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### Section 12.1 : 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?

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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.