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 \(6\alpha^3\): 
\[ 6\alpha^3 = 6 \times 2^3 = 6 \times 8 = 48. \]

Answer 2

Step 1: Write the equations of the given lines.
The two lines are:
\[ L_1: \frac{x+2}{2} = \frac{y+3}{3} = \frac{z-5}{4} \]
\[ L_2: \frac{x-3}{1} = \frac{y-2}{-3} = \frac{z+4}{2} \]
From this, the direction ratios (d.r’s) are:
For \(L_1\): \( \vec{a_1} = (2, 3, 4) \)
For \(L_2\): \( \vec{a_2} = (1, -3, 2) \)

Points on the lines:
On \(L_1\): \( P_1(-2, -3, 5) \)
On \(L_2\): \( P_2(3, 2, -4) \)

Step 2: Formula for the shortest distance between skew lines.
\[ \text{Shortest distance} = \frac{|(\vec{P_2P_1}) \cdot (\vec{a_1} \times \vec{a_2})|}{|\vec{a_1} \times \vec{a_2}|} \]
where \( \vec{P_2P_1} = P_1 - P_2 \).

Compute:
\[ \vec{P_2P_1} = (-2 - 3, -3 - 2, 5 - (-4)) = (-5, -5, 9) \]

Step 3: Find the cross product \( \vec{a_1} \times \vec{a_2} \).
\[ \vec{a_1} \times \vec{a_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & -3 & 2 \end{vmatrix} = \hat{i}(3\cdot2 - 4(-3)) - \hat{j}(2\cdot2 - 4(1)) + \hat{k}(2(-3) - 3(1)) \]
\[ = \hat{i}(6 + 12) - \hat{j}(4 - 4) + \hat{k}(-6 - 3) \]
\[ = (18, 0, -9) \]
So, \[ \vec{a_1} \times \vec{a_2} = (18, 0, -9) \]
and its magnitude is:
\[ |\vec{a_1} \times \vec{a_2}| = \sqrt{18^2 + 0^2 + (-9)^2} = \sqrt{405} = 9\sqrt{5}. \]

Step 4: Compute the scalar triple product.
\[ (\vec{P_2P_1}) \cdot (\vec{a_1} \times \vec{a_2}) = (-5)(18) + (-5)(0) + (9)(-9) = -90 - 81 = -171. \]
Hence, the shortest distance:
\[ d = \frac{| -171 |}{9\sqrt{5}} = \frac{171}{9\sqrt{5}} = \frac{19}{\sqrt{5}}. \]
Given that the shortest distance = \(\frac{38}{3\sqrt{5}} k\), equating:
\[ \frac{38}{3\sqrt{5}}k = \frac{19}{\sqrt{5}} \Rightarrow k = \frac{3}{2}. \]

Step 5: Given integration condition.
\[ \int_{0}^{k} \lfloor x^2 \rfloor dx = \alpha - \sqrt{\alpha} \]
Substitute \(k = \frac{3}{2}\).
We know \(x^2\) ranges from 0 to \(\frac{9}{4}\). Thus:
- For \(0 \le x < 1\), \(\lfloor x^2 \rfloor = 0\).
- For \(1 \le x < \sqrt{2}\), \(\lfloor x^2 \rfloor = 1\).
- For \(\sqrt{2} \le x < \frac{3}{2}\), \(\lfloor x^2 \rfloor = 2\).

Hence:
\[ \int_{0}^{3/2} \lfloor x^2 \rfloor dx = \int_0^1 0\,dx + \int_1^{\sqrt{2}} 1\,dx + \int_{\sqrt{2}}^{3/2} 2\,dx \]
\[ = (0) + (\sqrt{2} - 1) + 2\left(\frac{3}{2} - \sqrt{2}\right) \]
\[ = \sqrt{2} - 1 + 3 - 2\sqrt{2} = 2 - \sqrt{2}. \]
So:
\[ \alpha - \sqrt{\alpha} = 2 - \sqrt{2}. \]
Comparing, we get:
\[ \alpha = 2. \]

Step 6: Compute \(6\alpha^3\).
\[ 6\alpha^3 = 6(2^3) = 6 \times 8 = 48. \]

Final Answer:
\[ \boxed{48} \]