Find the distance from the origin to the plane x + 2y + 2z = -6.
A)1
B)2
C)6
D) $\frac { 1 } { 5 }$
E)15
F) $\frac { 1 } { 2 }$
G)4

H)5

Question

Find the distance from the origin to the plane x + 2y + 2z = -6.
A)1
B)2
C)6
D) $\frac { 1 } { 5 }$ 
E)15
F) $\frac { 1 } { 2 }$ 
G)4

H)5

Quizplus · Accepted Answer

The distance from the origin to the plane $Ax + By + Cz + D = 0$ is given by $\frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$. Here, $A = 1$, $B = 2$, $C = 2$, and $D = 6$. So, the distance is $\frac{6}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{6}{3} = 2$.

Find the Distance from the Origin to the Plane X 15\frac { 1 } { 5 }51​

Find the Distance from the Origin to the Plane X $\frac { 1 } { 5 }$