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Question Number 96089 by bebo last updated on 29/May/20

	Answered by mr W last updated on 30/May/20

	Commented by mr W last updated on 30/May/20

	$M E T H O D I$ $p o i n t A o n l i n e 1 :$ $x = 3 + 3 s$ $y = 8 - s$ $z = 3 + s$ $p o i n t B o n l i n e 2 :$ $x = - 3 - 3 t$ $y = - 7 + 2 t$ $z = 6 + 4 t$ $s q u a r e o f d i s t a n c e b e t w e e n A & B :$ $D = d^{2} = {(3 + 3 s + 3 + 3 t)}^{2} + {(8 - s + 7 - 2 t)}^{2} + {(3 + s - 6 - 4 t)}^{2}$ $D = 9 {(s + t + 2)}^{2} + {(s + 2 t - 15)}^{2} + {(s - 4 t - 3)}^{2}$ $s u c h t h a t d i s m i n i m u m :$ $\frac{\partial D}{\partial s} = 18 (s + t + 2) + 2 (s + 2 t - 15) + 2 (s - 4 t - 3) = 0$ $\Rightarrow 11 s + 7 t = 0 . . . (i)$ $\frac{\partial D}{\partial t} = 18 (s + t + 2) + 4 (s + 2 t - 15) - 8 (s - 4 t - 3) = 0$ $\Rightarrow 7 s + 29 t = 0 . . . (i i)$ $f r o m (i) a n d (i i) w e g e t :$ $s = 0, t = 0$ $\Rightarrow p o i n t o n l i n e 1 i s A (3, 8, 3)$ $\Rightarrow p o i n t o n l i n e 2 i s B (- 3, - 7, 6)$ $t h e d i s t a n c e b e t w e e n t h e m i s t h e$ $s h o r t e s t d i s t a n c e b e t w e e n b o t h l i n e s :$ $d_{m i n} = \sqrt{{(3 + 3)}^{2} + {(8 + 7)}^{2} + {(3 - 6)}^{2}} = 3 \sqrt{30}$ $e q n . o f t h e s h o r t e s t d i s t a n c e i s A B :$ $\frac{x - 3}{6} = \frac{y + 7}{15} = \frac{z - 6}{- 3}$ $o r$ $\frac{x - 3}{2} = \frac{y + 7}{5} = \frac{z - 6}{- 1}$
	Commented by mr W last updated on 30/May/20

	$M E T H O D I I$ $l_{1} = (3, - 1, 1)$ $l_{2} = (- 3, 2, 4)$ $l_{3} = l_{1} \times l_{2} = (- 6, - 15, 3)$ $P (3, 8, 3), Q (- 3, - 7, 6)$ $P Q = (6, 15, - 3)$ $d = \frac{P Q \cdot l_{3}}{∣ l_{3} ∣}$ $= \frac{(6, 15, - 3) \cdot (- 6, - 15, 3)}{\sqrt{6^{2} + 15^{2} + 3^{2}}} = 3 \sqrt{30}$
	Answered by Sourav mridha last updated on 30/May/20

	$\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4} this line is going$ $through the point (say) p (- 3, - 7, 6)$ $for the straight line - -$ $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}, any point on this line$ $described by a parameter t, like that$ $L (3 + 3 t, 8 - t, 3 + t) .$ $the shortest distance means the$ $perpendicular distance {bet}^{n} this two$ $lines, lets say ⊥_{r} distance is P L . (([$ $the DR of PL is$ $[6 + 3 t, 15 - t, - 3 + t]$ $PL is both ⊥_{r} to two lines so :$ $3 (6 + 3 t) - 1 (15 - t) + 1 (- 3 + t) = 0$ $\Rightarrow 18 + 9 t - 15 + t - 3 + t = 0$ $so t = 0$ $t h e p o i t L (3, 8, 3)$ $and we know p (- 3, - 7, 6)$ $so ∣ P L ∣= \sqrt{6^{2} + 15^{2} + 3^{2}} = \sqrt{270} = 3 \sqrt{30} unit$ $now the {eq}^{n} of P L in cartetian form$ $\frac{x + 3}{6} = \frac{y + 7}{15} = \frac{z - 6}{- 3}$ $o r i n v e c t o r f o r m$ $r = (- 3 \hat{i} - 7 \hat{j} + 6 \hat{k}) + t (6 \hat{i} + 15 \hat{j} - 3 \hat{k})$
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