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QLDSpecialist MathematicsSyllabus dot point

Topic 2: Vectors in two and three dimensions

Use vectors in three dimensions, compute scalar (dot) and vector (cross) products, find angles between vectors, scalar and vector projections, and apply these to geometric problems and the equations of lines and planes

A focused answer to the QCE Specialist Mathematics Unit 3 dot point on three-dimensional vectors. Covers component form, magnitude, the dot product and angle between vectors, scalar and vector projections, the cross product and its geometric meaning, and the vector and parametric equations of lines, with a verified worked example and the projection sign trap.

Generated by Claude Opus 4.76 min answer

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What this dot point is asking

QCAA wants you to work with vectors in three dimensions: write them in component form, find magnitudes and unit vectors, compute dot and cross products, find angles and projections, and use vectors to describe lines and to solve geometric problems. This is heavily assessed in IA2 and the external assessment, where setting up the right product for the question is the key decision.

The answer

Component form and magnitude

A vector in three dimensions is written a=a1i+a2j+a3k\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} or as a column. Its magnitude is

a=a12+a22+a32.|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}.

A unit vector in the direction of a\mathbf{a} is a^=aa\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|}. The vector from point AA to point BB is AB=ba\vec{AB} = \mathbf{b} - \mathbf{a}, the position vector of BB minus that of AA.

The dot (scalar) product

The dot product produces a scalar:

ab=a1b1+a2b2+a3b3=abcosθ,\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = |\mathbf{a}||\mathbf{b}|\cos\theta,

where θ\theta is the angle between the vectors. Rearranging gives the angle:

cosθ=abab.\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}.

If ab=0\mathbf{a}\cdot\mathbf{b} = 0 and neither vector is zero, the vectors are perpendicular. The dot product is the tool for angles and perpendicularity.

Scalar and vector projections

The scalar projection of a\mathbf{a} onto b\mathbf{b} is the signed length of the shadow of a\mathbf{a} along b\mathbf{b}:

scalar proj=abb.\text{scalar proj} = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}.

The vector projection multiplies this by the unit vector b^\hat{\mathbf{b}}:

vector proj=(abb2)b.\text{vector proj} = \left(\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\right)\mathbf{b}.

The scalar projection is negative when the angle is obtuse, which the sign of ab\mathbf{a}\cdot\mathbf{b} captures automatically.

The cross (vector) product

The cross product produces a vector perpendicular to both inputs:

a×b=ijka1a2a3b1b2b3.\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}.

Its magnitude is a×b=absinθ|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta, which equals the area of the parallelogram spanned by a\mathbf{a} and b\mathbf{b}. The direction follows the right-hand rule. The cross product is the tool for finding a normal vector and for areas.

Equations of lines

A line through point AA with position vector a\mathbf{a} and direction d\mathbf{d} has vector equation

r=a+td,tR.\mathbf{r} = \mathbf{a} + t\mathbf{d}, \quad t \in \mathbb{R}.

Splitting into components gives the parametric equations x=a1+td1x = a_1 + t d_1, y=a2+td2y = a_2 + t d_2, z=a3+td3z = a_3 + t d_3. Two lines are parallel when their direction vectors are scalar multiples, and they intersect when a single pair (t,s)(t, s) satisfies all three component equations.