In VCE Specialist Mathematics, vectors in three-dimensional space $\mathbb{R}^3$ are represented using the standard unit vectors $\mathbf{i}, \mathbf{j},$ and $\mathbf{k}$, which correspond to the $x, y,$ and $z$ axes respectively. A vector $\mathbf{a}$ is written as:
$$\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$$
The scalar product (or dot product) of two vectors results in a scalar value (a real number). It is fundamental for determining the angle between vectors and finding projections.
For vectors $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ and $\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}$:
$$\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3$$
The scalar product relates the magnitudes of the vectors and the cosine of the angle $\theta$ between them:
$$\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos(\theta)$$
where \$0 \le \theta \le \pi$.
EXAM TIP: When asked to find the angle between two vectors, always use the rearranged dot product formula: $\cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$. Ensure your calculator is in the correct mode (Degrees or Radians) as specified by the question.
Resolving a vector $\mathbf{a}$ in the direction of vector $\mathbf{b}$ involves finding how much of $\mathbf{a}$ points along $\mathbf{b}$ and how much points perpendicular to it.
The scalar resolute of $\mathbf{a}$ in the direction of $\mathbf{b}$ is the “length” of the projection of $\mathbf{a}$ onto $\mathbf{b}$. It can be positive, negative, or zero.
$$\text{Scalar Resolute} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}$$
The vector resolute of $\mathbf{a}$ in the direction of $\mathbf{b}$ (parallel component) is:
$$\mathbf{a}_{||} = \left( \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2} \right) \mathbf{b} = \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \right) \mathbf{b}$$
The vector resolute of $\mathbf{a}$ perpendicular to $\mathbf{b}$ is:
$$\mathbf{a}{\perp} = \mathbf{a} - \mathbf{a}{||}$$
COMMON MISTAKE: Students often forget that the vector resolute must be a vector. Always multiply the scalar component by the unit vector $\hat{\mathbf{b}}$ or use the $|\mathbf{b}|^2$ denominator to ensure the result is in the direction of $\mathbf{b}$.
The vector product (or cross product) of two vectors results in a vector that is perpendicular to both original vectors.
The cross product is most easily calculated using a \$3 \times 3$ determinant:
$$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}$$
Expanding the determinant:
$$\mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2)\mathbf{i} - (a_1b_3 - a_3b_1)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k}$$
The magnitude of the cross product is given by:
$$|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin(\theta)$$
The direction of $\mathbf{a} \times \mathbf{b}$ is determined by the Right-Hand Rule: curl your fingers from $\mathbf{a}$ to $\mathbf{b}$, and your thumb points in the direction of $\mathbf{a} \times \mathbf{b}$.
KEY TAKEAWAY: While the dot product tells us about “parallelness” (maximum when parallel, zero when perp), the cross product tells us about “perpendicularity” (maximum when perp, zero when parallel).
The cross product has significant applications in calculating areas in 3D space.
The magnitude $|\mathbf{a} \times \mathbf{b}|$ is equal to the area of the parallelogram formed by vectors $\mathbf{a}$ and $\mathbf{b}$.
$$\text{Area}_{\text{parallelogram}} = |\mathbf{a} \times \mathbf{b}|$$
A triangle formed by vectors $\mathbf{a}$ and $\mathbf{b}$ is exactly half of the parallelogram.
$$\text{Area}_{\text{triangle}} = \frac{1}{2}|\mathbf{a} \times \mathbf{b}|$$
The cross product $\mathbf{a} \times \mathbf{b}$ provides a vector that is normal (perpendicular) to the plane containing vectors $\mathbf{a}$ and $\mathbf{b}$. This is essential for finding the equations of planes in Unit 4.
VCAA FOCUS: Questions often require students to find a unit vector perpendicular to two given vectors. To do this, calculate $\mathbf{n} = \mathbf{a} \times \mathbf{b}$ and then find the unit vector $\hat{\mathbf{n}} = \frac{\mathbf{n}}{|\mathbf{n}|}$.
| Feature | Scalar (Dot) Product | Vector (Cross) Product |
|---|---|---|
| Notation | $\mathbf{a} \cdot \mathbf{b}$ | $\mathbf{a} \times \mathbf{b}$ |
| Result Type | Scalar (Number) | Vector |
| Formula | $ | \mathbf{a} |
| Commutative? | Yes: $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$ | No: $\mathbf{a} \times \mathbf{b} = -\mathbf{b} \times \mathbf{a}$ |
| Zero Case | $\mathbf{a} \perp \mathbf{b} \implies \mathbf{a} \cdot \mathbf{b} = 0$ | $\mathbf{a} \parallel \mathbf{b} \implies \mathbf{a} \times \mathbf{b} = \mathbf{0}$ |
| Main Use | Angles, Projections (Resolutes) | Areas, Normal vectors |
STUDY HINT: Practice calculating the cross product by hand using the determinant method. While CAS calculators can do this, technology-free Exam 1 frequently tests your ability to perform these algebraic manipulations accurately.
