Find the distance between the planes given below. $\begin{array} { l }
4 x - 4 y + 5 z - 104 = 0 \
20 x - 20 y + 25 z - 100 = 0
\end{array}$
A) $\frac { - 84 } { \sqrt { 57 } }$
B) $\frac { 84 } { \sqrt { 57 } }$
C) $\frac { 89 } { \sqrt { 57 } }$
D) $\frac { - 168 } { \sqrt { 57 } }$
E) $\frac { 168 } { \sqrt { 57 } }$

Question

Find the distance between the planes given below. $\begin{array} { l } 
4 x - 4 y + 5 z - 104 = 0 \
20 x - 20 y + 25 z - 100 = 0
\end{array}$
A) $\frac { - 84 } { \sqrt { 57 } }$
B) $\frac { 84 } { \sqrt { 57 } }$
C) $\frac { 89 } { \sqrt { 57 } }$
D) $\frac { - 168 } { \sqrt { 57 } }$
E) $\frac { 168 } { \sqrt { 57 } }$

Quizplus · Accepted Answer

The distance between two parallel planes $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$ is given by $\frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}}$. The second plane equation can be simplified by dividing all terms by 5, resulting in $4x - 4y + 5z - 20 = 0$. Thus, $D_1 = -104$ and $D_2 = -20$. Substituting $A = 4$, $B = -4$, $C = 5$, $D_1 = -104$, and $D_2 = -20$ into the formula gives the distance as $\frac{|-20 - (-104)|}{\sqrt{4^2 + (-4)^2 + 5^2}} = \frac{84}{\sqrt{57}}$.

Find the Distance Between the Planes Given Below 4x−4y+5z−104=020x−20y+25z−100=0\begin{array} { l } 4 x - 4 y + 5 z - 104 = 0 \\20 x - 20 y + 25 z - 100 = 0\end{array}4x−4y+5z−104=020x−20y+25z−100=0​

Find the Distance Between the Planes Given Below $\begin{array} { l } 4 x - 4 y + 5 z - 104 = 0 \\20 x - 20 y + 25 z - 100 = 0\end{array}$