Activity 9.1.4.
Remember that in two dimensions, the distance between \(P=(x_0, y_0)\) and \(Q=(x_1, y_1)\) is
\begin{equation*}
\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}
\end{equation*}
and is related to the pythagorean theorem applied to a right triangle that measures changes in each coordinate direction (as demonstrated in part 9.1.1.f).
Let \(P=(x_0, y_0, z_0)\) and \(Q=(x_1, y_1, z_1)\) be two points in \(\R^3\text{.}\) These two points form opposite vertices of a rectangular box whose sides are fundamental planes as illustrated in Figure 9.1.14, and the distance between \(P\) and \(Q\) is the length of the blue diagonal shown in Figure 9.1.14.
(a)
Consider the right triangle \(PRS\) in the base of the box whose hypotenuse is shown as the red line in Figure 9.1.14. What are the coordinates of \(R\) and \(S\text{?}\)
(b)
Give the equation of the fundamental plane that contains the right triangle \(PRS\text{.}\)
(c)
Since the right triangle \(PRS\) lies in a plane, we can use the Pythagorean Theorem to find a formula for the length of the hypotenuse of this triangle. Find the length of the segment \(PR\) in terms of \(x_0\text{,}\) \(y_0\text{,}\) \(x_1\text{,}\) and \(y_1\text{.}\)
(d)
Triangle \(PRQ\) has hypotenuse drawn with as blue segment connecting the points \(P\) and \(Q\text{.}\) Segment \(PR\text{,}\) which is the hypotenuse of triangle \(PRS\) that we considered earlier, is a leg of triangle \(PRQ\text{.}\) This triangle lies entirely in a plane, so we can again use the Pythagorean Theorem to find the length of its hypotenuse. Show that the length of \(PQ\) is
\begin{equation*}
\sqrt{(x_1-x_0)^2 + (y_1-y_0)^2 + (z_1-z_0)^2}.
\end{equation*}
