In VCE Specialist Mathematics, vectors provide a powerful framework for describing and solving problems involving lines and planes in three-dimensional space. This area of study bridges the gap between algebraic manipulation and geometric interpretation.
A vector in three dimensions is represented as $\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$, where $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are unit vectors in the directions of the $x, y,$ and $z$ axes respectively.
The scalar product is used to find the angle between vectors and to test for perpendicularity.
$$\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 = |\mathbf{a}||\mathbf{b}|\cos(\theta)$$
* If $\mathbf{a} \cdot \mathbf{b} = 0$, then $\mathbf{a} \perp \mathbf{b}$ (provided $\mathbf{a}, \mathbf{b} \neq \mathbf{0}$).
The vector product results in a vector that is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$.
$$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} = (a_2b_3 - a_3b_2)\mathbf{i} - (a_1b_3 - a_3b_1)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k}$$
* Magnitude: $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin(\theta)$.
* Geometric use: The area of a triangle with sides defined by vectors $\mathbf{a}$ and $\mathbf{b}$ is $\frac{1}{2}|\mathbf{a} \times \mathbf{b}|$.
EXAM TIP: Use the dot product to find angles and the cross product to find a vector perpendicular to a plane. If a question asks to show two vectors are perpendicular, always show that $\mathbf{a} \cdot \mathbf{b} = 0$.
A line in 3D space is uniquely determined by a point on the line and a direction vector.
Let $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ be the position vector of a point on the line, and $\mathbf{d} = d_1\mathbf{i} + d_2\mathbf{j} + d_3\mathbf{k}$ be the direction vector parallel to the line.
| Form | Equation |
|---|---|
| Vector Form | $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}, \lambda \in \mathbb{R}$ |
| Parametric Form | $x = a_1 + \lambda d_1, \quad y = a_2 + \lambda d_2, \quad z = a_3 + \lambda d_3$ |
| Cartesian Form | $\frac{x - a_1}{d_1} = \frac{y - a_2}{d_2} = \frac{z - a_3}{d_3}$ |
COMMON MISTAKE: When converting to Cartesian form, ensure the coefficients of $x, y,$ and $z$ are exactly $1$. For example, if you have $\frac{2x-4}{6}$, rewrite it as $\frac{x-2}{3}$ before identifying the direction vector component.
A plane is defined by a point $\mathbf{a}$ on the plane and a normal vector $\mathbf{n}$ (a vector perpendicular to the surface).
The set of all points $\mathbf{r}$ on the plane satisfies:
$$(\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0 \implies \mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$$
Where $\mathbf{a} \cdot \mathbf{n}$ results in a constant $k$.
If $\mathbf{n} = n_1\mathbf{i} + n_2\mathbf{j} + n_3\mathbf{k}$, the equation is:
$$n_1x + n_2y + n_3z = k$$
KEY TAKEAWAY: The coefficients of $x, y,$ and $z$ in the Cartesian equation of a plane are the components of the normal vector $\mathbf{n}$.
| Target | Method | Formula |
|---|---|---|
| Between two lines | Use direction vectors $\mathbf{d_1}, \mathbf{d_2}$ | $\cos(\theta) = \frac{ |
| Between two planes | Use normal vectors $\mathbf{n_1}, \mathbf{n_2}$ | $\cos(\theta) = \frac{ |
| Between line and plane | Use direction $\mathbf{d}$ and normal $\mathbf{n}$ | $\sin(\theta) = \frac{ |
VCAA FOCUS: The angle between a line and a plane is a frequent exam question. Remember to use sine instead of cosine if you are using the normal vector of the plane, or calculate the angle with the normal and subtract from $90^\circ$.
Vectors can be used to prove geometric properties:
1. Collinearity: Three points $A, B, C$ are collinear if $\vec{AB} = k\vec{BC}$ for some scalar $k$.
2. Coplanarity: Four points are coplanar if the volume of the parallelepiped they form is zero, or if one vector can be expressed as a linear combination of two others: $\mathbf{c} = \alpha\mathbf{a} + \beta\mathbf{b}$.
3. Isosceles Triangles: Show that two side vectors have equal magnitudes (e.g., $|\vec{AB}| = |\vec{AC}|$).
4. Intersection: To find the intersection of a line and a plane, substitute the parametric expressions for $x, y, z$ from the line into the Cartesian equation of the plane and solve for $\lambda$.
REMEMBER: To show that a line is perpendicular to a plane, the direction vector of the line $\mathbf{d}$ must be parallel to the normal vector of the plane $\mathbf{n}$ (i.e., $\mathbf{d} = k\mathbf{n}$). To show a line is parallel to a plane, $\mathbf{d} \cdot \mathbf{n} = 0$.
