Unit 6 · Plane Vectors　平面向量

By the end of this chapter you can:

1. Add, subtract, and scalar-multiply vectors using both the geometric (triangle/parallelogram) rule and coordinate arithmetic
2. Compute the dot product $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$ and apply the perpendicularity shortcut $\vec{a} \cdot \vec{b} = 0$
3. Determine whether two vectors are collinear using the scalar condition $\vec{a} = \lambda \vec{b}$ or the coordinate cross-test $x_1 y_2 = x_2 y_1$
4. Solve exam problems that mix coordinate calculation, angle-finding, and geometric proof with vectors

Exam weight on past CSCA papers: ~8% (3–4 of 48 MCQs).

6.1　Vector Operations　向量的运算

A vector (向量) is a quantity with both magnitude (size) and direction. We write $\vec{a}$ or $\overrightarrow{AB}$ (from point $A$ to point $B$). A plain number with no direction is a scalar (标量).

Key vocabulary

Addition: triangle rule and parallelogram rule

Triangle rule (三角形法则): Place the tail of $\vec{b}$ at the head of $\vec{a}$. The sum $\vec{a} + \vec{b}$ runs from the tail of $\vec{a}$ to the head of $\vec{b}$.

Parallelogram rule (平行四边形法则): Place both vectors at the same tail point. The diagonal of the resulting parallelogram is $\vec{a} + \vec{b}$.

$$\vec{a} + \vec{b} = \vec{b} + \vec{a} \qquad (\text{commutative})$$
$$(\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c}) \qquad (\text{associative})$$

⚠️ Direction matters. $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ (the middle point cancels). But $\overrightarrow{AB} + \overrightarrow{CB} \ne \overrightarrow{AC}$ — the arrow on $\overrightarrow{CB}$ runs the wrong way.

Subtraction

$$\vec{a} - \vec{b} = \vec{a} + (-\vec{b})$$

Geometrically: place both vectors at the same tail; the difference points from the head of $\vec{b}$ to the head of $\vec{a}$.

Scalar multiplication (数乘)

For scalar $\lambda \in \mathbb{R}$:

$$|\lambda \vec{a}| = |\lambda| \cdot |\vec{a}|$$

• $\lambda > 0$: same direction as $\vec{a}$
• $\lambda < 0$: opposite direction
• $\lambda = 0$: gives $\vec{0}$

Distributive laws:
$$\lambda(\vec{a} + \vec{b}) = \lambda \vec{a} + \lambda \vec{b}, \qquad (\lambda + \mu)\vec{a} = \lambda \vec{a} + \mu \vec{a}$$

Worked Example 6.1.A

In parallelogram $ABCD$, let $\overrightarrow{AB} = \vec{p}$ and $\overrightarrow{AD} = \vec{q}$. Express $\overrightarrow{AC}$ and $\overrightarrow{BD}$ in terms of $\vec{p}$ and $\vec{q}$.

Solution.

By the parallelogram rule, diagonal $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD} = \vec{p} + \vec{q}$.

For the other diagonal, use the triangle rule through $A$:
$$\overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{AD} = -\vec{p} + \vec{q} = \vec{q} - \vec{p}.$$

$$\boxed{\overrightarrow{AC} = \vec{p} + \vec{q}, \quad \overrightarrow{BD} = \vec{q} - \vec{p}}$$

🔑 Cancellation trick: In any chain $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} = \overrightarrow{AD}$, intermediate points cancel like a telescoping sum. This is the fastest way to simplify multi-leg vector paths.

6.2　Dot Product　数量积（内积）

The dot product of two vectors is a scalar (number), not a vector:

$$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$$

where $\theta$ is the angle between the vectors ($0 \le \theta \le \pi$).

Properties you must know

🔑 Perpendicularity shortcut: $\vec{a} \perp \vec{b} \iff \vec{a} \cdot \vec{b} = 0$. (Because $\cos 90° = 0$.) This is the single most useful fact in this chapter — expect it on every exam.

Finding the angle between two vectors

Rearrange the dot product formula:

$$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$$

The angle $\theta \in [0°, 180°]$, so check the sign of the dot product first:

• $\vec{a} \cdot \vec{b} > 0 \Rightarrow \theta$ is acute
• $\vec{a} \cdot \vec{b} = 0 \Rightarrow \theta = 90°$ (perpendicular)
• $\vec{a} \cdot \vec{b} < 0 \Rightarrow \theta$ is obtuse

Projection formula

The scalar projection of $\vec{b}$ onto $\vec{a}$ (the component of $\vec{b}$ in the direction of $\vec{a}$):

$$\text{proj}_{\vec{a}} \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$$

Equivalently, $|\vec{b}|\cos\theta$.

Worked Example 6.2.A

Vectors $\vec{a}$ and $\vec{b}$ satisfy $|\vec{a}| = 2$, $|\vec{b}| = 3$, and the angle between them is $60°$. Find $\vec{a} \cdot \vec{b}$ and $|\vec{a} + \vec{b}|$.

Solution.

