3D Geometry: Numerical Problems Set (13)

Question

Find the distance from the point $P(1, 2, 3)$ to the plane $2x - 3y + 6z = 5$.

Solution — Step by Step

Why This Works

The formula comes from projecting the vector from any point on the plane to $P$ onto the unit normal of the plane. The normal vector to $ax + by + cz = d$ is $(a, b, c)$, and dividing by its magnitude gives the unit normal.

The absolute value handles the case where the point is on either side of the plane — distance is always non-negative.

Alternative Method

Vector form. Plane: $\vec{r} \cdot \vec{n} = d$ with $\vec{n} = (2, -3, 6)$. Position vector of $P$: $\vec{p} = (1, 2, 3)$.

$D = \frac{|\vec{p}\cdot\vec{n} - d|}{|\vec{n}|} = \frac{|9|}{|7|} = \frac{9}{7}$

Same.

Common Mistake

Students often forget the absolute value and get a negative distance, then panic. Distance is a magnitude — always positive. The sign of $ax_1 + by_1 + cz_1 - d$ tells you which side of the plane the point is on, not the distance.

Another trap: writing the plane equation as $2x - 3y + 6z + 5 = 0$ (moving $5$ to the LHS with sign flip) and then plugging $d = 5$ — that double-counts. Either keep the form $ax + by + cz = d$ with $d = 5$, or use $ax + by + cz + d' = 0$ with $d' = -5$ and compute $|ax_1 + by_1 + cz_1 + d'|$.