Calculus II - The 3-D Coordinate System

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Section 12.1 : The 3-D Coordinate System

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2. Which of the points \(P = \left( {4, - 2,6} \right)\) and \(Q = \left( { - 6, - 3,2} \right)\) is closest to the \(yz\)-plane?

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The shortest distance between any point and any of the coordinate planes will be the distance between that point and its projection onto that plane.

Let’s call the projections of \(P\) and Q onto the \(yz\)-plane \(\overline{P}\) and \(\overline{Q}\) respectively. They are,

\[\overline{P} = \left( {0, - 2,6} \right)\hspace{0.75in}\hspace{0.25in}\overline{Q} = \left( {0, - 3,2} \right)\] Show Step 2

To determine which of these is closest to the \(yz\)-plane we just need to compute the distance between the point and its projection onto the \(yz\)-plane.

Note as well that because only the \(x\)-coordinate of the two points are different the distance between the two points will just be the absolute value of the difference between two \(x\) coordinates.

Therefore,

\[d\left( {P,\overline{P}} \right) = 4\hspace{0.75in}d\left( {Q,\overline{Q}} \right) = 6\]

Based on this is should be pretty clear that \(P = \left( {4, - 2,6} \right)\) is closest to the \(yz\)-plane.