Vector addition and the section formula

The two basic vector operations are addition and scalar multiplication. Geometrically, they correspond to "putting arrows head-to-tail" and "stretching/reversing arrows." Algebraically, both are componentwise.

Addition

Given $\vec{a}$ and $\vec{b}$, the sum $\vec{a} + \vec{b}$ is constructed by placing the tail of $\vec{b}$ at the head of $\vec{a}$. The arrow from the tail of $\vec{a}$ to the head of $\vec{b}$ is $\vec{a} + \vec{b}$ , the triangle law.

Equivalently, drawing both from the same tail and completing the parallelogram, the diagonal is $\vec{a} + \vec{b}$ , the parallelogram law.

Algebraically: $(a_1, a_2, a_3) + (b_1, b_2, b_3) = (a_1 + b_1, a_2 + b_2, a_3 + b_3)$.

Properties:

• $\vec{a} + \vec{b} = \vec{b} + \vec{a}$ (commutative)
• $(\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c})$ (associative)
• $\vec{a} + \vec{0} = \vec{a}$
• $\vec{a} + (-\vec{a}) = \vec{0}$

Subtraction

$\vec{a} - \vec{b} = \vec{a} + (-\vec{b})$. Geometrically: the vector from the head of $\vec{b}$ to the head of $\vec{a}$ (when both start at the same tail).

Scalar multiplication

For a scalar $\lambda \in \mathbb{R}$: $\lambda \vec{a} = (\lambda a_1, \lambda a_2, \lambda a_3).$ If $\lambda > 0$, $\lambda \vec{a}$ has the same direction as $\vec{a}$ and magnitude $\lambda|\vec{a}|$. If $\lambda < 0$, the direction reverses. If $\lambda = 0$, the result is $\vec{0}$.

Distributive: $\lambda(\vec{a} + \vec{b}) = \lambda \vec{a} + \lambda \vec{b}$.

The section formula

Given two points $A$ and $B$ with position vectors $\vec{a}$ and $\vec{b}$, a point $P$ on segment $AB$ divides it in ratio $m : n$ (internally) if $AP : PB = m : n$. The position vector of $P$ is $\vec{p} = \frac{m \vec{b} + n \vec{a}}{m + n}.$ For external division ($P$ outside the segment, with $AP : PB = m : n$ where one is "beyond"), the formula becomes $\vec{p} = \frac{m \vec{b} - n \vec{a}}{m - n}.$ Midpoint ($m = n = 1$, internal): $\vec{p} = \dfrac{\vec{a} + \vec{b}}{2}$.

Derivation of the section formula

Let $P$ divide $AB$ internally in ratio $m : n$, so $AP = \dfrac{m}{m+n}AB$. Then $\vec{p} = \vec{a} + \vec{AP} = \vec{a} + \frac{m}{m+n}(\vec{b} - \vec{a}) = \frac{n\vec{a} + m\vec{b}}{m+n}.$

Worked examples

Example 1. $\vec{a} = \hat{i} + 2\hat{j}$, $\vec{b} = 3\hat{i} - \hat{j}$. Find $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$.

$\vec{a} + \vec{b} = 4\hat{i} + \hat{j}$. $\vec{a} - \vec{b} = -2\hat{i} + 3\hat{j}$.

Example 2. Find the midpoint of $A = (1, 2, 3)$ and $B = (5, 6, -1)$.

Midpoint $= \dfrac{A + B}{2} = (3, 4, 1)$.

Example 3. Find the point dividing the segment from $A = (1, -2, 4)$ to $B = (7, 10, -2)$ in ratio $2 : 3$ internally.

By the section formula: $P = \dfrac{2B + 3A}{5} = \dfrac{(14, 20, -4) + (3, -6, 12)}{5} = \dfrac{(17, 14, 8)}{5} = (3.4, 2.8, 1.6)$.

Example 4. A point $P$ divides $AB$ externally in ratio $3 : 1$, where $A = (0, 0, 0)$ and $B = (4, 6, 8)$. Find $P$.

