The shortest distance between (1−x)2+(x−y)2+(y−z)2+z2=14 and 4x+2y+4z+7=0 is equal to ______ units.

(1−x)2+(x−y)2+(y−z)2+z2=14 (1−x)2+(x−y)2+(y−z)2+z24≥((1−x)+(x−y)+(y−z)+z4)2 ⇒(1−x)2+(x−y)2+(y−z)2+z2≥14Given (1−x)2+(x−y)2+(y−z)2+z2=14@ 1 – x = x – y = y – z = z =14,−14(34,12,14)(54,64,74)Distance to the plane 4x + 2y + 4z + 7 = 0d1=|3+1+1+736|, d2=|5+3+7+736| =|126|=2 =|226|=113 unitsShortest distance = 2 units

The shortest distance between (1−x)2+(x−y)2+(y−z)2+z2=14 and 4x+2y+4z+7=0 is equal to ______ units.

(1−x)2+(x−y)2+(y−z)2+z2=14 (1−x)2+(x−y)2+(y−z)2+z24≥((1−x)+(x−y)+(y−z)+z4)2 ⇒(1−x)2+(x−y)2+(y−z)2+z2≥14Given (1−x)2+(x−y)2+(y−z)2+z2=14@ 1 – x = x – y = y – z = z =14,−14(34,12,14)(54,64,74)Distance to the plane 4x + 2y + 4z + 7 = 0d1=|3+1+1+736|, d2=|5+3+7+736| =|126|=2 =|226|=113 unitsShortest distance = 2 units

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The shortest distance between ${(1 - x)}^{2} + {(x - y)}^{2} + {(y - z)}^{2} + z^{2} = \frac{1}{4}$ and $4 x + 2 y + 4 z + 7 = 0$ is equal to ______ units.

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$\frac{11}{3}$

answer is A.

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Detailed Solution

${(1 - x)}^{2} + {(x - y)}^{2} + {(y - z)}^{2} + z^{2} = \frac{1}{4} \frac{{(1 - x)}^{2} + {(x - y)}^{2} + {(y - z)}^{2} + z^{2}}{4} \geq {(\frac{(1 - x) + (x - y) + (y - z) + z}{4})}^{2} \Rightarrow {(1 - x)}^{2} + {(x - y)}^{2} + {(y - z)}^{2} + z^{2} \geq \frac{1}{4}$

Given ${(1 - x)}^{2} + {(x - y)}^{2} + {(y - z)}^{2} + z^{2} = \frac{1}{4}$ $@ 1 - x = x - y = y - z = z = \frac{1}{4}, \frac{- 1}{4}$

$(\frac{3}{4}, \frac{1}{2}, \frac{1}{4}) (\frac{5}{4}, \frac{6}{4}, \frac{7}{4})$

Distance to the plane 4x + 2y + 4z + 7 = 0
$d_{1} = | \frac{3 + 1 + 1 + 7}{\sqrt{36}} |, d_{2} = | \frac{5 + 3 + 7 + 7}{\sqrt{36}} |$
$= | \frac{12}{6} | = 2 = | \frac{22}{6} | = \frac{11}{3}$ units
Shortest distance = 2 units