The distance between the planes \[r\cdot \left( i+2j-2k \right)+5=0~\]and \[r\cdot \left( 2i+4j-4k \right)-16=0\]is
A 3
B 13/3
C 13
D -13

Hint: The plane whose general equation is given as \[r\text{ }\left( ai\text{ }+\text{ }bj\text{ }+\text{ }cz \right)\text{ }+\text{ }d\text{ }=\text{ }0\], where r is the position vector of a arbitrary point on the line and \[ai\text{ }+\text{ }bj\text{ }+\text{ }cz\]is the direction ratio of the normal line at the plane. To find the distance between two planes we need to convert the given equations of the plane in their Cartesian form or can say we need to modify the given two equations of the plane. For this, put \[r\text{ }=\text{ }xi\text{ }+\text{ }yj\text{ }+\text{ }zk\]in two given equations of the plane.

Formula Used: The formula to determine the distance between two planes is given as
\[\left| {{d}_{1}}\text{ }\text{ }{{d}_{2}} \right|/root\text{ }({{a}^{2}}+\text{ }{{b}^{2}}+\text{ }{{c}^{2}})\], where a, b, and c are the direction the ratio of the line perpendicular or normal to the plane and d1 is the perpendicular distance of the former plane from origin and d2 is the perpendicular distance of the later plane from the origin.

Complete step by step solution: The equation of the first place given in the question is given as
\[r\cdot \left( i+2j-2k \right)+5=0~\],
And
\[\left( xi\text{ }+\text{ }yj\text{ }+\text{ }zk \right).\text{ }\left( i\text{ }+\text{ }2j\text{ }-\text{ }2k \right)\text{ }+\text{ }5\text{ }=\text{ }0\]
Or
\[x\text{ }+\text{ }2y\text{ }\text{ }2z\text{ }+\text{ }5\text{ }=\text{ }0\] (a cartesian form of the plane equation)
The equation of another plane is given as
\[r\cdot \left( 2i+4j-4k \right)-16=0\]
Divide both sides of the equation by 2 such as
\[r.\left( i\text{ }+\text{ }2j\text{ }\text{ }2k \right)\text{ }\text{ }8\text{ }=\text{ }0\]
And
\[\left( xi\text{ }+\text{ }yj\text{ }+\text{ }zk \right).\text{ }\left( i\text{ }+\text{ }2j\text{ }-\text{ }2k \right)\text{ }\text{ }8\text{ }=\text{ }0\]
Or
\[x\text{ }+\text{ }2y\text{ }\text{ }2z\text{ }-\text{ }8\text{ }=\text{ }0\](A cartesian form of the plane equation)
Comparing both Cartesian equations with the general equation of a plane in the Cartesian form such as
\[ax\text{ }+\text{ }by\text{ }+\text{ }cz\text{ }\text{ }d\text{ }=\text{ }0\] (General equation of plane)
we will get
For former plane, \[{{a}_{1}}=1\], \[{{b}_{1}}=2\], \[{{c}_{1}}=-2\], and \[{{d}_{1}}=-5\]
For later plane, \[{{a}_{2}}=1\], \[{{b}_{2}}=2\], \[{{c}_{2}}=-2\], and \[{{d}_{2}}=8\]
So, \[a=1\], \[b=2\], and \[c=-2\]
The formula used to determine the distance between two planes is given as
\[\left| {{d}_{1}}\text{ }\text{ }{{d}_{2}} \right|/root\text{ }({{a}^{2}}+\text{ }{{b}^{2}}+\text{ }{{c}^{2}})\]
Putting the value of the solved above in the formula to find the distance between two planes we will get the desired result such as
\[\left| 5+8 \right|/\text{ }root\text{ }of\text{ }{{1}^{2}}+\text{ }{{2}^{2}}-{{2}^{2}}\]
\[\left| 13 \right|/3\text{ }=\text{ }13/3\]

Option ‘B’ is correct

Note: It is important to note that r is the position vector of a point passing through a line which is defined as the sum of xi, yj, and zk coordinates of the point where I, j, and k are unit vectors along the coordinates such as
\[r\text{ }=\text{ }xi\text{ }+\text{ }yj\text{ }+\text{ }zk\] or \[{{r}^{2}}\text{ }=\text{ }{{x}^{2}}\text{ }+\text{ }{{y}^{2}}\text{ }+\text{ }{{z}^{2}}.\]
In the equation of a plane, we need the value of r but not of \[{{r}^{2}}\]to convert the equation of a plane into its cartesian form.