If the shortest distance between the lines $$x+2$$λ$$=2y=$$−$$12z$$, $$x=y+4$$λ$$=6z$$−$$12$$λ is $$42$$ units, then a value of λ is

Lines $$x+2$$λ$$1=y1/2=z$$−$$1/12x1=y+4$$λ$$1=z$$−$$2$$λ$$1/6S$$.$$D=42$$; $$49$$λ$$2=98$$; λ$$=2$$

Equation of line in Straight lines

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Question

If the shortest distance between the lines $x + 2 λ = 2 y = - 12 z, x = y + 4 λ = 6 z - 12 λ$ is $4 \sqrt{2}$ units, then a value of $λ$ is

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Solution

Lines $\frac{x + 2 λ}{1} = \frac{y}{1 / 2} = \frac{z}{- 1 / 12}$
$\frac{x}{1} = \frac{y + 4 λ}{1} = \frac{z - 2 λ}{1 / 6}$
$S . D = 4 \sqrt{2}; 49 λ^{2} = 98; λ = \sqrt{2}$

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