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NSWMaths Extension 2Syllabus dot point

How are vectors in three dimensions described, how do we find lines and spheres, and how does the scalar product extend to space?

Represent three-dimensional vectors in component form, compute the scalar product and magnitude, and find vector equations of lines and the equation of a sphere

A focused answer to the HSC Maths Extension 2 dot point on 3D vectors. Component form, magnitude, the scalar product and angle between vectors, parametric vector equations of lines, and the equation of a sphere, with verified worked examples.

Generated by Claude Opus 4.76 min answer

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

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  1. What this dot point is asking
  2. Vectors in component form
  3. The scalar product in three dimensions
  4. Vector equation of a line
  5. The equation of a sphere

What this dot point is asking

NESA wants you to extend two-dimensional vectors into three dimensions. You must write vectors in component form using i\mathbf{i}, j\mathbf{j}, k\mathbf{k}, compute magnitude and the scalar (dot) product, find the angle between vectors, write vector equations of lines, and derive the equation of a sphere.

Vectors in component form

A point in space is described by an ordered triple (x,y,z)(x, y, z). The position vector of that point is

r=xi+yj+zk,\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k},

where i\mathbf{i}, j\mathbf{j}, k\mathbf{k} are mutually perpendicular unit vectors along the axes. Addition, subtraction and scalar multiplication act component by component, exactly as in two dimensions.

The magnitude (length) of a=a1i+a2j+a3k\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} is

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

which is the three-dimensional Pythagorean distance from the origin.

The scalar product in three dimensions

The scalar (dot) product is

a⋅b=a1b1+a2b2+a3b3=∣a∣ ∣b∣cos⁡θ,\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:

cos⁡θ=a⋅b∣a∣ ∣b∣.\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|}.

Two non-zero vectors are perpendicular exactly when a⋅b=0\mathbf{a}\cdot\mathbf{b} = 0. The scalar product is commutative and distributes over addition.

Vector equation of a line

A line through the point with position vector a\mathbf{a} and parallel to the direction vector d\mathbf{d} consists of all points

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

As the parameter λ\lambda runs over the real numbers, r\mathbf{r} traces the whole line. To find the line through two points AA and BB, take a=OA→\mathbf{a} = \overrightarrow{OA} and direction d=AB→=b−a\mathbf{d} = \overrightarrow{AB} = \mathbf{b} - \mathbf{a}.

To test whether a point lies on the line, solve for a single λ\lambda that reproduces all three components. If no consistent λ\lambda exists, the point is off the line.

The equation of a sphere

A sphere is the set of points at fixed distance RR (the radius) from a centre CC with position vector c\mathbf{c}:

∣r−c∣=R.|\mathbf{r} - \mathbf{c}| = R.

In Cartesian coordinates with c=(a,b,c)\mathbf{c} = (a, b, c) this is

(x−a)2+(y−b)2+(z−c)2=R2.(x - a)^2 + (y - b)^2 + (z - c)^2 = R^2.

This is the natural 3D analogue of the circle equation, and completing the square converts an expanded form back to centre-radius form.

Exam-style practice questions

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

2024 HSC2 marksFind the angle between the two vectors u = (1, 2, -2) and v = (4, -4, 7), giving your answer in radians, correct to 1 decimal place.
Show worked answer →

Use the scalar (dot) product: u . v = |u| |v| cos(theta).

Dot product: u . v = (1)(4) + (2)(-4) + (-2)(7) = 4 - 8 - 14 = -18.

Magnitudes: |u| = sqrt(1 + 4 + 4) = sqrt(9) = 3, and |v| = sqrt(16 + 16 + 49) = sqrt(81) = 9.

So cos(theta) = -18/(3 . 9) = -18/27 = -2/3.

theta = arccos(-2/3) = 2.3 radians (to 1 decimal place).

Mark notes: 1 mark for computing u . v and the magnitudes, 1 mark for cos(theta) = -2/3 giving theta = 2.3 radians.

2023 HSC2 marksFind a vector equation of the line through the points A(-3, 1, 5) and B(0, 2, 3).
Show worked answer →

A vector equation of a line needs a point on it and a direction vector.

Direction vector: AB = B - A = (0 - (-3), 2 - 1, 3 - 5) = (3, 1, -2).

Using A(-3, 1, 5) as the point, a vector equation is

r = (-3, 1, 5) + lambda (3, 1, -2), where lambda is a real parameter.

(Using B as the point, or any non-zero scalar multiple of the direction, is equally correct.)

Mark notes: 1 mark for a correct direction vector AB = (3, 1, -2), 1 mark for a complete vector equation with a point and the parameter lambda.

2024 HSC3 marksThe line L passes through the points A(3, 5, -4) and B(7, 0, 2). (i) Find a vector equation of the line L. (ii) Determine, giving reasons, whether the point C(10, 5, -2) lies on the line L.
Show worked answer →

Part (i). Direction vector: AB = B - A = (7 - 3, 0 - 5, 2 - (-4)) = (4, -5, 6).

A vector equation is r = (3, 5, -4) + lambda (4, -5, 6).

Part (ii). The point C lies on L only if there is a single lambda giving all three coordinates of C(10, 5, -2):

  • x: 10 = 3 + 4 lambda, so lambda = 7/4.
  • y: 5 = 5 - 5 lambda, so lambda = 0.

These values of lambda are inconsistent (7/4 is not 0), so no single lambda produces C.

Therefore C(10, 5, -2) does not lie on the line L.

Mark notes: part (i) 1 mark for a correct vector equation. Part (ii): 1 mark for substituting C and solving for lambda in at least one coordinate, 1 mark for showing the lambda values are inconsistent and concluding C is not on L.