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Question Number 52012 by adilmalik78623@gmail.com last updated on 02/Jan/19

Commented by mr W last updated on 02/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 02/Jan/19

$p o i n t B (3 r + 2, 4 r + 3, 5 r + 4) b e t h e f o o t o f p e r p e n d i c u l a r$ $d i r e c t i o n r a t i o b e t w e e n p o i n t A (1, 2, 3) a n d (3 r + 2, 4 r + 3, 5 r + 4)$ $i s (3 r + 1, 4 r + 1, 5 r + 1)$ $3 (3 r + 1) + 4 (4 r + 1) + 5 (5 r + 1) = 0$ $9 r + 16 r + 25 r = - 3 - 4 - 5$ $50 r = - 12$ $r = \frac{- 12}{50}$ $d . r (3 r + 1, 4 r + 1, 5 r + 1)$ $= \frac{- 36}{50} + 1, \frac{- 48}{50} + 1, \frac{- 60}{50} + 1$ $= \frac{14}{50}, \frac{2}{50}, \frac{- 10}{50}$ $= 7, 1, - 5$ $f o o t o f p e r p e n d i c u l a r$ $(\frac{- 36}{50} + 2, \frac{- 48}{50} + 3, \frac{- 60}{50} + 4)$ $(\frac{64}{50}, \frac{102}{50}, \frac{140}{50})$ $l e n g t h \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}}$ $\sqrt{{(\frac{64}{50} - 1)}^{2} + {(\frac{102}{50} - 2)}^{2} + {(\frac{140}{50} - 3)}^{2}}$ ${[\frac{14^{2} + 2^{2} + {(- 10)}^{2}}{50^{2}}]}^{\frac{1}{2}} = \sqrt{\frac{196 + 4 + 100}{2500}} = \sqrt{\frac{300}{2500}}$ $= \frac{\sqrt{3}}{5}$ $e q n \frac{x - 1}{7} = \frac{y - 2}{1} = \frac{z - 4}{5}$ $s i r l s c h e c k . . .$
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