Two forces, 3 N and 4 N. Pull the angle between them round and the resultant runs from 7 down to 1. The sum of the magnitudes stays at 7 the whole way, which is why adding magnitudes is right exactly once.
The two arrows keep their lengths all the way through. Only the angle changes, and the diagonal is the answer.
| Scalar: a size only | Vector: a size and a direction |
|---|---|
| Distance travelled, 5 km | Displacement, 5 km north |
| Speed, 20 m s⁻¹ | Velocity, 20 m s⁻¹ on a bearing of 070° |
| Mass, 70 kg | Weight, 687 N downwards |
| Temperature, 31°C | Force, 4 N to the right |
The distinction is not decoration. Walk 3 km then 4 km and the distance is always 7 km, because distances are scalars and do add. Your displacement is between 1 and 7 km, because displacements are vectors and do not.
A vector in two dimensions is written as a column or in base vectors:
a = (3, 4) = 3i + 4j
and in three dimensions the third base vector k joins in. The rules are:
Magnitudes do not add, but a scalar does pass through one: |3(1, 2, 2)| = 3 × 3 = 9. Stretching a vector by 3 stretches its length by 3, which is the one simplification that is safe.
And kv is always parallel to v, for any non-zero k: same direction if k is positive, opposite if negative, and the length multiplied by |k|. That is the test for parallel vectors, and it is how you build a vector with a given direction and a given size. A particle moving at speed 7 in the direction 3i + 4j has velocity
7 × (3, 4)/5 = 7(0.6, 0.8) = (4.2, 5.6)
which is normalise, then scale: the unit vector carries the direction and the 7 carries the size.
A unit vector has magnitude 1 and carries only the direction. Divide by the magnitude:
the unit vector along (3, 4) is (3, 4)/5 = (0.6, 0.8)
and a check costs one line: 0.6² + 0.8² = 0.36 + 0.64 = 1. Any answer to "normalise this vector" that does not pass that check is wrong, and you can test it yourself before anyone marks it.
Position vectors and the vector between two points. The position vector of a point is the arrow from the origin to it, so A at (1, 5) has a = (1, 5). The vector from A to B is
AB = b − a, destination minus start.
With A(1, 5) and B(4, 1): AB = (3, −4), magnitude 5, and BA = (−3, 4), also magnitude 5. Same length, opposite direction. Subtracting the wrong way round gives a vector pointing backwards, and since the magnitude is unchanged there is nothing in the number to warn you.
Put the 3 along the x-axis and the 4 at angle θ to it. Then
a + b = (3 + 4cosθ, 4sinθ)
and its magnitude, at the angles in the figure, is:
So 7 is not an estimate that is slightly out. At 180° it is seven times the true answer.
The zero vector is a vector. Two forces of 4 N directly opposed give (0, 0), not "no answer". It has magnitude 0 and no direction, and it is what equilibrium looks like in components. Writing the answer as 0 rather than 0 costs nothing in a calculation and everything in a question that asks what the object does next.
Both machines hold vectors and return magnitudes directly, which removes the arithmetic and leaves you the modelling. Entering a vector takes longer than squaring three numbers, so for a magnitude, by hand is often faster; for a resultant at an awkward angle the machine wins.
When you may use it. Applications. A calculator is allowed in every paper. Vectors are entered as one-column matrices on both machines, so the matrix skills from 3.9 carry straight over.
The mark people lose. Writing |a + b| = |a| + |b|. It is true only when the two point the same way, and it is the step that turns a 3.61 into a 7. The habit that prevents it: never add two magnitudes. Add the vectors, componentwise, and take the magnitude once at the end. On the Casio, note also that the command is Norm(, not Abs: Abs is for real and complex numbers and will not take a vector at all.
1. Find the magnitude of (3, 4).
2. Find the unit vector in the direction of (3, 4) and give its first component.
3. Two forces of 3 N and 4 N act at 90° to each other. Find the magnitude of the resultant.
4. The same two forces, now at 120° to each other. Find the resultant, to 2 decimal places.
5. When is |a + b| equal to |a| + |b|?
1 markThe components, resolved correctly.
1 markThe sum, componentwise.
1 markThe magnitude, with units.
A student who writes 7 has skipped the first two marks and lost the third. There is no partial credit for adding magnitudes, because no step of it is a step of the method.
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