r = (1, 2) + λ(2, −1) and r = (3, 1) + λ(2, −1) are the same line. So is r = (1, 2) + λ(4, −2). A line has infinitely many vector equations and no reason to prefer one.
Press through the three equations and watch the line not move. The hollow circle moves, and the step length changes, and the line is the same set of points every time.
r = a + λb
Read that list again and the headline follows. "Any point" and "any vector along it" are not unique, so the equation is not either. Replace a with another point on the line, or b with any multiple of itself, and you have a different-looking equation for exactly the same line.
For the 3D example r = (1, 2, 3) + λ(2, −1, 2):
| λ | The point |
|---|---|
| 0 | (1, 2, 3) |
| 1 | (3, 1, 5) |
| 2 | (5, 0, 7) |
| −1 | (−1, 3, 1) |
Since (3, 1, 5) is on the line, r = (3, 1, 5) + μ(2, −1, 2) is the same line. And since (4, −2, 4) points the same way, r = (1, 2, 3) + t(4, −2, 4) is too. If your answer differs from the book's, check whether its a is on your line and its b is a multiple of yours. If both, you are right.
Parametric. Just write the three components separately:
x = 1 + 2λ, y = 2 − λ, z = 3 + 2λ
Cartesian. Make λ the subject of each and set them equal:
(x − 1)/2 = (y − 2)/(−1) = (z − 3)/2
This form is the useful one for checking, because it has no parameter in it: a point is on the line exactly when its coordinates satisfy it. All three of the two-dimensional equations in the figure reduce to (x − 1)/2 = (y − 2)/(−1), which is the proof that they are one line.
Watch the minus signs. The denominators are the direction components, so a direction of −1 gives a denominator of −1, and the numerators subtract the point's coordinates. Both are easy to write with the wrong sign and the result still looks like a line equation.
Lines have no position in this question, only direction, so use the direction vectors and ignore a entirely. Then it is 3.13:
cos θ = (b1 · b2) / (|b1| |b2|)
For directions (2, −1, 2) and (1, 2, 2): the product is 4, both magnitudes are 3, so cos θ = 4/9 = 0.4444 and θ = 63.61°.
Take the acute angle. A direction vector can be reversed without changing the line, and reversing it changes the sign of the product: with (−1, −2, −2) the product is −4 and the angle comes out 116.39°, which is 180 minus the first. Those are the same two lines, so the answer is the acute one. In practice: take the modulus of the scalar product before dividing.
In radians, which is the unit examination papers assume unless a question indicates otherwise: 63.61° is 1.1102 radians, and the obtuse partner 116.39° is 2.0314, so the two still add to π. Check the angle unit before you start, because both numbers look like reasonable answers in either unit.
Kinematics, which is the same equation renamed. If λ is time, then:
For r = (1, 2, 3) + t(2, −1, 2) the speed is |(2, −1, 2)| = 3. After 4 seconds the particle is at (9, −2, 11) and has travelled 12 units, which is speed times time as it should be.
And this is where the non-uniqueness stops being free: r = (1, 2, 3) + t(4, −2, 4) is the same line but a different motion, at speed 6. As geometry the two equations are interchangeable; as kinematics they are not.
The useful job here is the one students most need: deciding whether the book's answer and theirs are the same line. That is two checks and the machine does both.
When you may use it. Analysis Paper 1 is non-calculator, and line work is mostly Paper 1 because the numbers are chosen to be whole. The angle between two lines is the part that needs a calculator, and it is Paper 2 work.
The mark people lose. Assuming a different equation is a wrong equation. Students who match the book's a and b exactly are fine; students who do not often rewrite perfectly good work. The habit: run the two checks. Is their a a point on your line, and is their b a multiple of your b? Both yes means the same line. Converting to Cartesian form is useful because the parameter is gone, but the Cartesian form is not unique either: it still carries the point and the scale you started from. The pair in question 5 below reduces to (x − 1)/2 = (y − 2)/(−1) = (z − 3)/2 and to (x − 3)/4 = (y − 1)/(−2) = (z − 5)/4. Those are the same line, written two ways. Waiting for the two forms to match character for character is the error, not the test.
Throughout: the line r = (1, 2, 3) + λ(2, −1, 2).
1. Find the z-coordinate of the point at λ = 2.
2. If λ is time in seconds, find the speed.
3. Find the angle between this line and the line with direction (1, 2, 2), in degrees to 2 decimal places.
4. How far does the particle travel in the first 4 seconds?
5. A classmate writes r = (3, 1, 5) + μ(4, −2, 4) for the same line. Are they right?
1 markA correct point on the line.
1 markA correct direction vector.
1 markThe equation written in the form r = a + λb.
Any valid point and any valid direction earn the first two marks, so there is no need to match a particular answer. The third mark is for the form: a direction written without the λ, or a point and direction listed without being assembled, does not earn it.
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