Distance of plane x+y+z=3 from origin is - Vidyadip

MCQ

Distance of plane $x+y+z=3$ from origin is

A
$3$
✓
$\sqrt{3}$
C
$\frac{1}{\sqrt{3}}$
D
$\frac{1}{3}$

✓

Answer

Correct option: B.

$\sqrt{3}$

(B)
Here, $a =1, b=1, c =1, d=-3$ and $x=y= z =0$
$\therefore \quad d=\left|\frac{-3}{\sqrt{1^2+1^2+1^2}}\right|=\sqrt{3}$

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