Calculus II - The 3-D Coordinate System

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Section 12.1 : The 3-D Coordinate System

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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.