Vector Algebra · Lesson 4 of 9

Vectors Between Points and the Section Formula

“Use position-vector differences to describe segments, divide them in a ratio and investigate shapes.”

Learning Objectives

• Find a directed vector joining two points and calculate its length. • Derive the internal section formula using vector addition. • Apply external division and midpoint formulas with correct endpoint weights. • Use segment vectors and lengths to establish geometric relationships.

From two positions to a displacement

A position vector starts at the origin, but a displacement between two points usually does not. Suppose $p = O P$ and $q = O Q$ . Travelling from $O$ to $P$ and then to $Q$ has the same resultant as travelling directly from $O$ to $Q$ .

Vector joining two pointsLaTeX

P Q = q - p = (x_{2} - x_{1}) \hat{i} + (y_{2} - y_{1}) \hat{j} + (z_{2} - z_{1}) \hat{k}

P = (x_{1}, y_{1}, z_{1})

is the initial point and

Q = (x_{2}, y_{2}, z_{2})

is the terminal point. Subtract initial coordinates from terminal coordinates.

Difference of position vectors— OP + PQ = OQ, so PQ is the terminal position minus the initial position.

The length of this displacement is the distance between the points. Reversing the order of subtraction gives the opposite vector, while the distance remains unchanged. The component formula also supplies the direction ratios of the directed segment.

Distance between two pointsLaTeX

∣ P Q ∣ = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}

The squares make the distance independent of whether we calculate from

P

Q

or from

Q

P

Example — A directed displacement in space

Problem
Find the vector from $P (2, 3, 0)$ to $Q (- 1, - 2, - 4)$ and its magnitude.

1.Use terminal minus initial: $P Q = (- 1 - 2) \hat{i} + (- 2 - 3) \hat{j} + (- 4 - 0) \hat{k}$ .
2.Thus $P Q = - 3 \hat{i} - 5 \hat{j} - 4 \hat{k}$ . Its magnitude is $9 + 25 + 16 = 50 = 52$ .
3.The reverse vector is $3 \hat{i} + 5 \hat{j} + 4 \hat{k}$ . Its magnitude is also $52$ , but it describes a different directed displacement.

Example — Direction cosines of a segment

Problem
Find the direction cosines of the vector from $A (1, 2, - 3)$ to $B (- 1, - 2, 1)$ .

1.The directed vector is $A B = - 2 \hat{i} - 4 \hat{j} + 4 \hat{k}$ .
2.Its magnitude is $4 + 16 + 16 = 6$ .
3.The direction cosines are $(- 1/3, - 2/3, 2/3)$ . Their squared sum is $1/9 + 4/9 + 4/9 = 1$ . Reversing the segment would reverse all three signs.

Dividing a segment internally

Let a point $R$ lie between $P$ and $Q$ with $P R : R Q = m : n$ , where $m, n$ are positive. Internal division asks where $R$ must be, using this ratio and the endpoint positions. The ratio compares segment lengths; it does not directly name the coefficients of the two endpoint vectors.

Internal division— PR:RQ = m:n. R lies between P and Q.

Write $r = O R$ . The arrows $P R = r - p$ and $R Q = q - r$ point in the same direction. Since their lengths are in the ratio $m : n$ , multiplying the first by $n$ and the second by $m$ makes them equal.

Derivation of internal divisionLaTeX

n (r - p) n r - n p (m + n) r r = m (q - r) = m q - m r = n p + m q = \frac{n p + m q}{m + n}

The endpoint

P

receives weight

n

and

Q

receives weight

m

. This ordering follows from the vector equation, not from a memorised matching of letters.

For example, if $P R : R Q = 2 : 1$ , the point is closer to $Q$ . Its position is $(p + 2 q) /3$ , so the nearer endpoint receives the larger weight. Another useful check is to write $r = p + \frac{m}{m + n} (q - p)$ : move the correct fraction of the way from $P$ towards $Q$ .

External division and the midpoint

A point dividing a segment externally lies on its extended line, beyond one endpoint. The ratio is still a ratio of positive distances, but the two segment vectors need different orientation choices. This produces a subtraction in both the numerator and denominator.

External divisionLaTeX

r = \frac{m q - n p}{m - n}, m \neq = n

For

P R : R Q = m : n

externally,

m, n > 0

. If

m > n

R

lies beyond

Q

; if

m < n

, it lies beyond

P

To verify this formula without relying on a sketch, observe that $P R$ and $QR$ point in the same direction for either external position. Thus $n (r - p) = m (r - q)$ . Rearranging gives $(m - n) r = m q - n p$ . Equal positive distances would place a point at the midpoint, not outside the segment, so external division in the ratio $1 : 1$ has no finite solution.

