Question

Find the distance between given planes 2x – y + 2z = 5 and 5x – 2.5y + 5z = 20

Answer 1

Now we know that the planes $ax+by+cz={{d}_{1}}$ and $ax+by+cz={{d}_{2}}$ are parallel.
So we will divide the equation 5x – 2.5y + 5z = 20 by 2.5 and hence we will have an equation of parallel planes.
Now we know that the distance between parallel planes $ax+by+cz={{d}_{1}}$ and $ax+by+cz={{d}_{2}}$ is given by $\left| \dfrac{{{d}_{2}}-{{d}_{1}}}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|$. Hence we can find the distance between two planes.

Complete step-by-step answer :
Now we are given two planes whose equations are 2x – y + 2z = 5 and 5x – 2.5y + 5z = 20. Let us say
$2x-y+2z=5................(1)$
$5x-2.5y+5z=20.............(2)$
Now dividing equation (2) by 2.5 we get
$\begin{aligned} & \dfrac{5}{2.5}x-\dfrac{2.5}{2.5}y+\dfrac{5}{2.5}z=\dfrac{20}{2.5} \\\ & 2x-y+2z=8.....................(3) \\\ \end{aligned}$
Hence Now if we check equation (2) and equation (3) they are in the form of $ax+by+cz={{d}_{1}}$ and $ax+by+cz={{d}_{2}}$
Also we know that the planes $ax+by+cz={{d}_{1}}$ and $ax+by+cz={{d}_{2}}$ are equations of parallel planes. Hence we get the given planes are parallel
Now consider equation (2) and equation (3)
We know that the distance between parallel planes $ax+by+cz={{d}_{1}}$ and $ax+by+cz={{d}_{2}}$ is given by $\left| \dfrac{{{d}_{2}}-{{d}_{1}}}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|$
Hence comparing the equation (2) and equation (3) we have a = 2 , b = -1, c = 2, ${{d}_{1}}=5$ and ${{d}_{2}}=8$
Let D be the distance between two planes
Now substituting the values of a, b, c, ${{d}_{1}}$ and ${{d}_{2}}$ we get
$D=\left| \dfrac{8-5}{\sqrt{{{2}^{2}}+{{(-1)}^{2}}+{{(2)}^{2}}}} \right|$
$=\left| \dfrac{3}{\sqrt{4+1+4}} \right|$
$=\left| \dfrac{3}{\sqrt{9}} \right|$
Now we know that the square root of 9 is 3. Hence we get
$=\left| \dfrac{3}{3} \right|=1$
Hence we get the value of D is 1.
Hence we have the distance between the two given planes is 1.

Note : If we don’t know if the planes are parallel we can use the result which says ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z={{d}_{1}}$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z={{d}_{2}}$ are parallel if $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$