External formula: $P = \dfrac{3 B - 1 \cdot A}{3 - 1} = \dfrac{3(4, 6, 8)}{2} = (6, 9, 12)$.

Example 5. Show that the medians of a triangle are concurrent at the centroid $G = \dfrac{\vec{a} + \vec{b} + \vec{c}}{3}$.

Let triangle $ABC$ have position vectors $\vec{a}, \vec{b}, \vec{c}$. Midpoint of $BC$ is $M_A = \dfrac{\vec{b} + \vec{c}}{2}$. The median from $A$ to $M_A$ has parametric form $\vec{a} + t(M_A - \vec{a})$ for $t \in [0, 1]$. At $t = 2/3$: $\vec{a} + \dfrac{2}{3}(M_A - \vec{a}) = \dfrac{\vec{a}}{3} + \dfrac{2}{3}M_A = \dfrac{\vec{a}}{3} + \dfrac{\vec{b} + \vec{c}}{3} = \dfrac{\vec{a} + \vec{b} + \vec{c}}{3}$. Symmetric for the other medians, so all three pass through $G$.

Example 6. If $P$ divides segment from $(2, 1)$ to $(8, 9)$ in ratio $1 : 2$, find $P$.

$P = \dfrac{1 \cdot (8, 9) + 2 \cdot (2, 1)}{3} = \dfrac{(8, 9) + (4, 2)}{3} = \dfrac{(12, 11)}{3} = (4, 11/3)$.

Try it yourself

1. Given $\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}$, $\vec{b} = \hat{i} + \hat{j} - 2\hat{k}$. Find $2\vec{a} - 3\vec{b}$.
2. Midpoint of $(2, -1, 5)$ and $(-4, 3, 1)$.
3. A point divides the segment from $(1, 2)$ to $(7, 11)$ in ratio $2 : 1$ internally. Find it.
4. A point divides the segment from $(1, 2)$ to $(7, 11)$ in ratio $2 : 1$ externally. Find it.
5. Find $\lambda$ if $\lambda(2\hat{i} + \hat{j}) = 4\hat{i} + 2\hat{j}$.
6. If $A = (1, 2)$, $B = (5, 6)$, and $C$ divides $AB$ in ratio $1 : 3$, find $C$.
7. The position vectors of $A, B$ are $\hat{i} + 2\hat{j}$ and $3\hat{i} - 4\hat{j}$. Find $\vec{AB}$.
8. Three forces $\vec{F}_1, \vec{F}_2, \vec{F}_3$ have position vectors $(1, 2)$, $(3, -1)$, $(-4, -1)$. Find their resultant.
9. The centroid of a triangle with vertices $(2, 3, 4)$, $(-1, 0, 5)$, $(2, 6, 0)$.
10. Vertices of a triangle: $A = (1, 1, 1)$, $B = (3, 3, 0)$, $C = (5, 5, 2)$. Show $A, B, C$ are collinear.
11. Show that the diagonals of a parallelogram bisect each other.
12. If $\vec{a}, \vec{b}, \vec{c}$ are position vectors of vertices of a triangle, find the position vector of the orthocentre.
13. Find $\vec{c}$ such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, where $\vec{a} = 2\hat{i} + \hat{j}$, $\vec{b} = \hat{i} - 3\hat{j}$.
14. Two ships are at positions $\vec{r}_1 = 3\hat{i} + 4\hat{j}$ and $\vec{r}_2 = -\hat{i} + 7\hat{j}$. Find the displacement from ship 1 to ship 2.

Pitfalls and tricks

• Section formula sign convention. For internal division $\vec{p} = \dfrac{m\vec{b} + n\vec{a}}{m+n}$; the larger part of the ratio multiplies the farther point.
• External division uses minus. Easy to confuse with internal; check whether $P$ is between $A$ and $B$ or outside.
• Vector addition is commutative, but the ordering matters for visualisation (head-to-tail).
• Three points collinear iff one of the position-vector differences is a scalar multiple of another.
• Centroid is the average of the three vertices' position vectors , handy in proofs.