Step 1 — dot product:
$$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos 60° = 2 \cdot 3 \cdot \tfrac{1}{2} = 3.$$

Step 2 — magnitude of sum (expand using $|\vec{v}|^{2} = \vec{v} \cdot \vec{v}$):
$$|\vec{a} + \vec{b}|^{2} = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^{2} + 2\vec{a} \cdot \vec{b} + |\vec{b}|^{2} = 4 + 6 + 9 = 19.$$
$$|\vec{a} + \vec{b}| = \sqrt{19}.$$

$$\boxed{\vec{a} \cdot \vec{b} = 3, \quad |\vec{a} + \vec{b}| = \sqrt{19}}$$

⚠️ Common mistake: Students write $|\vec{a} + \vec{b}| = |\vec{a}| + |\vec{b}|$. This is only true when both vectors point in exactly the same direction. Always expand $|\vec{a} + \vec{b}|^{2}$ using the dot product.

6.3　Collinear Vectors　共线向量（平行向量）

Two non-zero vectors are collinear (共线, also called parallel, 平行) if they lie along parallel lines — i.e., one is a scalar multiple of the other.

Scalar condition

$$\vec{a} \parallel \vec{b} \iff \exists\, \lambda \in \mathbb{R},\ \vec{a} = \lambda \vec{b} \qquad (\vec{b} \ne \vec{0})$$

The scalar $\lambda$ can be positive (same direction) or negative (opposite direction).

Coordinate collinearity test

If $\vec{a} = (x_1, y_1)$ and $\vec{b} = (x_2, y_2)$, then:

$$\vec{a} \parallel \vec{b} \iff x_1 y_2 - x_2 y_1 = 0 \iff x_1 y_2 = x_2 y_1$$

🔑 Memory aid: This is the cross-product condition. Think of it as the determinant $\begin{vmatrix} x_1 & y_1 \\ x_2 & y_2 \end{vmatrix} = 0$. If this determinant equals zero, the vectors are parallel.

⚠️ Don't confuse collinear vectors with collinear points. Three points $A$, $B$, $C$ are collinear iff $\overrightarrow{AB} \parallel \overrightarrow{AC}$ — apply the vector collinearity test to the vectors, not the coordinates directly.

Worked Example 6.3.A

If $\vec{a} = (2, -1)$ and $\vec{b} = (k, 3)$ are collinear, find $k$.

Solution.

Apply the collinearity condition $x_1 y_2 = x_2 y_1$:
$$2 \cdot 3 = k \cdot (-1) \implies 6 = -k \implies k = -6.$$

Check: $\vec{b} = (-6, 3) = -3(2, -1) = -3\vec{a}$. ✓

$$\boxed{k = -6}$$

6.4　Vectors in Coordinate Form　向量的坐标表示

In a coordinate plane with unit vectors $\vec{e}_1 = (1,0)$ and $\vec{e}_2 = (0,1)$, every vector can be written as:

$$\vec{a} = (a_1,\ a_2) = a_1 \vec{e}_1 + a_2 \vec{e}_2$$

The ordered pair $(a_1, a_2)$ is the component form of $\vec{a}$.

Point-to-vector conversion

If $A = (x_1, y_1)$ and $B = (x_2, y_2)$, then:

$$\overrightarrow{AB} = (x_2 - x_1,\ y_2 - y_1)$$

⚠️ Order matters: $\overrightarrow{AB} = B - A$ (head minus tail). $\overrightarrow{BA} = A - B = -\overrightarrow{AB}$.

Coordinate arithmetic

Let $\vec{a} = (a_1, a_2)$, $\vec{b} = (b_1, b_2)$, scalar $\lambda$.

The coordinate dot product $\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2$ follows directly from expanding $|\vec{a}||\vec{b}|\cos\theta$ — this is the formula you use in all numerical calculations.

Worked Example 6.4.A

Let $\vec{a} = (3, -2)$ and $\vec{b} = (-1, 4)$.

(i) Find $2\vec{a} - \vec{b}$.
(ii) Find $\vec{a} \cdot \vec{b}$.
(iii) Find the angle $\theta$ between $\vec{a}$ and $\vec{b}$.

Solution.

(i) $2\vec{a} - \vec{b} = 2(3,-2) - (-1,4) = (6,-4) - (-1,4) = (7,\ -8).$

(ii) $\vec{a} \cdot \vec{b} = (3)(-1) + (-2)(4) = -3 - 8 = -11.$

(iii)
$$|\vec{a}| = \sqrt{9 + 4} = \sqrt{13}, \qquad |\vec{b}| = \sqrt{1 + 16} = \sqrt{17}.$$
$$\cos\theta = \frac{-11}{\sqrt{13}\cdot\sqrt{17}} = \frac{-11}{\sqrt{221}}.$$

Since $\cos\theta < 0$, the angle is obtuse.

$$\boxed{\theta = \arccos\!\left(\frac{-11}{\sqrt{221}}\right) \approx 137.8°}$$

🔑 Perpendicularity in coordinates: $\vec{a} \perp \vec{b} \iff a_1 b_1 + a_2 b_2 = 0$. No magnitudes or trig needed — just multiply matching components and add.

