Question

Distance of the plane \(3x-4y+6z=11\) from the origin is

• A. \(\dfrac{3}{\sqrt{61}}\)
• B. \(\dfrac{11}{\sqrt{61}}\)
• C. \(\dfrac{6}{\sqrt{61}}\)
• D. \(\dfrac{4}{\sqrt{61}}\)

Hint:

Point–plane distance: $\displaystyle \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$; put plane as $Ax+By+Cz+D=0$ first.

Answer 1

The Correct Option is B

Write plane in \(Ax+By+Cz+D=0\) form: \(3x-4y+6z-11=0\). Distance from origin: \[ \frac{|D|}{\sqrt{A^{2}+B^{2}+C^{2}}} =\frac{|{-}11|}{\sqrt{3^{2}+(-4)^{2}+6^{2}}} =\frac{11}{\sqrt{9+16+36}} =\frac{11}{\sqrt{61}}. \]