Let the values of for which the shortest distance between the lines (x-1)/(2)=(y-2)/(3)=(z…

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 $(0,0),({\lambda }_{1},{\lambda }_{2})$ and $({\lambda }_{2},{\lambda }_{1})$ is

Accepted Answer

$$\begin{aligned}
&\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 \overrightarrow{\mathrm{q}}=\left|\[\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ 2 & 3 & 4 \ 3 & 4 & 5 \end{array}\end{aligned}\end{aligned}$$\right|=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}} \
\& \mathrm{~A} \equiv(1,2,3) \mathrm{B} \equiv(\lambda, 4,5) \
\& \text { Shortest Distance }=\left|\frac{\overrightarrow{\mathrm{AB}} \cdot(\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{q}})}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}\right| \
\& \frac{1}{\sqrt{6}}=\left|\frac{(\lambda-1) \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})}{\sqrt{6}}\right| \
\& \Rightarrow|-\lambda+1+4-2|=1 \Rightarrow|\lambda-3|=1 \
\& \Rightarrow \lambda=3 \pm 1=4,2
\
\&\text { Radius of circle passing through points }\
\&\begin{aligned}
& (0,0),(4,2) &(2,4) \
& =\frac{a b c}{4 \Delta}=\frac{\sqrt{20} \times \sqrt{20} \times \sqrt{8}}{4 \times \frac{1}{2}\left|$$\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}
\end{aligned}
\]

Solutions · 1