Simply 3D

Geometry Level 3

Consider a 3 D 3D figure represented as x y z 2 = 2 xyz^{2}=2 then its minimum distance from origin is


The answer is 2.

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Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

Shubhendra Singh
Jun 10, 2015

We know that the distance from origin = x 2 + y 2 + z 2 \sqrt{x^{2}+y^{2}+z^{2}}

By AM-GM

x 2 + y 2 + 2. z 2 2 4 ≥ ( x 2 × y 2 × z 4 4 ) 1 4 \dfrac{x^{2}+y^{2}+2. \dfrac{z^{2}}{2}}{4} \geq ( \dfrac{x^{2}×y^{2}×z^{4}}{4})^{\dfrac{1}{4}}

⟹ x 2 + y 2 + z 2 ≥ 4 \implies x^{2}+y^{2}+z^{2} \geq 4

So the min. Distance = 4 = 2 \sqrt{4}=2

Equality holds when x = ± 1 , y = ± 1 , z = ± 2 x=±1, y=±1, z=±\sqrt{2}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution. For sake of completeness, we should note that the minimum can be achieved at the point ( 1 , 1 , 2 ) , (1,1,\sqrt{2}), (as well as ( 1 , 1 , − 2 ) , ( − 1 , − 1 , 2 ) (1,1,-\sqrt{2}), (-1,-1,\sqrt{2}) and − 1 , − 1 , − 2 ) . -1,-1,-\sqrt{2}). )

Brian Charlesworth - 6 years ago

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Thanks sir, I have mentioned it.

Shubhendra Singh - 6 years ago

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