Question

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

The $xz-$plane

Answer 1

To find the distance:

For the $xz-$plane, the $y-$coordinate is $0$.

So, the distance from the given point $\left(4,-2,6\right)$ to $xz-$plane is the $y-$coordinate.

Therefore, the distance is $\left|-2\right|=2$.

Hence, the distance from the point $\left(4,-2,6\right)$ and the $xz-$plane is $\mathbf{2}$ units.