Topic 10 covers vectors in two and three dimensions: column and i, j, k notation, magnitude and direction, adding and scaling vectors, unit vectors, position vectors and the distance between two points in 3D, testing for parallel vectors and collinear points, dividing a line in a given ratio, and using vectors to solve geometric problems.
Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect β stars. Watch for the Exam trap notes β that is where the marks get thrown away. Press Start when you are ready.
Vectors Β· notation
Vectors in two and three dimensions
A vector has magnitude and direction; a scalar has magnitude only. In 3D we use the mutually perpendicular unit vectors i, j, k along the x, y and z axes.
A 3D vector and its magnitude β Pythagoras, applied twice.
a = xi + yj + zk Β· |a| = β(xΒ² + yΒ² + zΒ²)the column form is the same thing written vertically
Exam trap: the magnitude is never negative β square the components, then take the positive root. |2i β 3j + 6k| = β(4 + 9 + 36) = 7, not β(4 β 9 + 36).
Calculate
Your turn β magnitude in 2D
1Find the magnitude of the vector 3i β 4j.
Set-up
|3i β 4j| = β(9 + 16) = β25 = 5.
Hint: β(3Β² + (β4)Β²) = β25.
Calculate
Your turn β magnitude in 3D
2Find the magnitude of the vector a = 2i β 3j + 6k.
Set-up
|a| = β(4 + 9 + 36) = β49 = 7.
Hint: β(2Β² + 3Β² + 6Β²) = β49.
Quick check
Magnitude in 3D
?What is |xi + yj + zk|?
Vectors Β· algebra
Adding, scaling and unit vectors
Adding: add the components. (2i + j) + (3i β 4j) = 5i β 3j. Geometrically it is the triangle (or parallelogram) law.
Scaling: Ξ»a has the same direction as a (opposite if Ξ» < 0) and magnitude |Ξ»||a|.
Parallel: b is parallel to a exactly when b = Ξ»a for some scalar Ξ». So 4i β 2j + 6k = 2(2i β j + 3k) is parallel to 2i β j + 3k.
Unit vector: divide by the magnitude β Γ’ = a / |a|, which has magnitude 1.
Worked example β the unit vector in the direction of a = 2i β 3j + 6k
3Find the i-component of the unit vector in the direction of a = 2i β 3j + 6k. (Give a decimal to 4 d.p., or the fraction 2/7.)
Set-up
Γ’ = (1/7)(2i β 3j + 6k), so the i-component is 2/7 = 0.2857 (4 d.p.).
Hint: Divide each component by |a| = 7.
Quick check
Parallel vectors
?Which vector is parallel to 2i β j + 3k?
Vectors Β· direction
Magnitude and direction form
In 2D you can convert between component form and magnitudeβdirection form.
|v| = β(xΒ² + yΒ²) Β· tan ΞΈ = y/xΞΈ is measured anticlockwise from the positive x-axis β sketch it to get the correct quadrant
Worked example β v = 3i + 4j
|v| = β(9 + 16) = 5.
ΞΈ = arctan(4/3) = 53.1Β° (1 d.p.) above the positive x-axis. Both components are positive, so the vector really is in the first quadrant. β
Exam trap: arctan only ever returns an angle between β90Β° and 90Β°. For v = β3i β 4j the calculator still says 53.1Β°, but the true direction is 53.1 + 180 = 233.1Β°. Always sketch.
Calculate
Your turn β direction
4The vector v = 3i + 4j makes an angle ΞΈ with the positive x-axis. Find ΞΈ in degrees to 1 decimal place.
Set-up
tan ΞΈ = 4/3 = 1.3333
ΞΈ = arctan(1.3333) = 53.130β¦Β° = 53.1Β° (1 d.p.), in the first quadrant.
Β°
Hint: tan ΞΈ = 4/3.
Sort it
Sort by magnitude
Tap a vector, then its magnitude. Square the components and add β no calculator needed.
