Class 12: The Plane – Exercise 29.10

$\displaystyle \textbf{Question 1: }~\text{Find the distance between the parallel planes } \\ 2x-y+3z-4=0\text{ and }6x-3y+9z+13=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Multiplying the first equation of the plane by }3,\text{ we get}$
$\displaystyle 6x-3y+9z-12=0$
$\displaystyle 6x-3y+9z=12\ldots(1)$
$\displaystyle \text{The second equation of the plane is}$
$\displaystyle 6x-3y+9z=-13\ldots(2)$
$\displaystyle \text{We know that the distance between two planes }ax+by+cz=d_1\text{ and }ax+by+cz=d_2\text{ is}$
$\displaystyle \frac{|d_2-d_1|}{\sqrt{a^2+b^2+c^2}}$
$\displaystyle \text{So, the required distance}$
$\displaystyle =\frac{|-13-12|}{\sqrt{6^2+(-3)^2+9^2}}$
$\displaystyle =\frac{| -25 |}{\sqrt{36+9+81}}$
$\displaystyle =\frac{25}{\sqrt{126}}$
$\displaystyle =\frac{25}{3\sqrt{14}}\text{ units}$

$\displaystyle \textbf{Question 2: }~\text{Find the equation of the plane which passes through the point } \\ (3,4,-1)\text{ and is parallel to the plane }2x-3y+5z+7=0.\text{ Also, find the distance} \\ \text{between the two planes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the equation of a plane parallel to the given plane be}$
$\displaystyle 2x-3y+5z=k\ldots(1)$
$\displaystyle \text{This passes through }(3,4,-1).\text{ So,}$
$\displaystyle 2(3)-3(4)+5(-1)=k$
$\displaystyle \Rightarrow k=-11$
$\displaystyle \text{Substituting this in (1), we get}$
$\displaystyle 2x-3y+5z=-11\ldots(1),\text{ which is the equation of the required plane.}$
$\displaystyle \text{The equation of the given plane is}$
$\displaystyle 2x-3y+5z=-7\ldots(2)$
$\displaystyle \text{We know that the distance between two planes }ax+by+cz=d_1\text{ and }ax+by+cz=d_2\text{ is}$
$\displaystyle \frac{|d_2-d_1|}{\sqrt{a^2+b^2+c^2}}$
$\displaystyle \text{So, the required distance}$
$\displaystyle =\frac{|-7-(-11)|}{\sqrt{2^2+(-3)^2+5^2}}$
$\displaystyle =\frac{|-7+11|}{\sqrt{4+9+25}}$
$\displaystyle =\frac{4}{\sqrt{38}}\text{ units}$

$\displaystyle \textbf{Question 3: }~\text{Find the equation of the plane mid-parallel to the planes } \\ 2x-2y+z+3=0\text{ and }2x-2y+z+9=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that the distance between two planes }ax+by+cz=d_1\text{ and }ax+by+cz=d_2\text{ is} \frac{|d_2-d_1|}{\sqrt{a^2+b^2+c^2}}$
$\displaystyle \text{The equation of plane that is mid-parallel to the planes}$
$\displaystyle 2x-2y+z+3=0\ldots(1)$
$\displaystyle 2x-2y+z+9=0\ldots(2)$
$\displaystyle \text{is of the form }2x-2y+z+k=0\ldots(3)$
$\displaystyle \text{It means that the distance between (1) and (3) = distance between (1) and (2)}$
$\displaystyle \Rightarrow \frac{|k-3|}{\sqrt{4+4+1}}=\frac{|k-9|}{\sqrt{4+4+1}}$
$\displaystyle \Rightarrow |k-3|=|k-9|$
$\displaystyle \Rightarrow k-3=k-9\text{ or }k-3=-(k-9)$
$\displaystyle \Rightarrow 3=9\text{ (false)},\;k-3=-k+9$
$\displaystyle \Rightarrow 2k=12$
$\displaystyle \Rightarrow k=6$
$\displaystyle \text{Substituting this in (3), we get}$
$\displaystyle 2x-2y+z+6=0,\text{ which is the required equation of the plane.}$

$\displaystyle \textbf{Question 4: }~\text{Find the distance between the planes} \\ \overrightarrow{r}\cdot(\widehat{i}+2\widehat{j}+3\widehat{k})+7=0\text{ and }\overrightarrow{r}\cdot(2\widehat{i}+4\widehat{j}+6\widehat{k})+7=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given planes are}$
$\displaystyle \vec{r}\cdot(\hat{i}+2\hat{j}+3\hat{k})=-7$
$\displaystyle \Rightarrow x+2y+3z=-7$
$\displaystyle \text{Multiplying this equation of the plane by }2,\text{ we get}$
$\displaystyle 2x+4y+6z=-14\ldots(1)$
$\displaystyle \text{and}$
$\displaystyle \vec{r}\cdot(2\hat{i}+4\hat{j}+6\hat{k})=-7$
$\displaystyle \Rightarrow 2x+4y+6z=-7\ldots(2)$
$\displaystyle \text{We know that the distance between two planes }ax+by+cz=d_1\text{ and }ax+by+cz=d_2\text{ is} \frac{|d_2-d_1|}{\sqrt{a^2+b^2+c^2}}$
$\displaystyle \text{So, the required distance}$
$\displaystyle =\frac{|-7-(-14)|}{\sqrt{2^2+4^2+6^2}}$
$\displaystyle =\frac{|7|}{\sqrt{4+16+36}}$
$\displaystyle =\frac{7}{\sqrt{56}}\text{ units}$