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Question Number 24569 by ajfour last updated on 21/Nov/17

Commented by ajfour last updated on 21/Nov/17

$I n a g r a v i t y f r e e c u b i c a l b o x, a$ $b a l l b o u n c e s e l a s t i c a l l y . I f i t b e$ $p r o j e c t e d f r o m o r i g i n O t o w a r d s$ $A (4, 13, 16), a p o i n t o n t h e r o o f$ $p l a n e t h e n f i n d t h e s e c o n d a n d$ $t h i r d b o u n c e p o i n t s B, C .$ $T h e e d g e l e n g t h o f t h e c u b e i s$ $16 u n i t s .$
Commented by ajfour last updated on 22/Nov/17

$c o r r e c t . t h a n k s . p o i n t C (8, 6, 0) .$
Commented by jota last updated on 21/Nov/17

$B e s t a e n e l p l a n o y = \frac{13}{4} x$ $\Rightarrow 16 = \frac{13}{4} x_{B}$ $E n e s e p l a n o p o r s e m e j a n z a d e$ $t r i a n g u l o s$ $\frac{16}{13} = \frac{16 - z_{B}}{3}$ $\Rightarrow B (\frac{64}{13} 16 12 \frac{4}{13})$
	Answered by mrW1 last updated on 22/Nov/17

	$A (4, 13, 16)$ $O^{'} (0, 0, 32)$ $\frac{x_{B} - x_{A}}{x_{A} - x_{O^{'}}} = \frac{y_{B} - y_{A}}{y_{A} - y_{O^{'}}} = \frac{z_{B} - z_{A}}{z_{A} - z_{O^{'}}}$ $\frac{x_{B} - 4}{4 - 0} = \frac{16 - 13}{13 - 0} = \frac{z_{B} - 16}{16 - 32}$ $\Rightarrow x_{B} = 4 + 4 \times \frac{3}{13} = \frac{64}{13}$ $\Rightarrow y_{B} = 16$ $\Rightarrow z_{B} = 16 - 16 \times \frac{3}{13} = \frac{160}{13}$ $A^{'} (4, 19, 16)$ $\frac{x_{C} - x_{B}}{x_{B} - x_{A^{'}}} = \frac{y_{C} - y_{B}}{y_{B} - y_{A^{'}}} = \frac{z_{C} - z_{B}}{z_{B} - z_{A^{'}}}$ $\frac{x_{C} - \frac{64}{13}}{\frac{64}{13} - 4} = \frac{y_{C} - 16}{16 - 19} = \frac{0 - \frac{160}{13}}{\frac{160}{13} - 16}$ $\Rightarrow x_{C} = \frac{64}{13} + \frac{12}{13} \times (\frac{10}{3}) = \frac{104}{13} = 8$ $\Rightarrow y_{C} = 16 - 3 \times \frac{10}{3} = 6$ $\Rightarrow z_{C} = 0$ $\Rightarrow B (\frac{64}{13}, 16, \frac{160}{13})$ $\Rightarrow C (8, 6, 0)$
	Commented by ajfour last updated on 22/Nov/17

	$t h a n k y o u s i r!$
	Commented by mrW1 last updated on 22/Nov/17

	$R e l a t e d q u e s t i o n : w h i c h p o i n t (s) A$ $o n t h e r o o f p l a n e c a n b e s e l e c t e d$ $s u c h t h a t C = O ?$
	Commented by mrW1 last updated on 23/Nov/17

	$I t h i n k n o s u c h a p o i n t A e x i s t s$ $s o t h a t C = O .$
	Commented by ajfour last updated on 22/Nov/17

	$(8 + 4 \sqrt{2}, 8 + 4 \sqrt{2}, 16) . I s i t ?$ $B u t a n y w a y p o i n t B w i l l b e o n$ $t h e v e r t i c a l e d g e : x = 16, y = 16$ $G o d k n o w s i f i t w i l l r e a c h O!$
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