The distance from the origin to the plane.

Question

The question I am stuck on is as follows.

Give that a plane has the Cartesian equation being $3x+ 2y-6z=12$. Find the distance from the origin to the plane.

So far, what I have done is that I have solved the points where the plane meets $x, y,$ and $z$ axes at A, B and C respectively. I calculated that to be $A(4, 0, 0)$; $B(0,6,0)$ and $C (0,0,-2)$.

I have also calculated that the angle between the plane and the x-axis is $25.4 ^{\circ}$.

However, with these answers, I am not sure how I can find out the distance from the origin to the plane.

The answer suggested $4 \sin \theta$ where $\theta =25.4 ^{\circ}$, but I'm not sure how.

$\textbf{If I were to use the angle I calculated}$ to find the distance, Would anyone be able to advise me on how to do that?

Many thanks in advance for any ideas and Sorry for asking a trivial question, or any tag labeling errors.

Answer 1

According to the formula for distance from point to a plane:

$$d=\frac{|-12|}{\sqrt{9+4+36}}=\frac{12}{7}$$

Answer 2

Notice, in general, the normal distance of any point $(x_1, y_1, z_1)$ from the given plane: $ax+by+cz+d=0$ is given as $$\color{blue}{\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}}$$
hence, the distance of origin $(0, 0, 0)$ from the given plane: $3x+2y-6z=12$ is given as $$\frac{|3\cdot 0+2\cdot 0-6\cdot 0-12|}{\sqrt{3^2+2^2+(-6)^2}}=\frac{|-12|}{\sqrt {49}}=\frac{12}{7}$$