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].
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