Question

Find the distance from $\left(4,-2,6\right)$ to each of the following:

The $y-$axis

Answer 1

To find the distance:

Consider the given point.

$\left(4,-2,6\right)$

The $y-$axis means that the point is $\left(0,-2,0\right)$.

Let $\left(x_1,y_1,z_1\right)=\left(4,-2,6\right)$ and $\left(x_2,y_2,z_2\right)=\left(0,-2,0\right)$.

Using the distance formula and calculate the distance in below:

$\begin{array}{rcl}\text{d}& =& \sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}\left[\text{Distance formula}\right]\\ & \Rightarrow & \text{d}=\sqrt{{\left(0-4\right)}^2+{\left(-2-\left(-2\right)\right)}^2+{\left(0-6\right)}^2}\left[\text{Substitues all values}\right]\\ & \Rightarrow & \text{d}=\sqrt{{\left(-4\right)}^2+{\left(-2+2\right)}^2+{\left(-6\right)}^2}\left[\text{Simplifying}\right]\\ & \Rightarrow & \text{d}=\sqrt{16+0+36}\\ & \Rightarrow & \text{d}=\sqrt{52}\\ \therefore \text{d}& =& 2\sqrt{13}\end{array}$

Hence, the distance is $\mathbf{2}\sqrt{\mathbf{13}}$ units.