Skip to main content
ExamExplained
WA · Specialist Mathematics
Specialist Mathematics study scene
§-Syllabus dot point
WASpecialist MathematicsSyllabus dot point

How do vectors describe direction, length and angle in space?

Use 3D vectors with the dot product and cross product to find lengths, angles, projections and areas

WACE Specialist Unit 3 three-dimensional vectors: components, magnitude, unit vectors, the scalar (dot) product for angles and projections, and the vector (cross) product for perpendiculars and areas.

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. Components, magnitude and unit vectors
  3. The scalar (dot) product
  4. The vector (cross) product
  5. Choosing the right tool
  6. Position vectors and points in space

What this dot point is asking

SCSA wants confident vector algebra in three dimensions: adding and scaling vectors, computing magnitude and unit vectors, using the scalar product to find angles and scalar/vector projections, and using the vector product to find perpendiculars and the area of a triangle or parallelogram.

Components, magnitude and unit vectors

Write a=a1i+a2j+a3k\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}. Then 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^=1aa\hat{\mathbf{a}} = \tfrac{1}{|\mathbf{a}|}\mathbf{a}. The vector from point AA to point BB is AB=ba\overrightarrow{AB} = \mathbf{b} - \mathbf{a} (position vectors).

The scalar (dot) product

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.

This gives the angle between vectors via cosθ=abab\cos\theta = \tfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}. Two non-zero vectors are perpendicular exactly when ab=0\mathbf{a}\cdot\mathbf{b} = 0.

The scalar projection of a\mathbf{a} onto b\mathbf{b} is abb\tfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}, and the vector projection is abb2b\tfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\,\mathbf{b}.

The vector (cross) product

a×b=ijka1a2a3b1b2b3=(a2b3a3b2)i(a1b3a3b1)j+(a1b2a2b1)k.\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_2 b_3 - a_3 b_2)\mathbf{i} - (a_1 b_3 - a_3 b_1)\mathbf{j} + (a_1 b_2 - a_2 b_1)\mathbf{k}.

The result is perpendicular to both a\mathbf{a} and b\mathbf{b} (right-hand rule), with a×b=absinθ|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta. It is anticommutative: a×b=(b×a)\mathbf{a}\times\mathbf{b} = -(\mathbf{b}\times\mathbf{a}), and a×a=0\mathbf{a}\times\mathbf{a} = \mathbf{0}.

Choosing the right tool

The two products do different jobs, and SCSA questions reward picking the right one. Use the scalar product when the question asks for an angle, a test of perpendicularity, or a projection of one vector along another, because it returns a number tied to cosθ\cos\theta. Use the vector product when the question asks for a vector perpendicular to two given vectors, a normal to a plane, a test of whether vectors are parallel, or an area, because it returns a vector with magnitude tied to sinθ\sin\theta. A quick way to remember it: dot for direction comparison, cross for area and normals.

Position vectors and points in space

Geometry questions are usually phrased in terms of points, so the first move is almost always to convert points into vectors. The position vector of a point AA is the vector from the origin to AA, and the displacement from AA to BB is AB=ba\vec{AB} = \mathbf{b} - \mathbf{a}. To find an angle at a vertex, take the two edges leaving that vertex and dot them; to find an area, cross them. The midpoint of ABAB has position vector 12(a+b)\tfrac{1}{2}(\mathbf{a} + \mathbf{b}), and a point dividing ABAB in the ratio m:nm:n has position vector na+mbm+n\dfrac{n\mathbf{a} + m\mathbf{b}}{m + n}, both of which appear in collinearity and ratio problems.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20236 marksCalculator-assumed. Let a=(2,1,2)\mathbf{a} = (2, 1, -2) and b=(1,3,1)\mathbf{b} = (1, 3, 1). (a) Find a unit vector in the direction of a\mathbf{a}. (b) Find the scalar projection of b\mathbf{b} onto a\mathbf{a}.
Show worked answer →

Tests magnitude, unit vectors and projection.

(a) a=4+1+4=3|\mathbf{a}| = \sqrt{4 + 1 + 4} = 3. The unit vector is a^=13(2,1,2)=(23,13,23)\hat{\mathbf{a}} = \dfrac{1}{3}(2, 1, -2) = \left(\dfrac{2}{3}, \dfrac{1}{3}, -\dfrac{2}{3}\right).

(b) Dot product: ba=(1)(2)+(3)(1)+(1)(2)=2+32=3\mathbf{b} \cdot \mathbf{a} = (1)(2) + (3)(1) + (1)(-2) = 2 + 3 - 2 = 3. The scalar projection of b\mathbf{b} onto a\mathbf{a} is baa=33=1\dfrac{\mathbf{b} \cdot \mathbf{a}}{|\mathbf{a}|} = \dfrac{3}{3} = 1.

Markers reward the magnitude 33, the unit vector, the dot product, and dividing by a|\mathbf{a}| (not a2|\mathbf{a}|^2) for the scalar projection.

WACE 20205 marksCalculator-free. Points are A(2,1,0)A(2, 1, 0), B(4,1,3)B(4, 1, 3) and C(2,5,0)C(2, 5, 0). Find the area of triangle ABCABC.
Show worked answer →

A triangle-area-by-cross-product question.

Edge vectors from AA: AB=(2,0,3)\vec{AB} = (2, 0, 3) and AC=(0,4,0)\vec{AC} = (0, 4, 0). Cross product AB×AC\vec{AB} \times \vec{AC}: i\mathbf{i} component (0)(0)(3)(4)=12(0)(0) - (3)(4) = -12; j\mathbf{j} component [(2)(0)(3)(0)]=0-[(2)(0) - (3)(0)] = 0; k\mathbf{k} component (2)(4)(0)(0)=8(2)(4) - (0)(0) = 8. So AB×AC=(12,0,8)\vec{AB} \times \vec{AC} = (-12, 0, 8).

Magnitude =144+0+64=208=413= \sqrt{144 + 0 + 64} = \sqrt{208} = 4\sqrt{13}. The triangle area is half this, 2132\sqrt{13}.

Markers reward the two edge vectors, the cross product, the magnitude, and halving for the triangle.

ExamExplained