5οΈβ£ Magnitude 5
7οΈβ£ Magnitude 7
1οΈβ£3οΈβ£ Magnitude 13
Vectors Β· position
Position vectors and distance
The position vector of A is OA = a, measured from the origin. Then:
AB = b β a“destination minus start” β the single most useful line in the topic
The distance between two points in 3D is the magnitude of that vector:
d = β[(xββxβ)Β² + (yββyβ)Β² + (zββzβ)Β²]
Worked example β A(1, 2, 3) and B(4, 6, 15)
AB = b β a = (4 β 1)i + (6 β 2)j + (15 β 3)k = 3i + 4j + 12k.
|AB| = β(9 + 16 + 144) = β169 = 13.
Exam trap: AB = b β a, not a β b. Getting it backwards gives a vector of the right length pointing the wrong way β fine for a distance, fatal for a direction or a ratio.
Calculate
Your turn β distance in 3D
5A is the point (1, 2, 3) and B is (4, 6, 15). Find the exact distance AB.
The point P dividing AB in the ratio m : n has OP = a + [m/(m + n)](b β a).
Collinear points: A, B, C are collinear if AB is a scalar multiple of BC (parallel and sharing the point B).
Worked example β A(1, β2, 4), B(5, 2, β4), and P divides AB in the ratio 1 : 3
AB = b β a = 4i + 4j β 8k.
OP = a + ΒΌ(AB) = (1 + 1)i + (β2 + 1)j + (4 β 2)k = 2i β j + 2k, so P is (2, β1, 2).
|OP| = β(4 + 1 + 4) = β9 = 3.
Exam trap: the ratio 1 : 3 means P is one quarter of the way from A to B (1 part out of 1 + 3 = 4), not one third. Count the total parts.
Calculate
Your turn β ratio and magnitude
6A is (1, β2, 4) and B is (5, 2, β4). The point P divides AB in the ratio 1 : 3. Find the distance OP from the origin.
Set-up
b β a = 4i + 4j β 8k, so ΒΌ(b β a) = i + j β 2k.
OP = (i β 2j + 4k) + (i + j β 2k) = 2i β j + 2k, so P = (2, β1, 2).
|OP| = β(4 + 1 + 4) = β9 = 3.
Hint: OP = a + ΒΌ(b β a). Find P first, then its magnitude.
Quick check
Collinear points
?How do you show that the points A, B and C are collinear?
Match it
Vector flashcards
Tap a card on the left, then its meaning.
Object
Meaning
Vectors Β· in context
Using vectors in context
Vectors are the language of mechanics (Paper 3): forces, velocities and displacements all add as vectors.
Worked example β resultant force
Two forces act on a particle: Fβ = (3i + 4j) N and Fβ = (βi + 2j) N.
Resultant R = Fβ + Fβ = 2i + 6j N.
|R| = β(4 + 36) = β40 = 6.32 N, at arctan(6/2) = 71.6Β° from the i-direction.
A particle in equilibrium has resultant force 0, so the i-components sum to zero and the j-components sum to zero β two equations.
s = ut + Β½atΒ² works exactly the same way with vector u and a.
Not on 9MA0: the scalar (dot) product and the vector equation of a line belong to Further Mathematics. If a question seems to need the angle between two general vectors, you are looking at the wrong specification.
Recap
The big ideas to take away
Notation: a = xi + yj + zk in 3D; |a| = β(xΒ² + yΒ² + zΒ²)
Unit vector: Γ’ = a/|a| β magnitude 1
Parallel: b = Ξ»a Β· Collinear: AB = Ξ»BC through a shared point
Position vectors: AB = b β a; midpoint Β½(a + b); ratio m : n β a + m/(m+n) (b β a)
Distance in 3D: β[(xββxβ)Β² + (yββyβ)Β² + (zββzβ)Β²]
In context: forces and velocities add as vectors; equilibrium β each component sums to 0
Topic 10 is short but heavily examined β and it is the bridge into the Mechanics paper. Press Finish to see your score.
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