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Paul
February 18, 2026
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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Start SolutionFirst, 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 2To 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.