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HSC Maths Extension 1 vectors: deep dive (2026 guide)

A complete deep dive into vectors for HSC Mathematics Extension 1. Vector arithmetic, magnitude and unit vectors, the scalar product, projections, parametric vector equations of lines, and geometric proofs.

Generated by Claude OpusReviewed by Better Tuition Academy12 min readNESA-MATH-EXT1-V1

What this guide covers

Vectors were introduced into HSC Mathematics Extension 1 in the 2017 syllabus and are now examined every year. This guide covers:

  • Vector arithmetic, magnitude and unit vectors.
  • The scalar (dot) product, the angle between vectors, orthogonality.
  • Scalar and vector projection.
  • Parametric vector equations of lines and intersection problems.
  • Geometric proofs using vectors.

Vector arithmetic

A two-dimensional vector a=(a1,a2)\mathbf{a} = (a_1, a_2) can be written in column form, component form, or unit-vector form (a1i+a2ja_1 \mathbf{i} + a_2 \mathbf{j} with i=(1,0)\mathbf{i} = (1, 0), j=(0,1)\mathbf{j} = (0, 1)).

Addition, subtraction and scalar multiplication

a+b=(a1+b1,a2+b2),Ξ»a=(Ξ»a1,Ξ»a2).\mathbf{a} + \mathbf{b} = (a_1 + b_1, a_2 + b_2), \qquad \lambda \mathbf{a} = (\lambda a_1, \lambda a_2).

Geometrically: a+b\mathbf{a} + \mathbf{b} is the parallelogram-rule diagonal; λa\lambda \mathbf{a} scales magnitude by ∣λ∣|\lambda| and may reverse direction if λ<0\lambda < 0.

Magnitude and unit vector

∣a∣=a12+a22|\mathbf{a}| = \sqrt{a_1^2 + a_2^2}, and the unit vector a^=1∣a∣a\hat{\mathbf{a}} = \frac{1}{|\mathbf{a}|} \mathbf{a}.

Position vector versus displacement vector

If A=(a1,a2)A = (a_1, a_2) is a point, OA=a\mathbf{OA} = \mathbf{a} is its position vector (origin to AA). The displacement from AA to BB is AB=bβˆ’a\mathbf{AB} = \mathbf{b} - \mathbf{a}.

The scalar (dot) product

aβ‹…b=a1b1+a2b2=∣a∣∣b∣cos⁑θ.\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 = |\mathbf{a}| |\mathbf{b}| \cos \theta.

Angle between two vectors

cos⁑θ=aβ‹…b∣a∣∣b∣.\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}.

For non-zero vectors, θ∈[0,Ο€]\theta \in [0, \pi].

Orthogonality test

aβŠ₯b\mathbf{a} \perp \mathbf{b} iff aβ‹…b=0\mathbf{a} \cdot \mathbf{b} = 0.

This is the most common HSC use of the scalar product.

Properties

The dot product is commutative, bilinear, and aβ‹…a=∣a∣2\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2.

(a+b)β‹…(a+b)=∣a∣2+2aβ‹…b+∣b∣2.(\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) = |\mathbf{a}|^2 + 2 \mathbf{a} \cdot \mathbf{b} + |\mathbf{b}|^2.

This is the vector form of the cosine rule.

Projections

Scalar projection

projbscalara=aβ‹…b∣b∣.\text{proj}_{\mathbf{b}}^{\text{scalar}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}.

Signed length: positive if a\mathbf{a} has a component in the direction of b\mathbf{b}, negative if opposite.

Vector projection

projba=aβ‹…b∣b∣2b.\text{proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2} \mathbf{b}.

This is the scalar projection times the unit vector b^\hat{\mathbf{b}}.

Decomposition

Any vector a\mathbf{a} can be split into a component parallel to b\mathbf{b} (the vector projection) and a component perpendicular to b\mathbf{b}.

a=projba+aβŠ₯,aβŠ₯=aβˆ’projba.\mathbf{a} = \text{proj}_{\mathbf{b}} \mathbf{a} + \mathbf{a}_{\perp}, \quad \mathbf{a}_{\perp} = \mathbf{a} - \text{proj}_{\mathbf{b}} \mathbf{a}.

By construction aβŠ₯β‹…b=0\mathbf{a}_{\perp} \cdot \mathbf{b} = 0.

Parametric vector equations of lines

Point-direction form

A line through AA (position vector a\mathbf{a}) with direction d\mathbf{d}:

r=a+λd,λ∈R.\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}, \quad \lambda \in \mathbb{R}.

Line through two points

Direction AB=bβˆ’a\mathbf{AB} = \mathbf{b} - \mathbf{a}, so r=a+Ξ»(bβˆ’a)=(1βˆ’Ξ»)a+Ξ»b\mathbf{r} = \mathbf{a} + \lambda (\mathbf{b} - \mathbf{a}) = (1 - \lambda) \mathbf{a} + \lambda \mathbf{b}.

At Ξ»=0\lambda = 0: at AA. At Ξ»=1\lambda = 1: at BB.

Convert to Cartesian

If d=(d1,d2)\mathbf{d} = (d_1, d_2) with d1β‰ 0d_1 \neq 0, Ξ»=xβˆ’a1d1\lambda = \frac{x - a_1}{d_1}, so y=a2+xβˆ’a1d1d2y = a_2 + \frac{x - a_1}{d_1} d_2. Slope: d2d1\frac{d_2}{d_1}.

