If the point (x, y, z) lies on the plane x + y + z = 3 then $x^{2} + y^{2} + z^{2} ⩾ 3$ . Prove this fact.

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Solution

$E q^{n}$ of plane is $x + y + z = 3$
Comparing with $a x + b y + c z = D$
$a = 1, b = 1, c = 1, D = 3$
Distance of origin from base of perpendicular from origin or
in other words. shortest distance
from origin is given by
$d = ∣ ∣ ∣ \frac{a x + b y + c z - D}{\sqrt{x^{2} + b^{2} + c^{2}}} ∣ ∣ ∣$
$d = ∣ ∣ ∣ \frac{- 3}{\sqrt{3}} = \sqrt{3} ∣ ∣ ∣ .$
$∴$ Shortest distance from plane
to origin is $\sqrt{3}$ meaning that all points on plane are at distance
$\sqrt{3}$ or more from origin
Let $(x, y, z)$ be any pt on plane
$∴ \sqrt{(x - 0)^{2} + (y - 0)^{2} + (z - 0)^{2}} ⩾ \sqrt{3}$ (distance from origin)
$\sqrt{x^{2} + y^{2} + z^{2}} ⩾ \sqrt{3}$
Squaring both sides,
$x^{2} + y^{2} + z^{2} ⩾ 3$

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