Vector Proof I — Algebraic Techniques
Vectors turn geometry into algebra. A midpoint becomes an average of position vectors; collinearity becomes "one vector is a scalar multiple of another"; perpendicularity becomes "their dot product is zero". This lesson builds the four algebraic moves every Extension 2 vector proof depends on — and the habit of writing each line so the next one is forced.
If $M$ is the midpoint of segment $AB$, write the position vector of $M$ in terms of $\mathbf{a}$ and $\mathbf{b}$. Before checking — and how would you express the point that divides $AB$ in the ratio $2:3$ (from $A$ to $B$)?
Every vector proof rewards two habits: translate the geometry into position vectors (label every named point with a single letter), then pick the right algebraic test — midpoint average, scalar multiple for collinearity, dot product zero for perpendicularity. The right test makes the algebra collapse to one or two lines.
The translate-test-conclude reading: (1) write every named point as a position vector $\mathbf{a}, \mathbf{b}, \mathbf{c}, \ldots$, (2) compute the relevant vector ($\overrightarrow{XY} = \mathbf{y} - \mathbf{x}$), (3) apply the algebraic test for the geometric claim.
Midpoint: $\mathbf{m} = \tfrac{1}{2}(\mathbf{a}+\mathbf{b})$ · Collinear: $\overrightarrow{XY} = k\,\overrightarrow{XZ}$ · Perpendicular: $\overrightarrow{XY}\cdot\overrightarrow{XZ} = 0$
Key facts
- Midpoint formula: $\mathbf{m} = \tfrac{1}{2}(\mathbf{a}+\mathbf{b})$
- Division ratio $k:(1-k)$ from $A$: $(1-k)\mathbf{a} + k\mathbf{b}$
- Collinearity criterion: $\overrightarrow{XY} = k\,\overrightarrow{XZ}$
- Perpendicularity criterion: $\overrightarrow{XY}\cdot\overrightarrow{XZ} = 0$
Concepts
- Why position-vector algebra is equivalent to coordinate geometry
- Why scalar-multiple equivalence captures "parallel and on a common line"
- Why the dot product being zero captures perpendicularity
Skills
- Prove a midpoint coincidence by computing both as $\tfrac{1}{2}(\text{ends})$
- Prove collinearity by exhibiting a scalar $k$
- Prove perpendicularity by expanding $\overrightarrow{XY}\cdot\overrightarrow{XZ}$ and showing it is zero
Place the origin anywhere. Let $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$. The four facts below underwrite most algebraic proofs in vector geometry.
- Midpoint of $AB$: $\mathbf{m} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b})$.
- Ratio $\lambda:\mu$ from $A$ to $B$: $\mathbf{p} = \dfrac{\mu\mathbf{a} + \lambda\mathbf{b}}{\lambda+\mu}$ (section formula).
- Parallel vectors: $\mathbf{u} \parallel \mathbf{v}$ iff $\mathbf{u} = k\mathbf{v}$ for some scalar $k$.
- Collinear points: $X, Y, Z$ collinear iff $\overrightarrow{XY} = k\,\overrightarrow{XZ}$.
Worked through the hook: Triangle $ABC$ with $M$ midpoint of $AB$ and $N$ midpoint of $AC$.
- $\mathbf{m} = \tfrac{1}{2}(\mathbf{a}+\mathbf{b})$, $\mathbf{n} = \tfrac{1}{2}(\mathbf{a}+\mathbf{c})$.
- $\overrightarrow{MN} = \mathbf{n} - \mathbf{m} = \tfrac{1}{2}(\mathbf{c} - \mathbf{b}) = \tfrac{1}{2}\overrightarrow{BC}$.
- Conclusion: $MN \parallel BC$ and $|MN| = \tfrac{1}{2}|BC|$ — the midpoint theorem, proven in two lines.
Four core moves: midpoint average, section formula, scalar multiple = parallel, dot product 0 = perpendicular · Midpoint theorem proof: $\overrightarrow{MN} = \tfrac{1}{2}\overrightarrow{BC}$ in two lines · $\overrightarrow{XY} = \mathbf{y} - \mathbf{x}$ regardless of origin
Pause — copy the four core moves (midpoint, section formula, scalar multiple, dot product), the midpoint-theorem proof $\overrightarrow{MN} = \tfrac{1}{2}\overrightarrow{BC}$, and the formula $\overrightarrow{XY} = \mathbf{y}-\mathbf{x}$ into your book.
Quick check: Points $A$, $B$, $C$ have position vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$. $M$ is the midpoint of $BC$. Which position vector is $\overrightarrow{AM}$?
We just saw the four core proof moves: midpoint average, section formula, scalar multiple = parallel, dot product 0 = perpendicular, applied to the midpoint theorem $\overrightarrow{MN} = \tfrac{1}{2}\overrightarrow{BC}$. That raises a question: how do we test collinearity and perpendicularity rigorously using vectors? This card answers it → collinearity: find $k$ with $\overrightarrow{XY} = k\overrightarrow{XZ}$; perpendicularity: expand $\overrightarrow{XY}\cdot\overrightarrow{XZ}$ using bilinearity and show it equals 0.
