Question

In the following exercises find the distance of each of the given points from the corresponding given plane:

$\begin{array}{cc}point& Plane\\ \left(a\right)(0,0,0)& 3x-4y+12z=3\end{array}$

$\begin{array}{cc}point& Plane\\ \left(b\right)(3,-2,1)& 2x-y+2z+3=0\end{array}$

$\begin{array}{cc}point& Plane\\ \left(c\right)(2,3,-5)& x+2y-2z=9\end{array}$

$\begin{array}{cc}point& Plane\\ \left(d\right)(-6,0,0)& 2x-3y+6z-2=0\end{array}$

Answer 1

We know that the distance between a point $P(x_1,y_1,z_1)$ and a plane Ax+By+Cz=D is

$d=\frac{A{x}_1+B{y}_1+C{z}_1-D}{\sqrt{A^2+B^2+C^2}}$ ...(i)

(a) The given point is (0,0,0) and the plane is 3x -4y +12z-3=0

$\therefore$ From Eq. (i),

$d=\frac{|3\times 0-4\times 0+12\times 0-3|}{\sqrt{3^2+(-4{)}^2+(12{)}^2}}=\frac{|-3|}{\sqrt{9+16+144}}=\frac{3}{\sqrt{169}}=\frac{3}{13}units$

We know that the distance between a point $P(x_1,y_1,z_1)$ and a plane Ax+By+Cz=D is
$d=\frac{A{x}_1+B{y}_1+C{z}_1-D}{\sqrt{A^2+B^2+C^2}}$ ...(i)

(b) The given point is (3,-2,1) and the plane is 2x-y+2z+3=0

$\therefore$ From Eq.(i),

$d=\frac{|2\times 3+(-1)\times (-2)+2\times 1+3|}{\sqrt{2^2+(-1{)}^2+2^2}}=\frac{|6+2+2+3|}{\sqrt{4+1+4}}=\frac{13}{\sqrt{9}}=\frac{13}{3}units$

We know that the distance between a point $P(x_1,y_1,z_1)$ and a plane Ax+By+Cz=D is

$d=\frac{A{x}_1+B{y}_1+C{z}_1-D}{\sqrt{A^2+B^2+C^2}}$ ...(i)

The given points is (2,3,-5) and the plane is x+2y-2z-9=0

$\therefore$ From Eq.(i),

$d=\frac{|2+2\times 3-2\times (-5)-9|}{\sqrt{1^2+2^2+(-2^2)}}=\frac{|2+6+10-9|}{\sqrt{1+4+4}}=\frac{9}{\sqrt{9}}=\frac{9}{3}=3units$

We know that the distance between a point $P(x_1,y_1,z_1)$ and a plane Ax+By+Cz=D is

$d=\frac{A{x}_1+B{y}_1+C{z}_1-D}{\sqrt{A^2+B^2+C^2}}$ ...(i)

The given point is (-6,0,0) and the plane is 2x -3y+6z-2=0

$\therefore$ From Eq.(i)

$d=\frac{|2\times (-6)-3\times 0+6\times 0-2|}{\sqrt{2^2+(-3{)}^2+6^2}}=\frac{|-14|}{\sqrt{4+9+36}}=\frac{14}{7}=2units$

Note The distance is always positive.