Two vectors of length 3, at a right angle: their scalar product is 0 and the area they span is 9, the largest it ever gets. The zero is not a failed sum, it is the perpendicularity result, and the figure below shows the two readings trading places as the angle closes.
Both vectors keep their length of 3 throughout. Watch the two bars trade places.
Two ways to compute it, and the whole sub-topic lives in setting them equal.
From components: multiply matching components and add.
(1, 2, 2) · (2, 2, 1) = 2 + 4 + 2 = 8
From geometry:
v · w = |v| |w| cosθ
Here |v| = |w| = 3, so 8 = 9 cosθ, giving
cosθ = 8/9 = 0.889 and θ = 27.3°
The answer is a number, not a vector. That is what "scalar product" means, and an answer written as (2, 4, 2) has not finished the addition.
The sign tells you the shape of the angle before you compute it.
So a question asking "show that these are perpendicular" is asking for one dot product and the word "zero". Nothing else is needed, and nothing else earns the mark.
This one returns a vector, perpendicular to both of the originals:
(1, 2, 2) × (2, 2, 1) = (2×1 − 2×2, 2×2 − 1×1, 1×2 − 2×2) = (−2, 3, −2)
and a check costs two dot products: (−2, 3, −2) · (1, 2, 2) = −2 + 6 − 4 = 0, and against (2, 2, 1) it is −4 + 6 − 2 = 0. Perpendicular to both, as promised.
Its magnitude is what questions ask for:
|v × w| = √(4 + 9 + 4) = √17 = 4.12
and geometrically that is the area of the parallelogram the two vectors span, so the triangle with them as two sides has area
½ × 4.12 = 2.06
| Scalar product | Vector product | |
|---|---|---|
| Written | v · w | v × w |
| Gives | a number | a vector |
| Geometric form | |v||w| cosθ | magnitude |v||w| sinθ |
| Largest when | parallel | perpendicular |
| Zero when | perpendicular | parallel |
| Used for | the angle between two vectors or lines | areas, and a direction perpendicular to both |
| Order | v · w = w · v | w × v = −(v × w) |
The last row matters. Swapping the order of a vector product reverses it, so the two answers point opposite ways. The magnitude is the same either way, which is why area questions forgive the order and direction questions do not.
The syllabus asks for two more quantities, and both come straight from the products you already have. For v = (1, 2, 2) and w = (2, 2, 1):
Those two are at right angles to each other, so they have to satisfy Pythagoras:
2.667² + 1.374² = 9 = |v|²
That is the check, and it costs one line. If your two components do not square and add to |v|², one of them is wrong.
Which way does v × w point? Perpendicular to both, and of the two directions that are, the one given by the right-hand screw rule: curl the fingers of your right hand from v towards w and the thumb points along v × w. Dividing by the magnitude gives the unit normal, so the full statement of the vector product is
v × w = |v| |w| sinθ n
and here n = (−2, 3, −2)/4.123 = (−0.485, 0.728, −0.485), whose magnitude is 1 as it must be. Swap the order and the thumb reverses, which is why w × v = −(v × w).
Take the direction vectors, not the points, and dot them. For the lines
r = (1, 0, 4) + t(1, 2, 2) and r = (3, 1, 0) + s(2, 2, 1)
the angle is the same 27.3° as before, because the directions are the same two vectors. The starting points are irrelevant: two lines in three dimensions need not even meet and still have an angle between them.
Questions ask for the acute angle, so if a dot product comes out negative, take the angle and subtract it from 180°. A direction vector can be written either way round, and the sign of the dot product follows which way you wrote it, so the obtuse answer is an artefact of notation rather than geometry.
Three ways to lose the mark on a dot product.
Both products are single commands on both machines, so the arithmetic is free and the only remaining difficulty is knowing which product the question needs. Decide that on paper first: a number and an angle means dot, an area means cross.
When you may use it. Applications. A calculator is allowed in every paper. A three-component cross product by hand is error-prone and the machine is reliable, so this is one place to hand the work over without hesitation.
The mark people lose. Giving the parallelogram's area when the question asked for a triangle. |v × w| = 4.12 is the parallelogram; the triangle is 2.06. The two differ by a factor of two, which is large enough to be wrong and small enough to look right. Read the question twice and write "triangle, so halve it" on the page before you compute anything. And set degrees mode before the inverse cosine, or 27.3 arrives as 0.476.
1. Find (3, 4, 0) · (4, −3, 0).
2. Find (1, 2, 2) · (2, 2, 1).
3. Find the angle between (1, 2, 2) and (2, 2, 1), in degrees to 1 decimal place.
4. Find the area of the triangle with (1, 2, 2) and (2, 2, 1) as two of its sides, to 2 decimal places.
5. A scalar product comes out as exactly 0. What have you found?
1 markThe product itself, from components.
1 markThe two magnitudes.
1 markcosθ, then θ in degrees.
1 markFor an area, the halving if it is a triangle.
The commonest single lost mark in this sub-topic is the last line of all: 0.889 written where 27.3° belonged, or 4.12 where 2.06 belonged. Both are one step from correct and score as if they were nothing.
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