If the foot of the perpendicular from point (4, 3, 8) on the line L1:x−al=y−23=z−b4. is (3, 5, 7), then the shortest distance between the line L1 and line L2:x−23=y−44=z−55 is equal to:

The given line is L1:x−al=y−23=z−b4.Since the foot of the perpendicular of the point (4, 3, 8) on the above line is 3,5,7Hence, the line joining the above two points is perpendicular to the L1it gives l3−4+35−3+47−8=0−l+6−4=0l=2The point 3,5,7 lies on the line L1:x−al=y−23=z−b4.it gives a=1,b=3Hence the shortest distance between the lines L1:x−12=y−23=z−34. and the line L2:x−23=y−44=z−55 is 234345122ijk234345=28−10−36−5+46−4i15−16−j10−12+k8−9=−4−3+8−i+2j−k=11+4+1=16

If the foot of the perpendicular from point (4, 3, 8) on the line L1:x−al=y−23=z−b4. is (3, 5, 7), then the shortest distance between the line L1 and line L2:x−23=y−44=z−55 is equal to:

The given line is L1:x−al=y−23=z−b4.Since the foot of the perpendicular of the point (4, 3, 8) on the above line is 3,5,7Hence, the line joining the above two points is perpendicular to the L1it gives l3−4+35−3+47−8=0−l+6−4=0l=2The point 3,5,7 lies on the line L1:x−al=y−23=z−b4.it gives a=1,b=3Hence the shortest distance between the lines L1:x−12=y−23=z−34. and the line L2:x−23=y−44=z−55 is 234345122ijk234345=28−10−36−5+46−4i15−16−j10−12+k8−9=−4−3+8−i+2j−k=11+4+1=16

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If the foot of the perpendicular from point (4, 3, 8) on the line $L_{1} : \frac{x - a}{l} = \frac{y - 2}{3} = \frac{z - b}{4} .$ is (3, 5, 7), then the shortest distance between the line $L_{1}$ and line $L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ is equal to:

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$\sqrt{\frac{2}{3}}$

$\frac{1}{\sqrt{3}}$

$\frac{1}{2}$

$\frac{1}{\sqrt{6}}$

answer is D.

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Detailed Solution

$T h e g i v e n l i n e i s$ $L_{1} : \frac{x - a}{l} = \frac{y - 2}{3} = \frac{z - b}{4} .$

Since the foot of the perpendicular of the point (4, 3, 8) on the above line is $(3, 5, 7)$

Hence, the line joining the above two points is perpendicular to the $L_{1}$

it gives

$\begin{matrix} l (3 - 4) + 3 (5 - 3) + 4 (7 - 8) = 0 \\ - l + 6 - 4 = 0 \\ l = 2 \end{matrix}$

The point $(3, 5, 7)$ lies on the line $L_{1} : \frac{x - a}{l} = \frac{y - 2}{3} = \frac{z - b}{4} .$

it gives $a = 1, b = 3$

Hence the shortest distance between the lines $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} .$ and the line $L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ is

$\frac{|\begin{matrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 1 & 2 & 2 \end{matrix}|}{|\begin{matrix} i & j & k \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{matrix}|} = \frac{2 (8 - 10) - 3 (6 - 5) + 4 (6 - 4)}{|i (15 - 16) - j (10 - 12) + k (8 - 9)|} = \frac{- 4 - 3 + 8}{|- i + 2 j - k|} = \frac{1}{\sqrt{1 + 4 + 1}} = \frac{1}{\sqrt{6}}$