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Question Number 45575 by rahul 19 last updated on 14/Oct/18

Commented by rahul 19 last updated on 14/Oct/18

$M y d o u b t :$ $I f i w i l l t a k e b o t h p o i n t s o n s a m e s i d e$ $o f p l a n e (w r o n g m e t h o d) a n d w i l l$ $t a k e p o i n t s o n o p p o s i t e s i d e i w i l l g e t$ $s a m e a n s w e r . . . . i t i s a l w a y s t r u e ?$ $r i g h t ? ? ?$
Commented by MrW3 last updated on 14/Oct/18

Commented by MrW3 last updated on 14/Oct/18

$p r o j e c t i o n o f P_{1} P_{2} a n d P_{1} P_{2}^{'} i s t h e s a m e,$ $i . e . Q_{1} Q_{2 .}$
Commented by rahul 19 last updated on 14/Oct/18

$t h a n k s s i r!$
	Answered by MrW3 last updated on 14/Oct/18

	$p r o j e c t i o n o f p o i n t (5, - 1, 4) i s (x_{1}, y_{1}, z_{1}) :$ $x_{1} = 5 - \frac{1 \times (5 - 1 + 4 - 7)}{3} = 5 - \frac{1 \times 1}{3} = \frac{14}{3}$ $y_{1} = - 1 - \frac{1 \times (5 - 1 + 4 - 7)}{3} = - 1 - \frac{1 \times 1}{3} = - \frac{4}{3}$ $z_{1} = 4 - \frac{1 \times (5 - 1 + 4 - 7)}{3} = 4 - \frac{1 \times 1}{3} = \frac{11}{3}$ $p r o j e c t i o n o f p o i n t (4, - 1, 3) i s (x_{2}, y_{2}, z_{2}) :$ $x_{2} = 4 - \frac{1 \times (4 - 1 + 3 - 7)}{3} = 4 - \frac{1 \times (- 1)}{3} = \frac{13}{3}$ $y_{2} = - 1 - \frac{1 \times (4 - 1 + 3 - 7)}{3} = - 1 - \frac{1 \times (- 1)}{3} = - \frac{2}{3}$ $z_{2} = 3 - \frac{1 \times (4 - 1 + 3 - 7)}{3} = 3 - \frac{1 \times (- 1)}{3} = \frac{10}{3}$ $l e n g t h o f p r o j e c t i o n :$ $L = \frac{1}{3} \sqrt{{(14 - 13)}^{2} + {(- 4 + 2)}^{2} + {(11 - 10)}^{2}} = \sqrt{\frac{2}{3}}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18

	$d i s t a n c e b e t w e e n (5, - 1, 4) a n d (4, - 1, 3)$ $\sqrt{{(5 - 4)}^{2} + {(- 1 + 1)}^{2} + {(4 - 3)}^{2}} = \sqrt{2}$ $t h e s t l i n e j o i n i n g (5, - 1, 4) a n d (4, - 1, 3)$ $\frac{x - 5}{4 - 5} = \frac{y + 1}{0} = \frac{z - 4}{3 - 4} =\to \frac{x - 5}{1} = \frac{y + 1}{0} = \frac{z - 4}{1}$ $a n g l e b e t w e e n n o r m s l t o t h e p l a n e a n d t h e s t$ $l i n e c o s θ = \frac{1 \times 1 + 0 \times 1 + 1 \times 1}{\sqrt{1^{2} + 1^{2} + 1^{2}} \times \sqrt{1^{2} + o^{2} + 1^{2}}} = \frac{2}{\sqrt{3} \times \sqrt{2}}$ $s i n θ = \frac{\sqrt{2}}{\sqrt{6}}$ $p r o j e c t i o n = p$ $s i n θ = \frac{p}{d} p = d s i n θ = \sqrt{2} \times \frac{\sqrt{2}}{\sqrt{6}} = \frac{\sqrt{2}}{\sqrt{3}}$
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