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Mathematics Question on Shortest Distance between Two Lines

If the shortest distance between the lines x+22=y+33=z−54andx−31=y−2−3=z+42 \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 3835k\frac{38}{3\sqrt{5}} k and ∫0k⌊x2⌋dx=α−α,\int_{0}^{k} \lfloor x^2 \rfloor dx = \alpha - \sqrt{\alpha}, where [x][x] denotes the greatest integer function, then 6α36\alpha^3 is equal to _____

Answer

Given lines:
x+22=y+33=z−54andx−31=y−2−3=z+42.\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: d⃗1=2i^+3j^+4k^.\vec{d}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}.
The direction vector of the second line is: d⃗2=i^−3j^+2k^.\vec{d}_2 = \hat{i} - 3\hat{j} + 2\hat{k}.
Shortest Distance Formula:
The shortest distance between skew lines with direction vectors d⃗1\vec{d}_1 and d⃗2\vec{d}_2 is given by:
Distance=∣d⃗1×d⃗2×PQ⃗∣∣d⃗1×d⃗2∣,\text{Distance} = \frac{|\vec{d}_1 \times \vec{d}_2 \times \vec{PQ}|}{|\vec{d}_1 \times \vec{d}_2|},
where PQ⃗\vec{PQ} is the vector between points on the lines.

Cross Product:
Calculating d⃗1×d⃗2\vec{d}_1 \times \vec{d}_2:
d⃗1×d⃗2=∣i^j^k^ 234 1−32∣.\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ 2 & 3 & 4 \\\ 1 & -3 & 2 \end{vmatrix}.
Simplifying: d⃗1×d⃗2=i^(3×2−4×(−3))−j^(2×2−4×1)+k^(2×(−3)−3×1),\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),
d⃗1×d⃗2=i^(6+12)−j^(4−4)+k^(−6−3),\vec{d}_1 \times \vec{d}_2 = \hat{i}(6 + 12) - \hat{j}(4 - 4) + \hat{k}(-6 - 3),
d⃗1×d⃗2=18i^+0j^−9k^.\vec{d}_1 \times \vec{d}_2 = 18\hat{i} + 0\hat{j} - 9\hat{k}.
Magnitude:
∣d⃗1×d⃗2∣=182+02+(−9)2=324+81=405=35.|\vec{d}_1 \times \vec{d}_2| = \sqrt{18^2 + 0^2 + (-9)^2} = \sqrt{324 + 81} = \sqrt{405} = 3\sqrt{5}.
Given distance:
3835k=distance,so k=195.\frac{38}{3\sqrt{5}} k = \text{distance}, \quad \text{so } k = \frac{19}{\sqrt{5}}.
Integral Evaluation:
Consider: ∫0k[x2] dx=∫010 dx+∫121 dx+∫2k2 dx.\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 kk: ∫0k[x2] dx=2−1+2(32−2)=2−2.\int_{0}^{k} [x^2] \, dx = \sqrt{2} - 1 + 2\left(\frac{3}{2} - \sqrt{2}\right) = 2 - \sqrt{2}.
Comparing with α−α\alpha - \sqrt{\alpha}, we find: α=2.\alpha = 2.
Calculating 6α36\alpha^3:
6α3=6×23=6×8=48.6\alpha^3 = 6 \times 2^3 = 6 \times 8 = 48.