Browse Questions

# Choose the correct answer in the distance between the two planes: $2x + 3y + 4z = 4$ and $4x + 6y + 8z = 12$ is

$\begin{array} ((A)\, 2\; units \quad& (B)\, 4\; units \quad& (C)\, 8\; units \quad& (D) \, \frac{2} { \sqrt{29}} \; units \end{array}$

Toolbox:
• Distance between parallel plane is $D=\begin{vmatrix}\large\frac{d_1-d_2}{\sqrt{a^2+b^2+c^2}}\end{vmatrix}$
Step 1:
Equation of the planes are :
$2x+3y+4z=4$----(1)
$4x+6y+8z=12$----(2)
Divide equ(2) by 2 through out we get
$2x+3y+4z=6$
Hence this implies the planes are parallel to each other.
Step 2:
We know that the distance between two parallel planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$ is $\bigg| \large\frac{d_1-d_2}{\sqrt{a^2+b^2+c^2}} \bigg|$
$d_1=6$
$d_2=4$
$\Rightarrow D=\large\frac{6-4}{\sqrt{2^2+3^2+4^2}}$
$\Rightarrow \large\frac{2}{\sqrt{4+9+16}}$
$\Rightarrow D=\large\frac{2}{\sqrt{29}}$
Therefore the distance between two parallel planes =$\large\frac{2}{\sqrt{29}}$units.
Hence D is the correct answer.