Question

Distance between plane $3x+4y-20=0$ and point $(0,0,-7)$ is

• A. $4$ units
• B. $3$ units
• C. $2$ units
• D. $1$ units

Answer 1

The correct option is A $4$ units

As we know that the distance of the point $(x_1,y_1,z_1)$ from $ax+by+cz+d=0$ is given as-

$d=\frac{|a{x}_1+b{y}_1+c{z}_1+d|}{\sqrt{a^2+b^2+c^2}}$

Given that the point is $(0,0,-7)$ and the plane is $3x+4y-20=0$

Therefore,

$d=\frac{|3\left(0\right)+4\left(0\right)-20|}{\sqrt{{\left(3\right)}^2+{\left(4\right)}^2}}=\frac{|-20|}{\sqrt{25}}=\frac{20}{5}=4$

Hence the distance between the point and the plane is $4$ units.