The 3-D Coordinate System — Question 10

Question 10 – Solution

Strategy First combine the $x$- and $y$-changes into a planar diagonal; then combine that diagonal with the $z$-change.

Derivation The planar displacement has length $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. A second right triangle therefore gives $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}.$$

Application Set the squared distance equal to 36: $$(t-1)^2+(2t+2)^2+(-4)^2=36.$$ This becomes $5t^2+6t-15=0$, so $$\boxed{t=\frac{-3\pm 2\sqrt{21}}{5}}.$$

Verification Substitution into the quadratic recovers squared distance 36; both values are valid because distance is nonnegative.