Question

Find the distance of the origin from the plane which passes through (2, 1, 8) , (1, 0, 2) and (−3, 4, 6)

Answer 1

For find the distance from the point to the plane we must find the distance from the point to the projection of the point onto a plane.

Lets find the equation of the plane. We are using this formula:

$$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\x_2-x_1 & y_2-y_1 & z_2-z_1\\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix}=0$$

where (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) are given points

$$\begin{vmatrix} x-2 & y-1 & z-8 \\1-2 & 0-1 & 2-8\\ -3-2 & 4-1 & 6-8 \end{vmatrix}=0$$

$$\begin{vmatrix} x-2 & y-1 & z-8 \\-1 & -1 & -6\\ -5 & 3 & -2 \end{vmatrix}=0$$

$(x-2)((-1)(-2)-(-6)3)-(y-1)((-1)(-2)-(-6)(-5)) + (z -8)((-1)3-(-1)(-5)) = 0$

$$20(x-2)+28(y-1)+(-8)(z-8)=0$$

$$20x+28y-8z-4=0$$

After simplifying equation of the plane is $5x+7y-2z-1=0$

Distance of the origin O(0, 0, 0) from the plane

$Ax+By+Cz+D=0$ could be find according to formula

$$d=\frac{|A\cdot 0+B\cdot 0+C\cdot 0+D|}{\sqrt{A^2+B^2+C^2}}$$

In our case

$$d=\frac{|5\cdot 0+7\cdot 0-2\cdot 0-1|}{\sqrt{5^2+7^2+2^2}}=\frac{1}{\sqrt{78}}\approx 0.113$$

Answer: the distance of the origin from the plane which passes through (2, 1, 8) , (1, 0, 2) and (−3, 4, 6) is equal $\frac{1}{\sqrt{78}}\approx 0.113$ .