JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the shortest distance between the lines \(\frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}\) and \(\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}\) is \(\frac{38}{3 \sqrt{5}} \mathrm{k}\) and \(\int_0^{\mathrm{k}}\left[\mathrm{x}^2\right] \mathrm{dx}=\alpha-\sqrt{\alpha}\), where \([\mathrm{x}]\) denotes the greatest integer function, then \(6 \alpha^3\) is equal to ...........
- A \(45\)
- B \(49\)
- C \(50\)
- D \(48\)
Answer & Solution
Correct Answer
(D) \(48\)
Step-by-step Solution
Detailed explanation
\(\frac{38}{3 \sqrt{5}} \hat{\mathrm{k}}=\frac{(5 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-9 \hat{\mathrm{k}})}{\sqrt{5}}\) \(\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & -3 & 2\end{array}\right|\)…
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