Paul's Online Notes
Home / Calculus III / 3-Dimensional Space / The 3-D Coordinate System
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 1-1 : The 3-D Coordinate System

3. Which of the points $$P = \left( { - 1,4, - 7} \right)$$ and $$Q = \left( {6, - 1,5} \right)$$ is closest to the $$z$$-axis?

Show All Steps Hide All Steps

Start Solution

First, let’s note that the coordinates of any point on the $$z$$-axis will be $$\left( {0,0,z} \right)$$.

Also, the shortest distance from any point not on the $$z$$-axis to the $$z$$-axis will occur if we draw a line straight from the point to the $$z$$-axis in such a way that it forms a right angle with the z‑axis.

So, if we start with any point not on the $$z$$-axis, say $$\left( {{x_1},{y_1},{z_1}} \right)$$, the point on the $$z$$-axis that will be closest to this point is $$\left( {0,0,{z_1}} \right)$$.

Let’s call the point closest to $$P$$ and $$Q$$ on the $$z$$-axis closest to be $$\overline{P}$$ and $$\overline{Q}$$ respectively. They are,

$\overline{P} = \left( {0,0, - 7} \right)\hspace{0.75in}\hspace{0.25in}\overline{Q} = \left( {0,0,5} \right)$ Show Step 2

To determine which of these is closest to the $$z$$-axis we just need to compute the distance between the point and its projection onto the $$z$$-axis.

The distances are,

\begin{align*}d\left( {P,\overline{P}} \right) & = \sqrt {{{\left( { - 1 - 0} \right)}^2} + {{\left( {4 - 0} \right)}^2} + {{\left( { - 7 - \left( { - 7} \right)} \right)}^2}} = \sqrt {17} \\ d\left( {Q,\overline{Q}} \right) & = \sqrt {{{\left( {6 - 0} \right)}^2} + {{\left( { - 1 - 0} \right)}^2} + {{\left( {5 - 5} \right)}^2}} = \sqrt {37} \end{align*}

Based on this is should be pretty clear that $$P = \left( { - 1,4, - 7} \right)$$ is closest to the $$z$$-axis.