Let the values of λ for which the shortest distance between the lines x-1/2 = y-2/3 = z-3/4

Question

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

(1) \(\frac{5 \sqrt{2}}{3}\)

(2) 4

(3) \(\frac{\sqrt{2}}{3}\)

(4) 3 

Answer 1

Correct option is: (1) \(\frac{5 \sqrt{2}}{3}\)   

\(\vec{p}=2 \hat{i}+3 \hat{j}+4 \hat{k}, \vec{q}=3 \hat{i}+4 \hat{j}+5 \hat{k}\)

\(\Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\)

\(\mathrm{A} \equiv(1,2,3) \mathrm{B} \equiv(\lambda, 4,5)\)

Shortest Distance \(=\left|\frac{\overrightarrow{\mathrm{AB}} \cdot(\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{q}})}{|\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{q}}|}\right|\)

\(\frac{1}{\sqrt{6}}=\left|\frac{((\lambda-1) \hat{i}+2 \hat{j}+2 \hat{\mathrm{k}}) \cdot(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})}{\sqrt{6}}\right|\)

\(\Rightarrow|-\lambda+1+4-2|=1 \Rightarrow|\lambda-3|=1\)

\(\Rightarrow \lambda=3 \pm 1=4,2\)

Radius of circle passing through points

(0,0), (4, 2) \(\&\) (2, 4)

\(=\frac{\mathrm{abc}}{4 \Delta}=\frac{\sqrt{20} \times \sqrt{20} \times \sqrt{8}}{4 \times \frac{1}{2}\left|\begin{array}{lll}1 & 1 & 1 \\ 0 & 4 & 2 \\ 0 & 2 & 4\end{array}\right|}=\frac{20 \times 2 \sqrt{2}}{2 \times 12}\)

\(=\frac{5 \sqrt{2}}{3}\)