Find image of plane x−y+z−3=0 in plane 2x+y−z+4=0 ?


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Question Number 45417 by rahul 19 last updated on 12/Oct/18

$F i n d i m a g e o f p l a n e x - y + z - 3 = 0 i n$ $p l a n e 2 x + y - z + 4 = 0 ?$
	Answered by MrW3 last updated on 13/Oct/18

	$s i n c e b o t h p l a n e s a r e p e r p e n d i c u l a r$ $t o e a c h o t h e r, t h e i m a g e i s i t s e l f, i . e .$ $x - y + z - 3 = 0$ $b u t I^{'} l l s h o w y o u a g e n e r a l w a y t o f i n d$ $t h e i m a g e i n a p l a n e .$ $t h e i m a g e o f p o i n t (X, Y, Z) i n p l a n e$ $a x + b y + c z + d = 0$ $i s p o i n t (U, V, W) w i t h$ $U = X - \frac{2 a (a X + b Y + c Z + d)}{a^{2} + b^{2} + c^{2}}$ $V = Y - \frac{2 b (a X + b Y + c Z + d)}{a^{2} + b^{2} + c^{2}}$ $W = Z - \frac{2 c (a X + b Y + c Z + d)}{a^{2} + b^{2} + c^{2}}$ $n o w w e t a k e a p o i n t P (u, v, w) o n t h e p l a n e$ $x - y + z - 3 = 0$ $I t s i m a g e i n t h e p l a n e$ $2 x + y - z + 4 = 0$ $i s P^{'} (p, q, r) .$ $w e c a n a l s o s a y P (u, v, w) i s t h e i m a g e$ $o f P^{'} (p, q, r) . u s i n g f o r m u l a a b o v e w e g e t$ $u = p - \frac{2 \times 2 (2 p + q - r + 4)}{2^{2} + 1^{2} + {(- 1)}^{2}}$ $\Rightarrow u = p - \frac{2 (2 p + q - r + 4)}{3}$ $s i m i l a r l y$ $\Rightarrow v = q - \frac{1 (2 p + q - r + 4)}{3}$ $\Rightarrow w = r - \frac{(- 1) (2 p + q - r + 4)}{3}$ $s i n c e u - v + w - 3 = 0$ $p - \frac{2 (2 p + q - r + 4)}{3} - q + \frac{(2 p + q - r + 4)}{3} + r + \frac{(2 p + q - r + 4)}{3} - 3 = 0$ $\Rightarrow p - q + r - 3 = 0$ $o r$ $\Rightarrow x - y + z - 3 = 0$ $t h i s i s t h e e x p e c t e d r e s u l t .$
	Commented by rahul 19 last updated on 13/Oct/18
	Thank you so much sir ! ☺️�� In the formula you posted , if we need to find foot of perpendicular then this factor 2 will not come . right?
	Commented by MrW3 last updated on 13/Oct/18

	$y e s s i r .$ $f o o t p o i n t (U^{'}, V^{'}, W^{'})$ $U^{'} = \frac{X + U}{2} = \frac{X}{2} + \frac{X}{2} - \frac{a (a X + b Y + c Z + d)}{a^{2} + b^{2} + c^{2}}$ $\Rightarrow U^{'} = X - \frac{a (a X + b Y + c Z + d)}{a^{2} + b^{2} + c^{2}}$
	Commented by MrW3 last updated on 13/Oct/18

	$t h i s f o r m u l a i s e a s y t o r e m e m b e r$ $a n d b e u s e d f o r e x a m p l e$ $t o f i n d t h e d i s t a n c e o f p o i n t (X, Y, Z) t o$ $a p l a n e :$ $d = \sqrt{{(X - U^{'})}^{2} + {(Y - V^{'})}^{2} + {(Z - W^{'})}^{2}}$ $= \frac{∣ a X + b Y + c Z + d ∣}{a^{2} + b^{2} + c^{2}} \sqrt{a^{2} + b^{2} + c^{2}}$ $= \frac{∣ a X + b Y + c Z + d ∣}{\sqrt{a^{2} + b^{2} + c^{2}}}$
	Commented by rahul 19 last updated on 13/Oct/18

	$o k s i r .$
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