Higher Maths: Vectors in Three Dimensions

Vectors carry a whole block of the geometric skills in Higher Mathematics, and a multi-part vectors question is a regular feature of the question papers. This material works through that block, based on the geometric skills listed in the Higher Mathematics course specification: determining vector connections and working with vectors in three dimensions.

The first part is about connections between points. You express a vector as the resultant of a pathway, such as the vector from a vertex of a triangle to the midpoint of the opposite side, and add vectors written with i, j and k to find a resultant and its magnitude. You then prove that three points are collinear by showing that two vectors are parallel and share a point, state the ratio in which the middle point divides the line, find unknown coordinates that make three points collinear, and find the coordinates of a point that divides a line internally in a given ratio.

The second part is about the scalar product. You evaluate it both in component form and as $|\mathbf{a}||\mathbf{b}|\cos\theta$, calculate the angle between two vectors, and find an angle in a triangle, where the classic slip is to use a vector that points into the vertex instead of away from it. You use a zero scalar product to find an unknown that makes two vectors perpendicular, and apply the properties of the scalar product, such as a⋅a=∣a∣2 and the distributive law, to evaluate expressions from magnitudes and an angle alone. Unit vectors are covered too: finding one, and finding a missing component so that a vector has magnitude 1.

Several questions are marked non-calculator, as they would be in question paper 1, and use exact values such as cos⁡60∘=12 and cos⁡120∘=−12; the angle calculations need a calculator, as in question paper 2.

The material offers three formats. The quiz has 12 questions whose explanations show every step and name the slip behind each typical wrong option. The flashcards collect the facts you need to recall quickly: the section formula, the collinearity argument, the two forms of the scalar product and the perpendicular test. The printable written work has 8 longer tasks to set out by hand, including a pathway in a tetrahedron, a full collinearity proof, an angle between two vectors, finding $|\mathbf{a} + \mathbf{b}|$ by expanding a scalar product, and an explanation of why angle calculations must use vectors pointing away from the vertex.

It is aimed at S5 and S6 students preparing for Higher Mathematics and assumes National 5 work on vectors and magnitudes.

  • Express a vector as the resultant of a pathway and find its magnitude
  • Prove that three points are collinear and state the ratio in which a point divides a line
  • Find the coordinates of a point dividing a line internally in a given ratio
  • Find and use unit vectors, including the basis vectors i, j and k
  • Evaluate a scalar product in component form and from magnitudes and an angle
  • Calculate the angle between two vectors and an angle in a triangle
  • Use a zero scalar product and the properties of the scalar product to find unknowns

Practice material written by Zestly, based on the Qualifications Scotland (formerly SQA) Higher Mathematics course specification, version 3.0 (geometric skills: determining vector connections; working with vectors).

Sample question

In triangle $ABC$, $\overrightarrow{AB} = \mathbf{p}$ and $\overrightarrow{AC} = \mathbf{q}$. $M$ is the midpoint of $BC$. Which expression is $\overrightarrow{AM}$?

See the answer

$\frac{1}{2}(\mathbf{p} + \mathbf{q})$

Follow a pathway: $\overrightarrow{AM} = \overrightarrow{AB} + \frac{1}{2}\overrightarrow{BC}$ and $\overrightarrow{BC} = \mathbf{q} - \mathbf{p}$. So $\overrightarrow{AM} = \mathbf{p} + \frac{1}{2}(\mathbf{q} - \mathbf{p}) = \frac{1}{2}(\mathbf{p} + \mathbf{q})$. $\frac{1}{2}(\mathbf{q} - \mathbf{p})$ is only $\overrightarrow{BM}$, and $\mathbf{p} + \mathbf{q}$ is twice the answer.

← Higher Mathematics

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