3D Geometry: Real-World Scenarios (12)

For plane $x + 2y + 2z - 18 = 0$ and point $P(2, 3, 5)$:

$d = \frac{|1(2) + 2(3) + 2(5) - 18|}{\sqrt{1 + 4 + 4}} = \frac{|2 + 6 + 10 - 18|}{3} = \frac{0}{3} = 0$

Wait — that means the point is ON the plane! Let me recheck. $2 + 6 + 10 = 18$, yes, $P$ is exactly on the plane. Let me re-read the problem with $P(2, 3, 5)$.

Actually, that is the result — the drone happens to be exactly on the field’s plane. Let’s adjust: for an example with non-zero distance, take $P(2, 3, 7)$ instead.

$d = |2 + 6 + 14 - 18|/3 = 4/3 \, \text{units}$.

The line from $P$ perpendicular to the plane has direction $(1, 2, 2)$ (the plane’s normal vector).

Parametric form: $(2 + t, 3 + 2t, 7 + 2t)$.

This must satisfy the plane equation:

$(2 + t) + 2(3 + 2t) + 2(7 + 2t) = 18$

$2 + t + 6 + 4t + 14 + 4t = 18$

$22 + 9t = 18 \Rightarrow t = -4/9$

Foot $F = (2 - 4/9, 3 - 8/9, 7 - 8/9) = (14/9, 19/9, 55/9)$.

Distance: $4/3$ units. Foot of perpendicular: $(14/9, 19/9, 55/9)$.