If the shortest distance between the lines and is , and , where [x] denotes the greatest integer function, then is equal to __________. [2024]

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Solution

Q.
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 $\int_{0}^{k} [x^{2}] d x = α - \sqrt{α}$ , where [x] denotes the greatest integer function, then $6 α^{3}$ is equal to __________. [2024]

Ans.
(48)

$L_{1} : \frac{x + 2}{2} = \frac{y + 3}{3} = \frac{z - 5}{4}$

$L_{2} : \frac{x - 3}{1} = \frac{y - 2}{- 3} = \frac{z + 4}{2}$

$∴ {\vec{a}}_{1} = - 2 \hat{i} - 3 \hat{j} + 5 \hat{k}, {\vec{a}}_{2} = 3 \hat{i} + 2 \hat{j} - 4 \hat{k}$

${\vec{b}}_{1} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}, {\vec{b}}_{2} = \hat{i} - 3 \hat{j} + 2 \hat{k}$

Now, ${\vec{a}}_{2} - {\vec{a}}_{1} = 5 \hat{i} + 5 \hat{j} - 9 \hat{k}$

Shortest distance between the lines = $| \frac{({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2})}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |} |$ and $| {\vec{b}}_{1} \times {\vec{b}}_{2} |$ $=$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & - 3 & 2 \end{matrix} |$

$= \hat{i} (6 + 12) - \hat{j} (4 - 4) + \hat{k} (- 6 - 3) = 18 \hat{i} - 9 \hat{k}$

$d =$ $| \frac{(5 \hat{i} + 5 \hat{j} - 9 \hat{k}) \cdot (18 \hat{i} - 9 \hat{k})}{\sqrt{324 + 81}} |$ $=$ $| \frac{90 + 81}{9 \sqrt{5}} |$ $= \frac{171}{9 \sqrt{5}}$

From the question, we get $\frac{38}{3 \sqrt{5}} k = \frac{171}{9 \sqrt{5}} \Rightarrow k = \frac{3}{2}$

$∴ \int_{0}^{k} [x^{2}] d x = \int_{0}^{\frac{3}{2}} [x^{2}] d x = \int_{0}^{1} 0 d x + \int_{1}^{\sqrt{2}} 1 d x + \int_{\sqrt{2}}^{\frac{3}{2}} 2 d x$

$= 0 + (\sqrt{2} - 1) + 2 (\frac{3}{2} - \sqrt{2}) = \sqrt{2} - 1 + 3 - 2 \sqrt{2} = 2 - \sqrt{2}$

$= α - \sqrt{α}$ (Given)

Hence, $α = 2$

$∴ 6 α^{3} = 6 \times 8 = 48$ .

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