Find the distance between the point and the plane.
$$\begin{array} { l }
( - 3 , - 5 , - 2 ) \
4 x + 6 y + 6 z = - 18
\end{array}$$
A) 36
B) $\frac { 9 } { 22 }$
C) $\frac { 36 } { \sqrt { 22 } }$
D) 0
E) $\frac { 18 } { \sqrt { 22 } }$

Question

Find the distance between the point and the plane.
$$\begin{array} { l } 
( - 3 , - 5 , - 2 ) \
4 x + 6 y + 6 z = - 18
\end{array}$$
A) 36
B) $\frac { 9 } { 22 }$
C) $\frac { 36 } { \sqrt { 22 } }$
D) 0
E) $\frac { 18 } { \sqrt { 22 } }$

Quizplus · Accepted Answer

The distance $d$ from a point $(x_1, y_1, z_1)$ to a plane $Ax + By + Cz + D = 0$ is given by the formula: $$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$$. Substituting the given values: $A = 4$, $B = 6$, $C = 6$, $D = 18$, and the point $(-3, -5, -2)$, we get: $$d = \frac{|4(-3) + 6(-5) + 6(-2) + 18|}{\sqrt{4^2 + 6^2 + 6^2}} = \frac{18}{\sqrt{22}}$$.

Find the Distance Between the Point and the Plane (−3,−5,−2)4x+6y+6z=−18\begin{array} { l } ( - 3 , - 5 , - 2 ) \\4 x + 6 y + 6 z = - 18\end{array}(−3,−5,−2)4x+6y+6z=−18​

Find the Distance Between the Point and the Plane $\begin{array} { l } ( - 3 , - 5 , - 2 ) \\4 x + 6 y + 6 z = - 18\end{array}$