Vectors — IB Math AA (HL)

1. Vector Operations

Vectors in 3D are denoted $\mathbf{v} = \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}$ or $v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k}$. The fundamental operations are:

Key Vector Formulas

Magnitude: $|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$

Unit vector: $\hat{\mathbf{v}} = \dfrac{\mathbf{v}}{|\mathbf{v}|}$

Dot product: $\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 = |\mathbf{a}||\mathbf{b}|\cos\theta$

Cross product: $\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix}$, perpendicular to both $\mathbf{a}$ and $\mathbf{b}$, with $|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$

Worked Example 1

Find the angle between $\mathbf{a} = (1, 2, -1)$ and $\mathbf{b} = (3, -1, 2)$

Dot product: $\mathbf{a}\cdot\mathbf{b} = (1)(3) + (2)(-1) + (-1)(2) = 3 - 2 - 2 = -1$

Magnitudes: $|\mathbf{a}| = \sqrt{1+4+1} = \sqrt{6}$, $|\mathbf{b}| = \sqrt{9+1+4} = \sqrt{14}$

Angle: $\cos\theta = \dfrac{-1}{\sqrt{6}\cdot\sqrt{14}} = \dfrac{-1}{\sqrt{84}} \approx -0.1091$

Result: $\theta = \arccos\!\left(\dfrac{-1}{\sqrt{84}}\right) \approx 96.3°$

Key Insight: If $\mathbf{a}\cdot\mathbf{b} = 0$, the vectors are perpendicular. If $\mathbf{a}\times\mathbf{b} = \mathbf{0}$, the vectors are parallel. These two tests are fundamental to all line and plane problems.

2. Equations of Lines in 3D

Vector Equation of a Line

A line through point $A$ (position vector $\mathbf{a}$) with direction $\mathbf{b}$:

$$\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}, \quad \lambda \in \mathbb{R}$$

Parametric form: $x = a_1 + \lambda b_1$, $y = a_2 + \lambda b_2$, $z = a_3 + \lambda b_3$

Symmetric form: $\dfrac{x-a_1}{b_1} = \dfrac{y-a_2}{b_2} = \dfrac{z-a_3}{b_3} = \lambda$

Worked Example 2

Find the equation of the line through $A(1,0,-2)$ and $B(3,1,4)$

Direction vector: $\overrightarrow{AB} = B - A = (3-1, 1-0, 4-(-2)) = (2,1,6)$

Vector equation: $\mathbf{r} = \begin{pmatrix}1\\0\\-2\end{pmatrix} + \lambda\begin{pmatrix}2\\1\\6\end{pmatrix}$

Symmetric form: $\dfrac{x-1}{2} = \dfrac{y}{1} = \dfrac{z+2}{6}$

3. Equations of Planes

Equation of a Plane

Normal vector form: $\mathbf{n}\cdot(\mathbf{r} - \mathbf{a}) = 0$, i.e., $ax + by + cz = d$ where $\mathbf{n} = (a,b,c)$

The normal vector $\mathbf{n}$ is perpendicular to every vector lying in the plane.

Distance from point $P$ to plane $ax+by+cz=d$: $$\text{dist} = \frac{|ax_P + by_P + cz_P - d|}{\sqrt{a^2+b^2+c^2}}$$

Worked Example 3

Find where the line $\mathbf{r} = (1,2,0) + \lambda(1,-1,3)$ meets the plane $2x+y-z=5$

Parametric coords on line: $x = 1+\lambda$, $y = 2-\lambda$, $z = 3\lambda$

Substitute into plane equation: $2(1+\lambda) + (2-\lambda) - (3\lambda) = 5$

Solve: $2 + 2\lambda + 2 - \lambda - 3\lambda = 5 \Rightarrow 4 - 2\lambda = 5 \Rightarrow \lambda = -\dfrac{1}{2}$

Intersection point: $x = 1-\frac{1}{2} = \frac{1}{2}$, $y = 2+\frac{1}{2} = \frac{5}{2}$, $z = -\frac{3}{2}$. Point: $\left(\frac{1}{2}, \frac{5}{2}, -\frac{3}{2}\right)$

HL Geometry: To find the equation of a plane through three points $A$, $B$, $C$: compute $\overrightarrow{AB}$ and $\overrightarrow{AC}$, then the normal is $\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC}$. Use any of the three points to find $d$ in $\mathbf{n}\cdot\mathbf{r} = d$.

Practice Problems

Q1. Find the angle between the line $\mathbf{r} = (1,0,2) + \lambda(1,1,-1)$ and the plane $x + 2y + 2z = 6$.

Show Solution

The angle $\phi$ between line (direction $\mathbf{d}$) and plane (normal $\mathbf{n}$) satisfies $\sin\phi = \dfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$.

$\mathbf{d} = (1,1,-1)$, $\mathbf{n} = (1,2,2)$. $\mathbf{d}\cdot\mathbf{n} = 1+2-2=1$. $|\mathbf{d}|=\sqrt{3}$, $|\mathbf{n}|=3$.

$\sin\phi = \dfrac{1}{3\sqrt{3}}$, so $\phi = \arcsin\!\left(\dfrac{1}{3\sqrt{3}}\right) \approx 11.0°$

Q2. Find the distance from $P(1,1,1)$ to the plane $2x - 2y + z = 4$.

Show Solution

$\text{dist} = \dfrac{|2(1)-2(1)+(1)-4|}{\sqrt{4+4+1}} = \dfrac{|2-2+1-4|}{3} = \dfrac{3}{3} = 1$

Q3. Find the equation of the plane through $A(1,0,0)$, $B(0,2,0)$, $C(0,0,3)$.

Show Solution

$\overrightarrow{AB} = (-1,2,0)$, $\overrightarrow{AC} = (-1,0,3)$.

$\mathbf{n} = \overrightarrow{AB}\times\overrightarrow{AC} = (6,3,2)$.

Using point $A$: $6(x-1)+3y+2z=0 \Rightarrow 6x+3y+2z=6$.

4. Exam Tips

To test if two lines are parallel, check if their direction vectors are scalar multiples.
To test if two lines intersect, set up equations for $\lambda$ and $\mu$, solve two simultaneously, then check consistency with the third equation.
The cross product gives a vector perpendicular to both inputs — use it to find plane normals and line/plane distances.
Always verify intersection points by substituting back into both equations.