Vector subtraction
Scalars & Vectors - OCR A-Level Physics
Key Definition
Vector subtraction
To subtract vector $\mathbf{B}$ from vector $\mathbf{A}$, reverse the direction of $\mathbf{B}$ to get $-\mathbf{B}$, then add this to $\mathbf{A}$ using the triangle (nose-to-tail) method. Subtraction is addition of the negative.
To subtract vector $\mathbf{B}$ from vector $\mathbf{A}$, reverse the direction of $\mathbf{B}$ to get $-\mathbf{B}$, then add this to $\mathbf{A}$ using the triangle (nose-to-tail) method. Subtraction is addition of the negative.
$$\mathbf{A} - \mathbf{B} = \mathbf{A} + (-\mathbf{B})$$
Diagram pending
Two vectors $\mathbf{A}$ and $\mathbf{B}$ drawn separately, then the construction for $\mathbf{A} - \mathbf{B}$: $\mathbf{A}$ drawn from origin, $-\mathbf{B}$ (with nose and tail swapped from $\mathbf{B}$) drawn from the nose of $\mathbf{A}$, and the resultant arrow drawn from the tail of $\mathbf{A}$ to the nose of $-\mathbf{B}$, labelled $\mathbf{A} - \mathbf{B}$.
Will be replaced with a GeoGebra SVG in stream 2.
- $-\mathbf{B}$ has the same magnitude as $\mathbf{B}$ but points in the opposite direction. On a diagram, swap the nose and tail of the arrow.
- To find $\mathbf{A} - \mathbf{B}$: draw $\mathbf{A}$, then draw $-\mathbf{B}$ from the nose of $\mathbf{A}$ (nose-to-tail). The resultant runs from the tail of $\mathbf{A}$ to the nose of $-\mathbf{B}$.
- Common use: change in velocity $\Delta\mathbf{v} = \mathbf{v}_\text{final} - \mathbf{v}_\text{initial}$. A ball that hits a wall at $5 \text{ m s}^{-1}$ and bounces straight back at $-3 \text{ m s}^{-1}$ has $\Delta v = -3 - 5 = -8 \text{ m s}^{-1}$ (sign shows direction reversed).
- Common use: relative velocity $\mathbf{v}_{A \text{ rel } B} = \mathbf{v}_A - \mathbf{v}_B$. Two cars travelling east at $20 \text{ m s}^{-1}$ and $15 \text{ m s}^{-1}$ have a relative velocity of $5 \text{ m s}^{-1}$ (the faster one pulls ahead).
- For vectors at an angle, calculate by components: subtract the $x$-components and $y$-components separately, then recombine with Pythagoras and $\tan^{-1}$.
$$(\mathbf{A} - \mathbf{B})_x = A_x - B_x \qquad (\mathbf{A} - \mathbf{B})_y = A_y - B_y$$
Common Mistake
HIGH
Wrong: Forgetting that velocities are vectors when calculating change in velocity. For a ball bouncing off a wall at the same speed, writing $\Delta v = 5 - 5 = 0$.
Right: Treat velocity as a signed quantity. If $+5 \text{ m s}^{-1}$ is the incoming velocity and the ball rebounds at $5 \text{ m s}^{-1}$ in the opposite direction (so $-5 \text{ m s}^{-1}$), then $\Delta v = -5 - (+5) = -10 \text{ m s}^{-1}$. The magnitude is $10 \text{ m s}^{-1}$, not zero. This drives the impulse and force on the ball.
Right: Treat velocity as a signed quantity. If $+5 \text{ m s}^{-1}$ is the incoming velocity and the ball rebounds at $5 \text{ m s}^{-1}$ in the opposite direction (so $-5 \text{ m s}^{-1}$), then $\Delta v = -5 - (+5) = -10 \text{ m s}^{-1}$. The magnitude is $10 \text{ m s}^{-1}$, not zero. This drives the impulse and force on the ball.
Examiner Tips and Tricks
- State the positive direction explicitly before substituting numbers (e.g. "take east as positive"). Examiners deduct marks for unstated sign conventions.
- The change $\Delta\mathbf{v}$ for a rebound is much larger than students expect, because the direction reverses. This is essential for momentum and impulse questions.
- On the parallelogram of two vectors drawn from the same point, the "other" diagonal (the one connecting the noses) represents $\mathbf{B} - \mathbf{A}$, not $\mathbf{A} + \mathbf{B}$.