If the foot of the perpendicular from point (4,3,8) on the line L1:x−al=y−23=z−b4,l≠0 is (3,5,7) then the shortest distance between the line L1 and line L2:x−23=y−44=z−55 is equal to:

M (3, 5, 7) satisfies the line L1. 3−al=5−23=7−b4b=3 and a+1=3M is foot of perpendicular from P (4, 3, 8) on line L1.Also, =2i^+3j^+4k^ and v→2=3i^+4j^+5k^∴v→1×v→2=−i^+2j^−k^Shortest distance =|A→B→.(v→1×v→2)|v→1×v→2||=16

If the foot of the perpendicular from point (4,3,8) on the line L1:x−al=y−23=z−b4,l≠0 is (3,5,7) then the shortest distance between the line L1 and line L2:x−23=y−44=z−55 is equal to:

M (3, 5, 7) satisfies the line L1. 3−al=5−23=7−b4b=3 and a+1=3M is foot of perpendicular from P (4, 3, 8) on line L1.Also, =2i^+3j^+4k^ and v→2=3i^+4j^+5k^∴v→1×v→2=−i^+2j^−k^Shortest distance =|A→B→.(v→1×v→2)|v→1×v→2||=16

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If the foot of the perpendicular from point (4,3,8) on the line $L_{1} : \frac{x - a}{l} = \frac{y - 2}{3} = \frac{z - b}{4}, l \neq 0$ 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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$\sqrt{\frac{2}{3}}$

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

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

$\frac{1}{2}$

answer is B.

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Detailed Solution

M (3, 5, 7) satisfies the line $L_{1}$ .
$\frac{3 - a}{l} = \frac{5 - 2}{3} = \frac{7 - b}{4}$
b=3 and a+1=3
M is foot of perpendicular from P (4, 3, 8) on line $L_{1}$ .
Also, $= 2 \hat{i} + 3 \hat{j} + 4 \hat{k} and {\vec{v}}_{2} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}$
$∴ {\vec{v}}_{1} \times {\vec{v}}_{2} = - \hat{i} + 2 \hat{j} - \hat{k}$
Shortest distance $= | \frac{\vec{A} \vec{B} . ({\vec{v}}_{1} \times {\vec{v}}_{2})}{| {\vec{v}}_{1} \times {\vec{v}}_{2} |} | = \frac{1}{\sqrt{6}}$