How would i use the distance formula to determine the distance between the point (2,1,6) and the yz-plane?

Question

How would i use the distance formula to determine the distance between the point (2,1,6) and the yz-plane? Can someone provide the final answer and the steps if possible

Answer 1

The $yz$ plane has equation $x=0$ so just find the distance from the point $(2,1,6)$ to its nearest point in the $yz$ plane, namely, the point $(0,1,6)$, for which you do not even need the distance formula, but you may use it.

Answer 2

The distance of a point $\vec{r}$ with a plane through the origin with normal vector $\hat{n}$ is simply

$$ d = \vec{r} \cdot \hat{n} $$

_{where $\cdot$ is the dot (inner) product}

Answer 3

The distance of the point from the $yz$ plane is simply the absolute value of the $x$ coordinate of the point.

The equation of the plane is simply $x=0$.

Using the point-distance formula from $ax + by + cz + d = 0$ to point $(x_0, y_0, z_0)$

$$D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}},$$

We see that $b=c=d=0$ and $a=1$ so

$$D = \frac{|1(2) + 0(1) + 0(6) + 0|}{\sqrt{1^2 + 0^2 + 0^2}} = 2.$$