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### Section 12.1 : The 3-D Coordinate System

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.