If the foot of the perpendicular from point (4, 3, 8) on the line ${L_1}:{{x - a} \over l} = {{y - 2} \over 3} = {{z - b} \over 4}$, l $\ne$ 0 is (3, 5, 7), then the shortest distance between the line L₁ and line ${L_2}:{{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}$ is equal to :

(3, 5, 7) lie on given line L₁<br><br>${{3 - a} \over l} = {3 \over 3} = {{7 - b} \over 4}$<br><br>${{7 - b} \over 4} = 1 \Rightarrow b = 3$<br><br>${{3 - a} \over l} = 1 \Rightarrow 3 - a = l$<br><br>M(4, 3, 8)<br><br>N(3, 5, 7)<br><br>DR'S of MN = (1, $-$2, 1)<br><br>MN $\bot$ line L₁<br><br>(1)(l) + ($-$2)(3) + 4(1) = 0<br><br>$\Rightarrow$ l = 2<br><br>a = 1<br><br>a = 1, b = 3, l = 2<br><br>${{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over 4}$<br><br>${{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}$<br><br>$A = \, &lt; 1,2,3 &gt;$<br><br>$B = \, &lt; 2,4,5 &gt;$<br><br>$\overrightarrow {AB} = \, &lt; 1,2,2 &gt;$<br><br>$\overrightarrow p = 2\widehat i + 3\widehat j + 4\widehat k$<br><br>$\overrightarrow q = 3\widehat i + 4\widehat j + 5\widehat k$<br><br>$$\overrightarrow p \, \times \overrightarrow q = - \widehat i + 2\widehat j - \widehat k$$<br><br>Shortest distance = $$\left| {{{\overrightarrow {AB} \,.\,(\overrightarrow p \times \overrightarrow q )} \over {|\overrightarrow p \times \overrightarrow q |}}} \right|\, = {1 \over {\sqrt 6 }}$$