Question

The perpendicular distance of the plane $3x-6y+5z=12$ from origin is be:

• A. $\frac{-\sqrt{70}}{12}$
• B. $\frac{-12}{\sqrt{70}}$
• C. $\frac{12}{\sqrt{70}}$
• D. $\frac{\sqrt{70}}{12}$

Answer 1

Answer: (C)

$\frac{12}{\sqrt{70}}$

Perpendicular distance of a plane $ax+by+cz=d=0$ from the origin is given by $\frac{|d|}{\sqrt{a^2+b^2+c^2}}$

Similarly, the perpendicular distance here would be given by

$\frac{12}{\sqrt{3^2+(-6{)}^2+5^2}}$

$=\frac{12}{\sqrt{9+36+25}}$

$=\frac{12}{\sqrt{70}}$