Distance between two parallel planes - class-XII

We have the equation of plane is 

We have,

The equation of second plane can be rearrange as $x+2y+3z+\dfrac{7}{2}=0$. 

Given planes can be written as $4x + 6y - 7z - 1 = 0$ and $4x + 6y - 7z - 2 = 0$
Since both the planes are parallel so distance between them is given by,
$d =\left | \dfrac{(-1-2)}{\sqrt{4^2+6^2+7^2}}\right |=\dfrac{1}{\sqrt{101}}$  

Given planes can be written as $4x + 6y - 7z - 1 = 0$ and $4x + 6y - 7z - 2 = 0$
Since both the planes are parallel so distance between them is given by,
$d =\left |  \dfrac{(-1-2)}{\sqrt{4^2+6^2+7^2}}\right |=\dfrac{1}{\sqrt{101}}$ $\therefore N=101$
Hence $\dfrac{N(N+1)}{2}=101 \times 51 = 5151$

Given planes can be written as $4x + 6y - 7z - 1 = 0$ and $4x + 6y - 7z - 2 = 0$
Since both the planes are parallel so distance between them is given by,
$d =\left |\dfrac{(-1-2)}{\sqrt{4^2+6^2+7^2}}\right |=\dfrac{1}{\sqrt{101}}$

Since the planes are all parallel planes,

$\displaystyle { p } _{ 1 }=\dfrac { \left| 2-6 \right|  }{ \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 } }  } =\dfrac { 4 }{ \sqrt { 4+9+16 }  } =\dfrac { 4 }{ \sqrt { 29 }  } $

Equation of the plane $4x-6y+8z+3=0$ can be written as $2x-3y+4z+\displaystyle\dfrac { 3 }{ 2 } =0$

So, $\displaystyle { p } _{ 2 }=\dfrac { \left| 2-\dfrac { 3 }{ 2 }  \right|  }{ \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 } }  } =\dfrac { 1 }{ 2\sqrt { 29 }  } $

and $\displaystyle { p } _{ 3 }=\dfrac { \left| 2+6 \right|  }{ \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 } }  } =\dfrac { 8 }{ \sqrt { 29 }  } $

$\Rightarrow { p } _{ 1 }+8{ p } _{ 2 }-{ p } _{ 3 }=0$

Given planes are $4x -5y+ 3z - 5 = 0$ and $4x -5y+ 3z + 2 = 0$
Since both the planes are parallel so distance between them is,
$= \left| \dfrac{(-5-2)}{\sqrt{4^2+5^2+3^2}}\right|=\dfrac{7}{5\sqrt{2}}$

$\vec { r } .\left( 2\hat { i } -2\hat { j } +\hat { k }  \right) =-3$

Since the planes are parallel planes

$3r,4r,5r$ be the point that cuts the two plane. 

The planes are r.(2i - 3j + 6k) = 5 and r.(6i - 9j + 18k) = -20. Dividing the second equation by 3 gives r.(2i - 3j + 6k) = -20/3. The distance is |5 - (-20/3)| / sqrt(2^2 + (-3)^2 + 6^2) = |15/3 + 20/3| / sqrt(4 + 9 + 36) = (35/3) / 7 = 5/3.

Distance between parallel planes $ax+by+cz+d _1$ and $ax+by+cz+d _2$ is given by $\dfrac{|d _1-d _2|}{\sqrt{a^2+b^2+c^2}}$.