If the shortest distance between the lines (x+2)/(2)=(y+3)/(3)=(z-5)/(4) and (x-3)/(1)=(y-2)/(-3)=(z+4)/(2) is 38 3 5 k , and ∫_0^ k [x^2] d x=α- α , where [x] denotes the greatest integer function, then 6 α^3 is equal

Correct answer: 0 aligned & 38 3 5 k = (5 i +5 j -9 k ) 5 ·| array ccc i & j & k 2 & 3 & 4 1 & -3 & 2 array | & 38 3 5 k = 19 5 & k= 19 5 & k=(3)/(2) & ∫_0^ 3 / 2 [x^2]=∫_0^1 0+∫_1^ 2 1+∫_ 2 ^ 3 / 2 2 & = 2 -1+2((3)/(2)- 2 ) & =2- 2 & α=2 & 6 α^3=48 aligned

JEE Main 20244 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2024 · 4 April · Morning Shift · Mathematics · Three Dimensional Geometry

If the shortest distance between the lines...

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}

\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________

Correct answer

Step-by-step solution

\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}\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_0^{3 / 2}\left[\mathrm{x}^2\right]=\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}\right) \\ & =2-\sqrt{2} \\ & \alpha=2 \\ & \Rightarrow 6 \alpha^3=48\end{aligned}

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