Question

Find the distance between the two planes $2x + 3y + 4z = 4$ and $4x + 6y + 8z = 12$.

Answer 1

Hint : If we have two planes $Ax + By + Cz = {d_1}$ and $Ax + By + Cz = {d_2}$ then distance between these two planes is given by $d = \left| {\dfrac{{{d_1} - {d_2}}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}} \right|$. We will use this information to find the required distance.

Complete step-by-step answer :
In this problem, we have the following equation of planes.
$2x + 3y + 4z = 4 \cdots \cdots \left( 1 \right) \\\ 4x + 6y + 8z = 12 \cdots \cdots \left( 2 \right) \\\$
Let us divide by $2$ on both sides of equation $\left( 2 \right)$. So, we get $2x + 3y + 4z = 6 \cdots \cdots \left( 3 \right)$. Note that here we have two parallel planes and we need to find the distance between them.
Let us compare the equation $\left( 1 \right)$ with $Ax + By + Cz = {d_1}$. So, we can write
$A = 2,B = 3,C = 4,{d_1} = 4$
Now let us compare the equation $\left( 3 \right)$ with $Ax + By + Cz = {d_2}$. So, we can write
$A = 2,B = 3,C = 4,{d_2} = 6$
Now we are going to find the distance between given planes by using the formula
$d = \left| {\dfrac{{{d_1} - {d_2}}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}} \right|$. Substitute the values of $A,B,C,{d_1}$ and ${d_2}$ in above formula, we get
$d = \left| {\dfrac{{4 - 6}}{{\sqrt {{2^2} + {3^2} + {4^2}} }}} \right| \\\ \Rightarrow d = \left| {\dfrac{{ - 2}}{{\sqrt {4 + 9 + 16} }}} \right| \\\ \Rightarrow d = \left| { - \dfrac{2}{{\sqrt {29} }}} \right| \\\ \Rightarrow d = \dfrac{2}{{\sqrt {29} }} \\\$
Hence, the required distance is $\dfrac{2}{{\sqrt {29} }}$.

Note : If two planes do not intersect then those planes are called parallel planes. Two planes which are not parallel always intersect in a line. In this problem, the coordinates of the normal vector of planes will be the coefficients of $x,y,z$. If ${n_1}$ is the normal vector of the first plane, ${n_2}$ is the normal vector of the second plane and ${n_1} = k{n_2}$ or ${n_2} = k{n_1}$ then these two planes are called parallel planes.