Worked Example · Vectors in Two Dimensions · Paper 2 · 5 marks

Expressing a Vector as a Combination of Two Others

Rig, founder of IGCSE Add Math Malaysia

Written by Rig, our founder

8 years teaching IGCSE & SPM maths · Updated 16 August 2026

To write one vector as a combination of two others, compare components. The i\mathbf{i}-parts must match and the j\mathbf{j}-parts must match, giving two simultaneous equations in the scalars λ\lambda and μ\mu.

The vectors a=i+j\mathbf{a} = \mathbf{i} + \mathbf{j}, b=ij\mathbf{b} = \mathbf{i} - \mathbf{j} and c=5i+j\mathbf{c} = 5\mathbf{i} + \mathbf{j} are given. Find scalars λ\lambda and μ\mu such that c=λa+μb\mathbf{c} = \lambda\mathbf{a} + \mu\mathbf{b}. [5]

The working

Step 1, write out λa+μb\lambda\mathbf{a} + \mu\mathbf{b} and group the components: λ(i+j)+μ(ij)=(λ+μ)i+(λμ)j(M1)\lambda(\mathbf{i} + \mathbf{j}) + \mu(\mathbf{i} - \mathbf{j}) = (\lambda + \mu)\mathbf{i} + (\lambda - \mu)\mathbf{j} \quad \text{(M1)}

Step 2, compare with c=5i+j\mathbf{c} = 5\mathbf{i} + \mathbf{j}, matching i\mathbf{i} and j\mathbf{j} components: λ+μ=5(i-components)\lambda + \mu = 5 \qquad (\mathbf{i}\text{-components}) λμ=1(j-components)(M1, A1)\lambda - \mu = 1 \qquad (\mathbf{j}\text{-components}) \quad \text{(M1, A1)}

Step 3, solve simultaneously. Add the two equations (the μ\mu cancels): 2λ=6    λ=3(M1)2\lambda = 6 \;\Rightarrow\; \lambda = 3 \quad \text{(M1)}

Substitute back into λ+μ=5\lambda + \mu = 5: 3+μ=5    μ=2(A1)3 + \mu = 5 \;\Rightarrow\; \mu = 2 \quad \text{(A1)}

So c=3a+2b\mathbf{c} = 3\mathbf{a} + 2\mathbf{b}. Check: 3(i+j)+2(ij)=5i+j=c3(\mathbf{i}+\mathbf{j}) + 2(\mathbf{i}-\mathbf{j}) = 5\mathbf{i} + \mathbf{j} = \mathbf{c} ✓.

Where the marks are won and lost

  • Comparing components is the method: two vectors are equal only if both components agree, giving one equation each.
  • Keep the i\mathbf{i}-equation and j\mathbf{j}-equation straight: λ+μ=5\lambda + \mu = 5 (from i\mathbf{i}) and λμ=1\lambda - \mu = 1 (from j\mathbf{j}). Swapping them scrambles the solution.
  • The check by substitution is quick insurance and is often rewarded.

Common mistakes

  • Failing to expand μ(ij)\mu(\mathbf{i} - \mathbf{j}) to μiμj\mu\mathbf{i} - \mu\mathbf{j} (sign on the j\mathbf{j}).
  • Setting up only one equation and guessing the other scalar.
  • Arithmetic slips solving the simultaneous pair.

Full method: Vector Problem-Solving notes. Topic home: Vectors pillar.

Common questions

How do I find the scalars in c = λa + μb?
Write the right-hand side out and compare components: the i-components must match and the j-components must match. That gives two simultaneous equations in λ and μ, which you solve as usual. Comparing components works because two vectors are equal only when both of their components are equal. Check your scalars by substituting back into λa + μb and confirming you recover c. Mixing up which equation is the i and which is the j comparison is the main pitfall.

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