To prove three named points are collinear or two segments perpendicular, write the relevant displacement vectors and apply the appropriate algebraic test:
- Collinearity of $X, Y, Z$: compute $\overrightarrow{XY}$ and $\overrightarrow{XZ}$. Show $\overrightarrow{XY} = k\,\overrightarrow{XZ}$ by extracting a common factor. If $k$ exists, the points are collinear.
- Perpendicularity of $XY$ and $XZ$ (or any other pair): compute the dot product $\overrightarrow{XY}\cdot\overrightarrow{XZ}$, expand using bilinearity, simplify, and show it equals $0$.
Collinearity: find $k$ such that $\overrightarrow{XY} = k\,\overrightarrow{XZ}$ · Perpendicularity: expand $\overrightarrow{XY}\cdot\overrightarrow{XZ}$ using bilinearity, show = 0 · Dot product algebra: $(\mathbf{u}+\mathbf{v})\cdot(\mathbf{w}+\mathbf{x}) = \mathbf{u}\cdot\mathbf{w} + \mathbf{u}\cdot\mathbf{x} + \mathbf{v}\cdot\mathbf{w} + \mathbf{v}\cdot\mathbf{x}$
Pause — copy the collinearity test ($\overrightarrow{XY} = k\overrightarrow{XZ}$ for some $k$), the perpendicularity test (expand $\overrightarrow{XY}\cdot\overrightarrow{XZ}$ using bilinearity and show $= 0$), and the full expansion law into your book.
Did you get this? True or false: three distinct points $X, Y, Z$ are collinear if and only if there exists a scalar $k$ (possibly $k = 0$ or $k$ negative) such that $\overrightarrow{XY} = k\,\overrightarrow{XZ}$.
Worked examples · 3 in a row, reveal as you go
$ABCD$ is a quadrilateral. Let $P$, $Q$, $R$, $S$ be midpoints of $AB$, $BC$, $CD$, $DA$ respectively. Prove that the midpoint of $PR$ coincides with the midpoint of $QS$.
Points $X$, $Y$, $Z$ have position vectors $\mathbf{x} = (1, 2, 1)$, $\mathbf{y} = (3, 5, 4)$, $\mathbf{z} = (7, 11, 10)$. Prove that $X$, $Y$, $Z$ are collinear.
In a rhombus $ABCD$ with $|\overrightarrow{AB}| = |\overrightarrow{AD}|$, prove that the diagonals $AC$ and $BD$ are perpendicular.
Fill the gap: The point that divides interval $AB$ in the ratio $\lambda:\mu$ from $A$ to $B$ has position vector $\mathbf{p} = \dfrac{\mu\mathbf{a} + \lambda\mathbf{b}}{\lambda + \mu}$. In particular, the midpoint corresponds to $\lambda = \mu = $ , giving $\mathbf{m} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})$, the of the endpoints.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: if $\mathbf{u}\cdot\mathbf{v} = 0$ and $\mathbf{u}, \mathbf{v}$ are both nonzero, then $\mathbf{u}$ and $\mathbf{v}$ are perpendicular.
Activities · practice with the ideas
In triangle $ABC$, let $M$ be the midpoint of $BC$. Prove that $\overrightarrow{AM} = \tfrac{1}{2}(\overrightarrow{AB} + \overrightarrow{AC})$.
Points $X = (1, 0, 2)$, $Y = (3, 2, 6)$, $Z = (6, 5, 12)$. Determine whether $X$, $Y$, $Z$ are collinear.
$P$ divides $AB$ in the ratio $1:3$ from $A$. Express $\overrightarrow{OP}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
In a square $ABCD$, let $\mathbf{u} = \overrightarrow{AB}$ and $\mathbf{v} = \overrightarrow{AD}$ with $|\mathbf{u}| = |\mathbf{v}|$ and $\mathbf{u}\cdot\mathbf{v} = 0$. Prove that the diagonals $AC$ and $BD$ are equal in length.
Prove the centroid of a triangle $ABC$ has position vector $\mathbf{g} = \tfrac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})$, and lies $2/3$ of the way along the median from $A$.
Odd one out: Three of these correctly express the position vector of the point $P$ that divides interval $AB$ internally in the ratio $\lambda:\mu$ from $A$. Which one is NOT?
Earlier you computed $\overrightarrow{MN}$ and $\overrightarrow{BC}$ where $M$, $N$ are midpoints of $AB$, $AC$.