Internal and external positions— For P at 0 and Q at 6, the ratio 2:1 gives an internal point at 4 and an external point at 12.

For a midpoint, the two internal distances are equal. Set the two weights equal in the internal formula. Each coordinate of the midpoint is consequently the average of the corresponding endpoint coordinates.

Midpoint position vectorLaTeX

r = \frac{p + q}{2}

This is the internal section formula with

m = n

Example — The same ratio in two different locations

Problem
If $p = 3 a - 2 b$ and $q = a + b$ , find the position dividing $P Q$ in the ratio 2:1 internally and externally.

1.Internally, $r = [p + 2 q] /3 = [3 a - 2 b + 2 a + 2 b] /3 = 5 a /3$ .
2.Externally, $s = [2 q - p] / (2 - 1) = 2 a + 2 b - 3 a + 2 b = - a + 4 b$ .
3.These formulas locate different points. The internal point is between the endpoints, while the external point lies beyond Q because the first ratio number is larger.

Using point vectors to investigate a shape

Coordinates let us investigate a geometric claim by calculating the relevant directed sides. If dot products have not yet been introduced, a right angle can be established by the converse of Pythagoras. Make sure to identify which side is longest, because that identifies the vertex of the right angle.

Example — A right angle from three position vectors

Problem
Show that $A (2, - 1, 1)$ , $B (1, - 3, - 5)$ and $C (3, - 4, - 4)$ form a right-angled triangle.

1.Calculate $A B = (- 1, - 2, - 6)$ , $B C = (2, - 1, 1)$ and $C A = (- 1, 3, 5)$ , with triples denoting components.
2.The squared lengths are $A B^{2} = 1 + 4 + 36 = 41$ , $B C^{2} = 4 + 1 + 1 = 6$ and $C A^{2} = 1 + 9 + 25 = 35$ .
3.Since $B C^{2} + C A^{2} = A B^{2}$ , the converse of Pythagoras gives a right angle opposite $A B$ , at $C$ . All three points are distinct and the two shorter sides have positive length.

Common mistake

For a vector joining two points, subtract initial position from terminal position. In the internal section formula, the weights cross over: the first distance ratio does not multiply the first endpoint position. Always check whether the requested point lies inside or outside the segment.

Quick check

If $P = (2, 1, 0)$ and $Q = (- 1, 5, 2)$ , then $P Q$ is

Quick check

If $P R : R Q = 3 : 2$ internally, the position of $R$ is

Quick check

Which external division ratio has no finite point?

Quick check

The midpoint of $(2, 3, 4)$ and $(4, 1, - 2)$ is

Quick check

If the squared side lengths are 6, 35 and 41, the right angle is

Practice Problems

Find the scalar and vector components of the displacement from $(2, 1)$ to $(- 5, 7)$ .
Find a unit vector directed from $P (1, 2, 3)$ to $Q (4, 5, 6)$ .
For $p = \hat{i} + 2 \hat{j} - \hat{k}$ and $q = - \hat{i} + \hat{j} + \hat{k}$ , find internal and external division positions for the ratio 2:1.
Find the midpoint of the segment joining $(2, 3, 4)$ and $(4, 1, - 2)$ . Verify it by taking differences from both endpoints.
Show that $A (1, - 2, - 8)$ , $B (5, 0, - 2)$ and $C (11, 3, 7)$ are collinear and determine $A B : B C$ .
Using $P = (0, 0, 0)$ and $Q = (6, 0, 0)$ , find the external division point for ratio 1:2. Explain why it lies beyond $P$ .
Explain why the vector identity for a closed three-point route does not, by itself, prove that the points form a nondegenerate triangle.

Key Takeaways

The vector from P to Q is the terminal position vector minus the initial one. Its magnitude is the distance between the points. Internal division uses a weighted average; external division uses a difference of weighted positions. The midpoint is the average of the endpoint position vectors. Vector calculations should finish with the geometric conclusion and the correct vertex or direction.

Previous · Lesson 3

Vector Components and Direction

Next · Lesson 5

The Dot Product and Angles

From two positions to a displacement

Dividing a segment internally

External division and the midpoint

Using point vectors to investigate a shape

Quiz

Practice Problems

Key Takeaways