Mathematics · Ch 8 — Vectors

Position Vector of a Point Dividing a Segment (Section Formula)

Internal Division

Let $A$ and $B$ be two points with position vectors $\vec a$ and $\vec b$ (relative to a fixed origin $O$ ), and let $R$ be the point on segment $AB$ that divides it internally in the ratio $m:n$ (so $AR:RB=m:n$ , with $R$ between $A$ and $B$ ).

Derivation. Since $R$ lies on $AB$ with $AR:RB=m:n$ , the vector $\vec{AR}$ is the fraction $\dfrac{m}{m+n}$ of the way along $\vec{AB}$ :

\vec{AR}=\frac{m}{m+n}\,\vec{AB}=\frac{m}{m+n}\left(\vec b-\vec a\right).

The position vector of $R$ is then

\vec r=\vec{OA}+\vec{AR}=\vec a+\frac{m}{m+n}(\vec b-\vec a)=\frac{(m+n)\vec a+m\vec b-m\vec a}{m+n}=\frac{m\vec b+n\vec a}{m+n}.

\vec r=\frac{m\vec b+n\vec a}{m+n}

Midpoint as a Special Case

Taking $m=n=1$ (ratio $1:1$ ) gives the midpoint of $AB$ :

\vec r=\frac{\vec a+\vec b}{2}.

External Division

If $R$ divides $AB$ externally in the ratio $m:n$ (so $R$ lies on line $AB$ produced, outside the segment, with $AR:RB=m:n$ but $R$ not between $A$ and $B$ ), the same reasoning with a subtraction in place of the internal split gives

\vec r=\frac{m\vec b-n\vec a}{m-n}\qquad(m\neq n)

External division can be remembered as "the internal-division formula with the sign of $n$ (and the denominator) flipped" -- algebraically, dividing externally in ratio $m:n$ is the same as dividing internally in ratio $m:(-n)$ .

In Coordinate (Component) Form

If $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ , the point $R$ dividing $AB$ internally in ratio $m:n$ has coordinates

R=\left(\frac{mx_2+nx_1}{m+n},\ \frac{my_2+ny_1}{m+n},\ \frac{mz_2+nz_1}{m+n}\right),

obtained by applying the vector section formula separately to each of the $\hat i,\hat j,\hat k$ components.

Important

The order matters: $m$ (the part nearer $B$ ) multiplies $\vec b$ , and $n$ (the part nearer $A$ ) multiplies $\vec a$ -- it is easy to swap these by mistake. A quick check is that when $m=n$ the formula must reduce to the simple average (midpoint) of $\vec a$ and $\vec b$ .

Applications …

Start your 14-day free trial to unlock the full solution →

← Previous · 5Addition of Vectors and Scalar Multiplication Next · 7 →Scalar (Dot) Product of Vectors