a = (3, 4) and b = (−4, 3) are both 5 long. a + b = (−1, 7), which is 7.0711 long, not 10. Direction is half of what a vector is, and adding the lengths throws it away.
|a| + |b| never moves off 10. |a + b| goes everywhere from 10 down to 0. They agree in exactly one position.
A vector has a magnitude and a direction, and nothing else. Two arrows of the same length pointing the same way are the same vector, wherever they are drawn. That is the point of the word.
In components, with base vectors i, j and k along the three axes:
v = v1i + v2j + v3k, written as a column of three numbers
and its magnitude is Pythagoras in three dimensions:
|v| = √(v1² + v2² + v3²)
For p = (1, 2, 2) that is √(1 + 4 + 4) = √9 = 3 exactly, which is why that vector appears so often in exam questions.
| Kind | What it means | Written |
|---|---|---|
| Position | Where a point is, measured from the origin. | The position vector of A is a. |
| Displacement | How to get from one point to another. | AB = b − a |
AB = b − a, and the order catches people: it is the destination minus the start, which is the opposite order to the letters. For A(1, 2, 2) and B(4, −2, 3), AB = (3, −4, 1), whose magnitude is √26 = 5.099. BA is (−3, 4, −1), the same length pointing the other way.
Move the slider from one end to the other. |a| and |b| never change, so |a| + |b| sits at 10 the whole time. But |a + b| starts at 10, falls through 8.6603 at 60°, 7.0711 at 90°, 5 at 120°, and reaches 0 when b points straight back along a.
So the rule |a + b| = |a| + |b| is right in exactly one position out of all of them: when the two point the same way. That is the general fact, and it has a name:
|a + b| ≤ |a| + |b|, the triangle inequality
with equality only when the vectors are parallel and pointing the same way. The name tells you the picture: the two vectors and their sum form a triangle, and one side of a triangle is never longer than the other two.
Scalar multiples DO scale the length. This is the distinction worth being precise about. For a = (3, 4), 3a = (9, 12) and |3a| = 15, which really is 3 × |a|. Multiplying by a number stretches without turning, so the length behaves simply. Adding brings a second direction in, and that is where the simple arithmetic fails.
In general |kv| = |k||v|, with the modulus on the k because a negative scalar reverses the direction but cannot give a negative length.
A unit vector has magnitude 1. To get one pointing along v, divide v by its own length:
v̂ = v / |v|
For AB = (3, −4, 1) with |AB| = 5.099, the unit vector is (0.5883, −0.7845, 0.1961), and its magnitude is 1 to every decimal place the calculator shows.
Check your unit vector by squaring and adding the components: it must come to 1. A student who divides by the wrong number gets a vector whose components square to something else, and that is a five-second check on a step that is easy to fumble.
Proofs with vectors. The guide asks for geometrical properties proved vectorially, and the method is always the same: write every displacement in terms of position vectors and let the algebra do the geometry.
For instance, if M is the midpoint of AB then m = (a + b)/2, because AM = ½AB = ½(b − a), so m = a + ½(b − a) = ½(a + b). No coordinates, no diagram, three lines. That is what makes vector proofs worth the trouble: they work in any number of dimensions and they never depend on how the picture happens to be drawn.
Both machines hold a vector as a short list or a matrix and will compute magnitudes and sums, which makes a long component calculation checkable. They will not draw the picture, and the picture is where the understanding is.
When you may use it. Analysis Paper 1 is non-calculator, and vector work is heavily Paper 1, because the numbers are chosen to come out whole. On Paper 2 the machine is a check on arithmetic.
The mark people lose. Writing AB = a − b. It is the right idea with the subtraction the wrong way round, and it gives a vector of the correct length pointing backwards, so the magnitude is right and everything with a direction in it is wrong. The habit: say "destination minus start" out loud. AB goes to B, so B's position comes first.
Throughout: a = (3, 4, 0), b = (−4, 3, 0), A(1, 2, 2) and B(4, −2, 3).
1. Find |a|.
2. Find |a + b|, to 4 decimal places.
3. Find |AB|, to 3 decimal places.
4. Find |3a|.
5. When is |a + b| equal to |a| + |b|?
1 markThe displacement written as destination minus start.
1 markThe components, correct and in order.
1 markThe magnitude, from the sum of squares.
On a proof question the marks are for the vector reasoning, so write each displacement in terms of position vectors before doing any arithmetic. A proof that falls back on coordinates usually scores less than one that stays in vectors, because the question was testing the method.
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