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Question Number 97901 by pranesh last updated on 10/Jun/20

Commented by pranesh last updated on 10/Jun/20

$s o l v e f a s t p l e a s e$
Commented by prakash jain last updated on 10/Jun/20

$Question is not clear . Please edit image$ $and upload again ?$
	Answered by mr W last updated on 11/Jun/20

	$M E T H O D I$ $s a y p l a n e t o u c h i n g t h e s u r f a c e i s$ $2 x + y + 2 z + 2 k = 0$ $\Rightarrow y = - 2 (x + z + k)$ ${(1 - x)}^{2} + {(x - y)}^{2} + {(y - z)}^{2} + z^{2} = \frac{1}{4}$ ${(1 - x)}^{2} + {(3 x + 2 z + 2 k)}^{2} + {(2 x + 3 z + 2 k)}^{2} + z^{2} - \frac{1}{4} = 0$ $1 - 2 x + x^{2} + 9 x^{2} + 4 z^{2} + 4 k^{2} + 12 x z + 12 k x + 8 k z + 4 x^{2} + 9 z^{2} + 4 k^{2} + 12 x z + 8 k x + 12 k z + z^{2} - \frac{1}{4} = 0$ $14 x^{2} + 14 z^{2} + 24 x z + 2 (10 k - 1) x + 20 k z + 8 k^{2} + \frac{3}{4} = 0$ $\Rightarrow 7 x^{2} + 7 z^{2} + 12 x z + (10 k - 1) x + 10 k z + 4 k^{2} + \frac{3}{8} = 0$ $7 z^{2} + (12 x + 10 k) z + 7 x^{2} + (10 k - 1) x + 4 k^{2} + \frac{3}{8} = 0$ $Δ = {(12 x + 10 k)}^{2} - 4 \times 7 [7 x^{2} + (10 k - 1) x + 4 k^{2} + \frac{3}{8}] = 0$ ${(6 x + 5 k)}^{2} - 7 [7 x^{2} + (10 k - 1) x + 4 k^{2} + \frac{3}{8}] = 0$ $\Rightarrow 13 x^{2} + (10 k - 7) x + 3 k^{2} + \frac{21}{8} = 0$ $Δ = {(10 k - 7)}^{2} - 4 \times 13 (3 k^{2} + \frac{21}{8}) = 0$ $\Rightarrow 112 k^{2} + 280 k + 175 = 0$ $\Rightarrow k = - \frac{5}{4}$ $d i s t a n c e b e t w e e n t h e p l a n e s$ $4 x + 2 y + 4 z - 5 = 0$ $4 x + 2 y + 4 z + 7 = 0$ $i s$ $d = \frac{7 - (- 5)}{\sqrt{4^{2} + 2^{2} + 4^{2}}} = 2$ ${t h a t}^{'} s t h e s h o r t e s t d i s t a n c e .$
	Answered by mr W last updated on 11/Jun/20

	$M E T H O D I I$ $d = \frac{4 x + 2 y + 4 z + 7}{\sqrt{4^{2} + 2^{2} + 4^{2}}} = \frac{1}{3} (2 x + y + 2 z + \frac{7}{2})$ ${i t}^{'} s t o f i n d t h e m i n i m u m o f$ $f (x, y, z) = 2 x + y + 2 z w i t h$ ${(1 - x)}^{2} + {(x - y)}^{2} + {(y - z)}^{2} + z^{2} - \frac{1}{4} = 0$ $F = 2 x + y + 2 z + λ [{(1 - x)}^{2} + {(x - y)}^{2} + {(y - z)}^{2} + z^{2} - \frac{1}{4}]$ $\frac{\partial F}{\partial x} = 2 + λ [- 2 (1 - x) + 2 (x - y)] = 0$ $\Rightarrow 1 + λ (- 1 + 2 x - y) = 0$ $\Rightarrow 2 x - y - 1 = - \frac{1}{λ} . . . (i)$ $\frac{\partial F}{\partial y} = 1 + λ [- 2 (x - y) + 2 (y - z)] = 0$ $\Rightarrow 1 + 2 λ (- x + 2 y - z) = 0$ $\Rightarrow 2 (- x + 2 y - z) = - \frac{1}{λ} . . . (i i)$ $\frac{\partial F}{\partial z} = 2 + λ [- 2 (y - z) + 2 z] = 0$ $\Rightarrow 1 + λ (- y + 2 z) = 0$ $\Rightarrow - y + 2 z = - \frac{1}{λ} . . . (i i i)$ $y = \frac{6 z + 1}{5}$ $x = z + \frac{1}{2}$ ${(\frac{1}{2} - z)}^{2} + {(\frac{3}{10} - \frac{z}{5})}^{2} + {(\frac{z}{5} + \frac{1}{5})}^{2} + z^{2} - \frac{1}{4} = 0$ $4 z^{2} - 2 z + \frac{1}{4} = 0$ ${(2 z - \frac{1}{2})}^{2} = 0$ $\Rightarrow z = \frac{1}{4}$ $\Rightarrow x = \frac{3}{4}$ $\Rightarrow y = \frac{1}{5} (\frac{6}{4} + 1) = \frac{1}{2}$ $d = \frac{1}{3} (2 \times \frac{3}{4} + \frac{1}{2} + 2 \times \frac{1}{4} + \frac{7}{2}) = 2$
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