Magnitude and the Unit Vector

Two staple vector skills sit in this question: the magnitude (a Pythagoras calculation on the components) and the unit vector (the same direction, scaled to length $1$ by dividing through by the magnitude).

The vector $\mathbf{a} = 3\mathbf{i} - 4\mathbf{j}$. (i) Find $|\mathbf{a}|$. [2] (ii) Find the unit vector in the direction of $\mathbf{a}$. [2]

The working

(i) The magnitude is $\sqrt{(\text{i-component})^2 + (\text{j-component})^2}$: $|\mathbf{a}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \quad \text{(M1, A1)}$

(ii) A unit vector is the vector divided by its magnitude: $\hat{\mathbf{a}} = \frac{1}{|\mathbf{a}|}\mathbf{a} = \frac{1}{5}(3\mathbf{i} - 4\mathbf{j}) = \frac{3}{5}\mathbf{i} - \frac{4}{5}\mathbf{j} \quad \text{(M1, A1)}$

As a decimal, $\hat{\mathbf{a}} = 0.6\mathbf{i} - 0.8\mathbf{j}$. Check: $\sqrt{0.6^2 + 0.8^2} = \sqrt{0.36 + 0.64} = 1$ ✓.

Where the marks are won and lost

• The sign inside the square does not matter, $(-4)^2 = 16$. The magnitude is always non-negative.
• The unit vector divides by $|\mathbf{a}|$ (here $5$), not by $|\mathbf{a}|^2$. Dividing by $25$ is a common slip.
• Keep the direction: both components are scaled by the same $\frac15$, so the $\mathbf{j}$-component stays negative.

Common mistakes

• Writing $|\mathbf{a}| = \sqrt{3^2 - 4^2}$ (subtracting instead of adding squares).
• Dividing by $25$ instead of $5$ for the unit vector.
• Losing the negative sign on the $\mathbf{j}$-component.

Full method: Vector Notation & Magnitude notes. Topic home: Vectors pillar.