Question

Find the perpendicular distance between the planes 12x-6y+4z-21=0 and 6x-3y+2z-1=0.

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

Answer:

Hence perpendicular distance between two plane  is 19\14 units.

Step-by-step explanation:

Formula for perpendicular distance between two plane is follows:

$D=\dfrac{\left|ax_1+by_1+cz_1+d\right|}{\sqrt{ a^2+b^2+c^2}}$

12x-6y+4z-21=0........(1) and  6x-3y+2z-1=0..........(2) {These are the parallel  equation in 3 variables)

12x-6y+4z-21=0...........(1) [Given equation]

2(6x-3y+2z)=21

6x-3y+2z=21\2............(3)

Put the equation (2) in the formula,we get:

$D=\dfrac{\left|6x-3y+2z-1\right|}{\sqrt{ 6^2+(-3)^2+2^2}}..........(4)$

After putting the value of 6x-3y+2z from equation(3) into equation (4).We get:

$D=\dfrac{\left| \dfrac{21}{2}-1\right|} {\sqrt{49}}$

$D=\dfrac{19}{14}\ units$

Hence perpendicular distance between two plane  is 19\14 units.