Ratio theorem

Ratio theorem

A line segment can be divided internally or externally by a point in a given ratio.

Derivation of the ratio theorem for internal division

Let A and B have position vectors $\mathbf{a}$ and $\mathbf{b}$, respectively. Let $\mathbf{r}$ be the position vector of point P dividing AB internally in the ratio $\lambda : \mu$.

From $\triangle OAP$: $\overrightarrow{OA} + \overrightarrow{AP} = \overrightarrow{OP}$, so $\overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{OA} = \mathbf{r} - \mathbf{a}$ (2.1)

From $\triangle OPB$: $\overrightarrow{OP} + \overrightarrow{PB} = \overrightarrow{OB}$, so $\overrightarrow{PB} = \overrightarrow{OB} - \overrightarrow{OP} = \mathbf{b} - \mathbf{r}$ (2.2)

Taking the ratio of equations (2.1) and (2.2):

$\frac{\overrightarrow{AP}}{\overrightarrow{PB}} = \frac{\mathbf{r} - \mathbf{a}}{\mathbf{b} - \mathbf{r}} = \frac{\lambda}{\mu}$

$\mu(\mathbf{r} - \mathbf{a}) = \lambda(\mathbf{b} - \mathbf{r})$

$\mu\mathbf{r} - \mu\mathbf{a} = \lambda\mathbf{b} - \lambda\mathbf{r}$

$\mu\mathbf{r} + \lambda\mathbf{r} = \lambda\mathbf{b} + \mu\mathbf{a}$

$(\lambda + \mu)\mathbf{r} = \mu\mathbf{a} + \lambda\mathbf{b}$

$\mathbf{r} = \frac{\mu\mathbf{a} + \lambda\mathbf{b}}{\lambda + \mu}$

The ratio theorem for internal division states that the position vector of a point P dividing AB internally in the ratio $\lambda : \mu$ is $\mathbf{r} = \frac{\mu\mathbf{a} + \lambda\mathbf{b}}{\lambda + \mu}$, where $\mathbf{a}$ and $\mathbf{b}$ are the position vectors of A and B, respectively.

The key difference between internal and external division is that in internal division, the point P lies on the line segment AB. In external division, P lies on the extension of the line segment AB.

Ratio theorem examples

Example 1

Two points A and B have position vectors $\mathbf{a} = -3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$ and $\mathbf{b} = 4\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}$, respectively. Find the position vector of a point that divides AB internally in the ratio 1:4.

Solution:

Given $\mathbf{a} = -3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$, $\mathbf{b} = 4\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}$, and $\lambda : \mu = 1:4$, the position vector $\mathbf{r}$ of the dividing point is:

$\mathbf{r} = \frac{\mu\mathbf{a} + \lambda\mathbf{b}}{\lambda + \mu} = \frac{4(-3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) + 1(4\mathbf{i} + 3\mathbf{j} - 4\mathbf{k})}{1 + 4}$

$\mathbf{r} = \frac{-12\mathbf{i} + 8\mathbf{j} + 12\mathbf{k} + 4\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}}{5} = \frac{-8\mathbf{i} + 11\mathbf{j} + 8\mathbf{k}}{5}$

$\mathbf{r} = -\frac{8}{5}\mathbf{i} + \frac{11}{5}\mathbf{j} + \frac{8}{5}\mathbf{k}$

Example 2

Let $\mathbf{a} = -7\mathbf{i} - 2\mathbf{j}$ and $\mathbf{b} = -2\mathbf{i} + 3\mathbf{j} + \mathbf{k}$ be the position vectors of points A and B, respectively. Find the position vector of a point that divides line segment AB externally in the ratio 5:2.

Solution:

Given $\mathbf{a} = -7\mathbf{i} - 2\mathbf{j}$, $\mathbf{b} = -2\mathbf{i} + 3\mathbf{j} + \mathbf{k}$, and $\lambda : \mu = 5:2$, the position vector $\mathbf{r}$ for external division is:

$\mathbf{r} = \frac{\mu\mathbf{a} - \lambda\mathbf{b}}{\mu - \lambda} = \frac{2(-7\mathbf{i} - 2\mathbf{j}) - 5(-2\mathbf{i} + 3\mathbf{j} + \mathbf{k})}{2 - 5}$

$\mathbf{r} = \frac{-14\mathbf{i} - 4\mathbf{j} + 10\mathbf{i} - 15\mathbf{j} - 5\mathbf{k}}{-3} = \frac{-4\mathbf{i} - 19\mathbf{j} - 5\mathbf{k}}{-3}$

$\mathbf{r} = \frac{4}{3}\mathbf{i} + \frac{19}{3}\mathbf{j} + \frac{5}{3}\mathbf{k}$

Example 3

The position vectors of points P and Q are $\mathbf{p}$ and $\mathbf{q}$, respectively.

(a) Find the position vector of point R that divides line segment PQ internally in the ratio 3:1.

(b) Find the position vector of point S that divides line segment PQ externally in the ratio 3:1.

(c) Show that $4\overrightarrow{RS} = 3\overrightarrow{PQ}$.

Solution:

(a) Internal division: $\mathbf{r} = \frac{1\mathbf{p} + 3\mathbf{q}}{3 + 1} = \frac{\mathbf{p} + 3\mathbf{q}}{4}$

(b) External division: $\mathbf{s} = \frac{1\mathbf{p} - 3\mathbf{q}}{1 - 3} = \frac{\mathbf{p} - 3\mathbf{q}}{-2} = \frac{3\mathbf{q} - \mathbf{p}}{2}$

(c) $\overrightarrow{RS} = \mathbf{s} - \mathbf{r} = \frac{3\mathbf{q} - \mathbf{p}}{2} - \frac{\mathbf{p} + 3\mathbf{q}}{4} = \frac{6\mathbf{q} - 2\mathbf{p} - \mathbf{p} - 3\mathbf{q}}{4} = \frac{3\mathbf{q} - 3\mathbf{p}}{4} = \frac{3}{4}(\mathbf{q} - \mathbf{p})$

$4\overrightarrow{RS} = 3(\mathbf{q} - \mathbf{p}) = 3\overrightarrow{PQ}$

Example 4

Show that the medians of a triangle trisect one another.

Solution:

Let the vertices of the triangle be A, B, and C with position vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$, respectively. Let M be the midpoint of BC. The position vector of M is $\mathbf{m} = \frac{\mathbf{b} + \mathbf{c}}{2}$.

Let G be the point that divides AM internally in the ratio 2:1. The position vector of G is:

$\mathbf{g} = \frac{1\mathbf{a} + 2\mathbf{m}}{2 + 1} = \frac{\mathbf{a} + 2(\frac{\mathbf{b} + \mathbf{c}}{2})}{3} = \frac{\mathbf{a} + \mathbf{b} + \mathbf{c}}{3}$

Since this expression is symmetric in $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$, the same point G would be obtained if we considered the medians from B or C. Therefore, G lies on all three medians and divides each in the ratio 2:1, meaning the medians trisect each other.