Cube ABCD.EFGH with length side 2 cm. Point P is the center of ABFE plane. The distance of HP line and the BG line is _


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Question Number 105404 by bemath last updated on 28/Jul/20

$C u b e A B C D . E F G H w i t h$ $l e n g t h s i d e 2 c m . P o i n t P i s t h e$ $c e n t e r o f A B F E p l a n e . T h e$ $d i s t a n c e o f H P l i n e a n d t h e B G$ $l i n e i s__$
	Answered by john santu last updated on 28/Jul/20

	Commented by john santu last updated on 28/Jul/20

	$n o r m a l v e c t o r o f p l a n e B G P^{'}$ $B (2, 2, 0); P^{'} (2, 3, 1); G (0, 2, 2)$ $w i t h B G = (- 2, 0, 2), B P^{'} = (0, 1, 1)$ $\vec{n} = \| \begin{matrix} - 2 0 2 \\ 0 1 1 \end{matrix} \| = (- 2, 2, - 2)$ $e q u a t i o n o f p l a n e B P^{'} G$ $- 2 x + 2 y - 2 z = 0;$ $p u t p o i n t H (0, 0, 2)$ $d i s t a n c e l i n e o f H P t o l i n e o f$ $B G = ∣ \frac{- 4}{\sqrt{4 + 4 + 4}} ∣ = \frac{4}{2 \sqrt{3}} = \frac{2}{\sqrt{3}}$
	Commented by john santu last updated on 28/Jul/20

	$i f p u t p o i n t P (2, 1, 1)$ $d i s t a n c e = ∣ \frac{- 4 + 2 - 2}{2 \sqrt{3}} ∣ = \frac{2}{\sqrt{3}}$
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