Intersection of two lines

Set parametric equations equal:

a1+Ξ»d1=a2+ΞΌd2.\mathbf{a}_1 + \lambda \mathbf{d}_1 = \mathbf{a}_2 + \mu \mathbf{d}_2.

Solve for Ξ»\lambda and ΞΌ\mu. If a unique solution, lines meet at a point. If the system is inconsistent, lines are parallel and disjoint.

Geometric proofs using vectors

The recipe

  1. Assign position vectors to labelled points.
  2. Express the relevant displacement vectors algebraically.
  3. Compute the geometric relation (equality, scalar multiple, dot product zero, etc.).
  4. Conclude the geometric statement.

Worked example: midpoint connector

In triangle ABCABC, let PP and QQ be midpoints of ABAB and ACAC. Show PQPQ is parallel to BCBC and half its length.

Position vectors a,b,c\mathbf{a}, \mathbf{b}, \mathbf{c}.

P=a+b2P = \frac{\mathbf{a} + \mathbf{b}}{2}, Q=a+c2Q = \frac{\mathbf{a} + \mathbf{c}}{2}.

PQ=Qβˆ’P=cβˆ’b2=12BC\mathbf{PQ} = Q - P = \frac{\mathbf{c} - \mathbf{b}}{2} = \frac{1}{2} \mathbf{BC}.

So PQ\mathbf{PQ} is parallel to BC\mathbf{BC} (same direction) and half the magnitude.

Worked example: parallelogram diagonals

ABCDABCD is a parallelogram, so AB=DC\mathbf{AB} = \mathbf{DC}, i.e., a+c=b+d\mathbf{a} + \mathbf{c} = \mathbf{b} + \mathbf{d}.

Midpoint of ACAC is a+c2\frac{\mathbf{a} + \mathbf{c}}{2}; midpoint of BDBD is b+d2\frac{\mathbf{b} + \mathbf{d}}{2}.

These are equal, so the diagonals bisect each other.

Worked example: perpendicular diagonals of a rhombus

If a parallelogram has perpendicular diagonals, the sides are equal in length (rhombus).

Let AB=u\mathbf{AB} = \mathbf{u} and AD=v\mathbf{AD} = \mathbf{v}. Diagonals: AC=u+v\mathbf{AC} = \mathbf{u} + \mathbf{v}, BD=vβˆ’u\mathbf{BD} = \mathbf{v} - \mathbf{u}.

Perpendicular: (u+v)β‹…(vβˆ’u)=∣v∣2βˆ’βˆ£u∣2=0(\mathbf{u} + \mathbf{v}) \cdot (\mathbf{v} - \mathbf{u}) = |\mathbf{v}|^2 - |\mathbf{u}|^2 = 0, so ∣u∣=∣v∣|\mathbf{u}| = |\mathbf{v}|.

Common exam questions

Question type A: angle and orthogonality

Find the angle between two given vectors. Find kk so that two vectors are perpendicular. These are 2-3 mark items.

Question type B: projections

Find the scalar or vector projection. Decompose a vector into parallel and perpendicular components. These are 3-4 mark items.

Question type C: parametric vector equation of a line

Write the vector equation, convert to Cartesian, or find an intersection. 3-4 marks.

Question type D: geometric proof using vectors

Show two lines are parallel, two are perpendicular, or a figure is a parallelogram. 4-5 marks.

Question type E: applications

Resolve a force into components along a given direction. Find collision conditions for two moving particles. 4-5 marks.

Common traps

Trap: PQ=Qβˆ’P\mathbf{PQ} = Q - P, not IMATH_80

The displacement from PP to QQ is the position vector of QQ minus the position vector of PP. Reversing gives the opposite vector.

Trap: Variance vs. standard deviation in dot-product squaring

∣a∣2=aβ‹…a|\mathbf{a}|^2 = \mathbf{a} \cdot \mathbf{a}, so ∣a+b∣2=∣a∣2+2aβ‹…b+∣b∣2|\mathbf{a} + \mathbf{b}|^2 = |\mathbf{a}|^2 + 2 \mathbf{a} \cdot \mathbf{b} + |\mathbf{b}|^2. This expansion is essential for the vector cosine rule.

Trap: ∣b∣|\mathbf{b}| vs IMATH_88

Scalar projection has ∣b∣|\mathbf{b}| in the denominator. Vector projection has ∣b∣2|\mathbf{b}|^2.

Trap: Parallel parameters

Two lines with the same direction vector are parallel. They may or may not coincide. Check if a point on one is also on the other.

Trap: Different parameters for different lines

When intersecting two lines, use Ξ»\lambda for one and ΞΌ\mu for another. Reusing the same parameter creates confusion.

Exam strategy

Vector questions typically run 2-5 marks each. Allocate 3βˆ’63-6 minutes per question. Steps:

  1. Read carefully: scalar or vector projection? parallel or equal?
  2. Write down position or displacement vectors.
  3. Compute the relevant quantity algebraically.
  4. State the answer with units (if appropriate) and direction (positive, negative, parallel, perpendicular).

For geometric proofs, work the algebra one line at a time and cite each step. Markers reward clear structure.

Linked dot points

For more detail, see the dot point pages at vector arithmetic and magnitude, scalar product, vector projection, parametric vector equations of lines, and geometric proofs with vectors.

For NESA past papers and marking guidelines, refer to educationstandards.nsw.edu.au.

  • vectors
  • scalar-product
  • projection
  • hsc-maths-extension-1
  • 2026