cube ABCD.EFGH side 4 cm point P is center BF and Q center AD. distance E to PQG


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Question Number 126609 by abdullahquwatan last updated on 22/Dec/20

$cube ABCD . EFGH s i d e 4 c m p o i n t P i s c e n t e r B F a n d Q c e n t e r A D . d i s t a n c e E t o P Q G$
	Answered by liberty last updated on 22/Dec/20

	$t h e e q u a t i o n o f p l a n e p a s s e s t r o u g h t$ $P (4, 4, 2), Q (2, 0, 0), G (0, 4, 4)$ $\Rightarrow \| \begin{matrix} x - 2 y z \\ - 2 4 4 \\ 2 4 2 \end{matrix} \| = 0$ $\Rightarrow (x - 2) (8 - 16) - y (- 4 - 8) + z (- 8 - 8) = 0$ $\Rightarrow - 8 (x - 2) + 12 y - 16 z = 0$ $\Rightarrow - 2 x + 3 y - 4 z + 4 = 0$ $\Rightarrow 2 x - 3 y + 4 z - 4 = 0$ $t h e d i s t a n c e p o i n t E (4, 0, 4) t o P Q G i s$ $\frac{∣ 8 - 0 + 16 - 4 ∣}{\sqrt{4 + 9 + 16}} = \frac{20}{\sqrt{29}} c m$
	Commented by abdullahquwatan last updated on 22/Dec/20

	$t h a n k y o u s i r$
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