3) - 3 \times 1), \]
\[ \vec{d}_1 \times \vec{d}_2 = \hat{i}(6 + 12) - \hat{j}(4 - 4) + \hat{k}(-6 - 3), \]
\[ \vec{d}_1 \times \vec{d}_2 = 18\hat{i} + 0\hat{j} - 9\hat{k}. \] 
Magnitude: 
\[ |\vec{d}_1 \times \vec{d}_2| = \sqrt{18^2 + 0^2 + (-9)^2} = \sqrt{324 + 81} = \sqrt{405} = 3\sqrt{5}. \] 
Given distance: 
\[ \frac{38}{3\sqrt{5}} k = \text{distance}, \quad \text{so } k = \frac{19}{\sqrt{5}}. \] 
Integral Evaluation: 
Consider: \[ \int_{0}^{k} [x^2] \, dx = \int_{0}^{1} 0 \, dx + \int_{1}^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{k} 2 \, dx. \]
Substituting the value of \(k\): \[ \int_{0}^{k} [x^2] \, dx = \sqrt{2} - 1 + 2\left(\frac{3}{2} - \sqrt{2}\right) = 2 - \sqrt{2}. \]
Comparing with \(\alpha - \sqrt{\alpha}\), we find: \[ \alpha = 2. \] 
Calculating \(6\alpha^3\): 
\[ 6\alpha^3 = 6 \times 2^3 = 6 \times 8 = 48. \]