Section formula in 3D

A point $R$ that divides the segment from $P$ to $Q$ in a given ratio sits at known coordinates , just average each coordinate weighted by the ratio. The 3D section formula is the natural extension of the 2D one.

Internal division

If $R$ divides the line segment from $P(x_1, y_1, z_1)$ to $Q(x_2, y_2, z_2)$ internally in the ratio $m : n$, then $\boxed{R = \left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n}, \frac{m z_2 + n z_1}{m + n}\right).}$

This means $\overrightarrow{PR} : \overrightarrow{RQ} = m : n$.

External division

If $R$ divides $PQ$ externally in ratio $m : n$ (so $R$ lies on the line, outside the segment), replace $n$ with $-n$: $R = \left(\frac{m x_2 - n x_1}{m - n}, \frac{m y_2 - n y_1}{m - n}, \frac{m z_2 - n z_1}{m - n}\right).$

(Requires $m \ne n$.)

Midpoint

Setting $m = n = 1$ in the internal formula: $\boxed{M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right).}$

Centroid of a triangle

For triangle with vertices $A(x_1, y_1, z_1), B(x_2, y_2, z_2), C(x_3, y_3, z_3)$: $G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right).$

The centroid divides every median in the ratio $2 : 1$ (vertex to midpoint of opposite side).

Worked examples

Example 1. Find the coordinates of the point dividing the segment from $(1, 2, 3)$ to $(7, 14, 21)$ in the ratio $1 : 2$ internally.

$R = \left(\dfrac{1 \cdot 7 + 2 \cdot 1}{3}, \dfrac{1 \cdot 14 + 2 \cdot 2}{3}, \dfrac{1 \cdot 21 + 2 \cdot 3}{3}\right) = \left(3, 6, 9\right)$.

Example 2. Midpoint of $(2, -3, 4)$ and $(6, 5, -2)$.

$M = (4, 1, 1)$.

Example 3. Find the centroid of the triangle with vertices $(1, 2, 3), (4, -1, 2), (7, 5, -2)$.

$G = \left(\dfrac{12}{3}, \dfrac{6}{3}, \dfrac{3}{3}\right) = (4, 2, 1)$.

Example 4. In what ratio does the $xy$-plane divide the segment from $(2, 3, -4)$ to $(6, -1, 8)$?

The $xy$-plane has $z = 0$. The intersection point has $z = 0$, so by section formula in $z$: $0 = \dfrac{m \cdot 8 + n \cdot (-4)}{m + n} \Rightarrow 8 m = 4 n \Rightarrow m : n = 1 : 2$.

Example 5. Find $R$ dividing $A(1, 0, 1)$ and $B(4, 3, -2)$ externally in ratio $2 : 1$.

$R = \left(\dfrac{2 \cdot 4 - 1 \cdot 1}{1}, \dfrac{2 \cdot 3 - 1 \cdot 0}{1}, \dfrac{2 \cdot (-2) - 1 \cdot 1}{1}\right) = (7, 6, -5)$.

Try it yourself

1. Point dividing $(0, 0, 0)$ and $(6, 9, 12)$ in ratio $1 : 2$.
2. Midpoint of $(-1, 3, 5)$ and $(5, -3, -3)$.
3. Centroid of $(2, 4, 6), (0, 0, 0), (-2, -4, -6)$.
4. Find the ratio in which $y$-axis divides the segment from $(2, 3, 4)$ to $(-3, 5, 6)$. (Note: $y$-axis means $x = 0$ and $z = 0$ , only intersection at one point, if any.)
5. Find the ratio in which the $xy$-plane divides the line joining $(1, -2, 3)$ and $(3, 4, -5)$.
6. The midpoint of a segment is $(3, 4, 5)$. One endpoint is $(1, 2, 3)$. Find the other.
7. Three vertices of a parallelogram are $A(3, -1, 2), B(1, 2, -4), C(-1, 1, 2)$. Find $D$. (Hint: diagonals bisect each other.)
8. The centroid of a triangle is $(1, 1, 1)$. Two vertices are $(2, 3, 1)$ and $(-1, 0, 2)$. Find the third.
9. Find the coordinates of the point trisecting the segment from $(1, 1, 1)$ to $(7, 7, 7)$ (closer to first endpoint).
10. Find the point on the $xz$-plane that divides the segment from $(2, 3, -1)$ to $(4, -3, 5)$.
11. If the points $(1, 2, 3), (3, 4, 5), (5, 6, 7)$ are vertices of a triangle, find its centroid.
12. In what ratio does the point $(-1, 0, 2)$ divide the segment from $(1, 2, 4)$ to $(-3, -2, 0)$?

Pitfalls / Tricks

• Order matters: the formula gives the point that divides from $P$ to $Q$ in the ratio $m : n$. Switching $P$ and $Q$ swaps $m$ and $n$.
• For external division, $m \ne n$ (else the formula has zero denominator).
• The centroid is the average of vertices , extends naturally from 2D.
• Insight. Section in 3D is "do the 2D formula three times". Once you internalise the 2D, 3D is just bookkeeping.