Dividing Lines in a Given Ratio

Video lesson · from 26:101 worked example

One lesson video covers all of Vectors, so it opens at 26:10 for this topic — not from the beginning.

If a point T divides a line segment PR in the ratio $m:n$ , we can use position vectors to find the coordinates of T.

From the ratio, $\vec{PT} = \frac{m}{m+n} \vec{PR}$ or we can use the ratio of parts: $n\vec{PT} = m\vec{TR}$ .

Expect this in three dimensions: finding an internal division point in 3D is a known weak spot, and it is where this topic is usually examined.
Ratio parts, not the ratio itself: for $m:n$ the fraction along the line is $\frac{m}{m+n}$ . Using $\frac{m}{n}$ is the standard error.
Which end are you starting from? $T$ dividing $PR$ in $2:1$ is not the same point as $1:2$ . Check the order of the letters in the question.
Add the position vector back on: $\vec{PT}$ is a step, not a position. The coordinates of $T$ come from $\vec{OT} = \vec{OP} + \vec{PT}$ .

Example 1

P is the point $(-2, 4, -1)$ and R is the point $(8, -1, 19)$ . The point T divides PR in the ratio $2:3$ . Find the coordinates of T.

Since T divides PR in ratio $2:3$ , we have $\frac{PT}{TR} = \frac{2}{3} \implies 3\vec{PT} = 2\vec{TR}$ .

\begin{aligned} 3(\vec{t} - \vec{p}) &= 2(\vec{r} - \vec{t}) \\ 3\vec{t} - 3\vec{p} &= 2\vec{r} - 2\vec{t} \\ 5\vec{t} &= 2\vec{r} + 3\vec{p} \end{aligned}

\begin{aligned} 5\vec{t} &= 2\begin{pmatrix} 8 \\ -1 \\ 19 \end{pmatrix} + 3\begin{pmatrix} -2 \\ 4 \\ -1 \end{pmatrix} \\ &= \begin{pmatrix} 16 \\ -2 \\ 38 \end{pmatrix} + \begin{pmatrix} -6 \\ 12 \\ -3 \end{pmatrix} \\ &= \begin{pmatrix} 10 \\ 10 \\ 35 \end{pmatrix} \end{aligned}

\vec{t} = \begin{pmatrix} 2 \\ 2 \\ 7 \end{pmatrix}

The coordinates of T are (2, 2, 7).