6.5　Vector Applications　向量应用

Vectors let us solve geometric problems algebraically — lengths, angles, midpoints, and parallelism proofs all reduce to vector arithmetic.

Midpoint formula via vectors

If $M$ is the midpoint of segment $AB$, then:

$$\overrightarrow{OM} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB})$$

In coordinates: $M = \left(\dfrac{x_A + x_B}{2},\ \dfrac{y_A + y_B}{2}\right)$.

Distance formula via vectors

$$|AB| = |\overrightarrow{AB}| = \sqrt{(x_B - x_A)^{2} + (y_B - y_A)^{2}}$$

Linear combination and basis

Any vector $\vec{v}$ in the plane can be written as a linear combination of two non-collinear vectors $\vec{e}_1$ and $\vec{e}_2$:

$$\vec{v} = x\,\vec{e}_1 + y\,\vec{e}_2$$

where $x, y$ are unique scalars. This is frequently used to match coefficients in proof problems.

Worked Example 6.5.A

In triangle $ABC$, let $\overrightarrow{AB} = \vec{m}$ and $\overrightarrow{AC} = \vec{n}$. Point $D$ is on $BC$ such that $BD:DC = 1:2$. Express $\overrightarrow{AD}$ in terms of $\vec{m}$ and $\vec{n}$.

Solution.

$D$ divides $BC$ in ratio $1:2$, so:
$$\overrightarrow{BD} = \frac{1}{1+2}\overrightarrow{BC} = \frac{1}{3}\overrightarrow{BC}.$$

Now express $\overrightarrow{BC}$ using triangle law:
$$\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC} = -\vec{m} + \vec{n}.$$

Therefore:
$$\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BD} = \vec{m} + \frac{1}{3}(-\vec{m} + \vec{n}) = \vec{m} - \frac{1}{3}\vec{m} + \frac{1}{3}\vec{n} = \frac{2}{3}\vec{m} + \frac{1}{3}\vec{n}.$$

$$\boxed{\overrightarrow{AD} = \tfrac{2}{3}\vec{m} + \tfrac{1}{3}\vec{n}}$$

🔑 Section formula: If $D$ divides $BC$ in ratio $m:n$ (from $B$ to $C$), then $\overrightarrow{BD} = \dfrac{m}{m+n}\overrightarrow{BC}$. Memorise this — it appears in nearly every vector geometry question.

⚠️ Exam trap: When the problem asks "find the angle $\theta$ between vectors $\vec{a}$ and $\vec{b}$", the answer must satisfy $0 \le \theta \le 180°$ (or $0 \le \theta \le \pi$). Vectors do not have a "signed" angle like slopes do.

Try it!　自测练习

Q1. In parallelogram $ABCD$, $\overrightarrow{AB} = \vec{a}$ and $\overrightarrow{BC} = \vec{b}$. Express $\overrightarrow{AC} - \overrightarrow{BD}$ in terms of $\vec{a}$ and $\vec{b}$.

Q2. Vectors $|\vec{p}| = 5$, $|\vec{q}| = 4$, and $\vec{p} \cdot \vec{q} = -10$. Find the angle between $\vec{p}$ and $\vec{q}$.

Q3. $\vec{a} = (1, t)$ and $\vec{b} = (2, 6)$. For what value of $t$ are they collinear? For what value of $t$ are they perpendicular?

Q4. $A = (1, 2)$, $B = (4, -1)$, $C = (5, 3)$. Find $\overrightarrow{AB} \cdot \overrightarrow{AC}$.

Q5. In triangle $OAB$, let $\overrightarrow{OA} = \vec{a}$ and $\overrightarrow{OB} = \vec{b}$. $M$ is the midpoint of $AB$. Show that $\overrightarrow{OM} = \dfrac{1}{2}(\vec{a} + \vec{b})$.

📌 Chapter summary

Topic	Key skill
6.1 Vector Operations	Triangle/parallelogram rules; cancellation $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$; scalar mult changes magnitude and direction
6.2 Dot Product	$\vec{a}\cdot\vec{b}=\|\vec{a}\|\|\vec{b}\|\cos\theta$; perpendicular $\Leftrightarrow$ dot product $=0$; expand $|\vec{a}+\vec{b}|^{2}$ by dot product
6.3 Collinear Vectors	$\vec{a}=\lambda\vec{b}$ (scalar form); coordinate test $x_1 y_2 = x_2 y_1$
6.4 Coordinate Form	Component-wise $+,-,\times\lambda$; $|\vec{a}|=\sqrt{a_1^2+a_2^2}$; $\vec{a}\cdot\vec{b}=a_1 b_1+a_2 b_2$
6.5 Applications	Section formula for ratio division; midpoint $=\tfrac{1}{2}(\vec{a}+\vec{b})$; match linear-combination coefficients in proofs

What's next → Unit 7 uses vector ideas directly: the slope of a line is related to a direction vector, and the perpendicular-line condition $k_1 k_2 = -1$ mirrors the dot-product perpendicularity rule $\vec{a} \cdot \vec{b} = 0$.