JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the values of \(\lambda\) for which the shortest distance between the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x-\lambda}{3}=\frac{y-4}{4}=\frac{z-5}{5}\) is \(\frac{1}{\sqrt{6}}\) be \(\lambda_1\) and \(\lambda_2\). Then the radius of the circle passing through the points \((0,0),\left(\lambda_1, \lambda_2\right)\) and \(\left(\lambda_2, \lambda_1\right)\) is
- A \(\frac{5 \sqrt{2}}{3}\)
- B 4
- C \(\frac{\sqrt{2}}{3}\)
- D 3
Answer & Solution
Correct Answer
(A) \(\frac{5 \sqrt{2}}{3}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \overrightarrow{\mathrm{p}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}, \overrightarrow{\mathrm{q}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}} \\ & \Rightarrow \overrightarrow{\mathrm{p}} \times…
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