$\overrightarrow{MN} = \mathbf{n} - \mathbf{m} = \tfrac{1}{2}(\mathbf{a}+\mathbf{c}) - \tfrac{1}{2}(\mathbf{a}+\mathbf{b}) = \tfrac{1}{2}(\mathbf{c} - \mathbf{b}) = \tfrac{1}{2}\overrightarrow{BC}$. The algebra forces two conclusions: $MN \parallel BC$ (scalar multiple) and $|MN| = \tfrac{1}{2}|BC|$ (magnitude of the scalar). This is the midpoint theorem — and the entire proof is two lines of vector algebra, with no diagram-chasing required. That economy is why vector proofs scale so well to 3D and harder problems.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Points $A, B, C$ have position vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$. $M$ is the midpoint of $AB$. Write $\overrightarrow{CM}$ in terms of $\mathbf{a}, \mathbf{b}, \mathbf{c}$. (2 marks)
Q2. Points $X = (1, 1, 1)$, $Y = (4, 3, 5)$, $Z = (10, 7, 13)$. Prove that $X$, $Y$, $Z$ are collinear and state in what ratio $Y$ divides $XZ$. (3 marks)
Q3. $OABC$ is a parallelogram with $\overrightarrow{OA} = \mathbf{u}$ and $\overrightarrow{OC} = \mathbf{v}$. Show that $\overrightarrow{OB}\cdot\overrightarrow{AC} = |\mathbf{v}|^2 - |\mathbf{u}|^2$. Hence give a condition on $\mathbf{u}, \mathbf{v}$ for the diagonals to be perpendicular. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $\mathbf{m} = \tfrac{1}{2}(\mathbf{b}+\mathbf{c})$. $\overrightarrow{AM} = \mathbf{m} - \mathbf{a} = \tfrac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a} = \tfrac{1}{2}((\mathbf{b} - \mathbf{a}) + (\mathbf{c} - \mathbf{a})) = \tfrac{1}{2}(\overrightarrow{AB} + \overrightarrow{AC})$.
2. $\overrightarrow{XY} = (2, 2, 4)$, $\overrightarrow{XZ} = (5, 5, 10) = \tfrac{5}{2}\overrightarrow{XY}$. All component ratios equal $5/2$, so $\overrightarrow{XZ}$ is a scalar multiple of $\overrightarrow{XY}$ and $X, Y, Z$ are collinear.
3. Using the section formula with $\lambda:\mu = 1:3$: $\mathbf{p} = \dfrac{3\mathbf{a} + 1\mathbf{b}}{1 + 3} = \dfrac{3\mathbf{a} + \mathbf{b}}{4}$.
4. $|\overrightarrow{AC}|^2 = (\mathbf{u}+\mathbf{v})\cdot(\mathbf{u}+\mathbf{v}) = |\mathbf{u}|^2 + 2\mathbf{u}\cdot\mathbf{v} + |\mathbf{v}|^2$. $|\overrightarrow{BD}|^2 = (\mathbf{v}-\mathbf{u})\cdot(\mathbf{v}-\mathbf{u}) = |\mathbf{v}|^2 - 2\mathbf{u}\cdot\mathbf{v} + |\mathbf{u}|^2$. Square gives $\mathbf{u}\cdot\mathbf{v} = 0$, so both equal $|\mathbf{u}|^2 + |\mathbf{v}|^2$. Hence $|\overrightarrow{AC}| = |\overrightarrow{BD}|$.
5. $M = \tfrac{1}{2}(\mathbf{b}+\mathbf{c})$. $G$ divides $AM$ in ratio $2:1$ from $A$, so $\mathbf{g} = \mathbf{a} + \tfrac{2}{3}(\mathbf{m} - \mathbf{a}) = \tfrac{1}{3}\mathbf{a} + \tfrac{2}{3}\mathbf{m} = \tfrac{1}{3}\mathbf{a} + \tfrac{1}{3}(\mathbf{b}+\mathbf{c}) = \tfrac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})$.
Q1 (2 marks): $\mathbf{m} = \tfrac{1}{2}(\mathbf{a}+\mathbf{b})$ [1]; $\overrightarrow{CM} = \mathbf{m} - \mathbf{c} = \tfrac{1}{2}(\mathbf{a}+\mathbf{b}-2\mathbf{c})$ [1].
Q2 (3 marks): $\overrightarrow{XY} = (3, 2, 4)$, $\overrightarrow{XZ} = (9, 6, 12) = 3\overrightarrow{XY}$ [1]; so $\overrightarrow{XZ}$ is a scalar multiple of $\overrightarrow{XY}$ — points collinear [1]. Since $\overrightarrow{XY} = \tfrac{1}{3}\overrightarrow{XZ}$, $Y$ divides $XZ$ in the ratio $1:2$ from $X$ [1].
Q3 (3 marks): $\overrightarrow{OB} = \mathbf{u} + \mathbf{v}$ (parallelogram rule) and $\overrightarrow{AC} = \mathbf{v} - \mathbf{u}$ [1]. $\overrightarrow{OB}\cdot\overrightarrow{AC} = (\mathbf{u}+\mathbf{v})\cdot(\mathbf{v}-\mathbf{u}) = \mathbf{u}\cdot\mathbf{v} - |\mathbf{u}|^2 + |\mathbf{v}|^2 - \mathbf{v}\cdot\mathbf{u} = |\mathbf{v}|^2 - |\mathbf{u}|^2$ [1]. Diagonals perpendicular $\Leftrightarrow$ $\overrightarrow{OB}\cdot\overrightarrow{AC} = 0$ $\Leftrightarrow$ $|\mathbf{u}| = |\mathbf{v}|$, i.e. the parallelogram is a rhombus [1].
Five timed questions on midpoints, section formulas, collinearity and perpendicularity proofs. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
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