Resultant of Two Vectors and Its Magnitude

Adding vectors is done component by component, then the magnitude of the resultant follows from Pythagoras. It’s a foundational skill that appears inside longer vector problems.

Two vectors are $\mathbf{a} = 3\mathbf{i} + 2\mathbf{j}$ and $\mathbf{b} = \mathbf{i} - 5\mathbf{j}$. (i) Find the resultant $\mathbf{a} + \mathbf{b}$. [2] (ii) Find the magnitude of the resultant. [2]

The working

(i) Add the $\mathbf{i}$-components and the $\mathbf{j}$-components separately: $\mathbf{a} + \mathbf{b} = (3 + 1)\mathbf{i} + (2 + (-5))\mathbf{j} = 4\mathbf{i} - 3\mathbf{j} \quad \text{(M1, A1)}$

(ii) The magnitude uses Pythagoras on the resultant’s components: $|\mathbf{a} + \mathbf{b}| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \quad \text{(M1, A1)}$

Where the marks are won and lost

• Add like components: $\mathbf{i}$ with $\mathbf{i}$, $\mathbf{j}$ with $\mathbf{j}$. Mind the signs: $2 + (-5) = -3$.
• The magnitude is $\sqrt{(\text{i-part})^2 + (\text{j-part})^2}$; the sign inside disappears on squaring, $(-3)^2 = 9$.
• $\sqrt{16 + 9} = \sqrt{25} = 5$ (a clean whole number here, a hint the working is right).

Common mistakes

• Adding an $\mathbf{i}$-component to a $\mathbf{j}$-component.
• Sign slip: $2 - 5 = -3$, not $+3$ or $-7$.
• Subtracting the squares instead of adding them for the magnitude.

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