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 $\frac{38}{3\sqrt{5}}k$, and $\int_{0}^{k}[x^{2}]dx=\alpha-\sqrt{\alpha}$, where $[x]$ denotes the greatest integer function, then $6\alpha^{3}$ is equal to ____.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Three Dimensional Geometry

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 5} k

, and

\int_{0}^{k} [x^{2}] d x = α - α

, where

[x]

denotes the greatest integer function, then

6 α^{3}

is equal to ____.

Answer: 48

Solution

For the two lines the shortest distance works out to

\frac{38}{3 5} \cdot k

with

k = \frac{3}{2}

.

Then

\int_{0}^{3/2} [x^{2}] d x

[x^{2}] = 0

[0, 1)

= 1

[1, 2)

= 2

[2, \frac{3}{2})

, so

\int_{0}^{3/2} [x^{2}] d x = (2 - 1) + 2 (\frac{3}{2} - 2) = 2 - 2 .

Matching

α - α = 2 - 2

gives

α = 2

. Hence

6 α^{3} = 6 \cdot 8 = 48

.
Answer: 48.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Integer answer

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