Vectors in the Plane — Topic Review — Solutions
15 exam-style questions covering all lessons in this topic. Click Show Answer to reveal the full worked solution.
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Fluency
Q1 — Magnitude and Unit Vector
Given v = 5i − 12j, find: (a) |v| (b) the unit vector v̂ (c) a vector of magnitude 3 in the direction of v.
(a) |v| = √(52 + (−12)2) = √(25 + 144) = √169 = 13
(b) v̂ = v/|v| = (5/13)i − (12/13)j
(c) 3v̂ = 3 × (5/13)i − 3 × (12/13)j = (15/13)i − (36/13)j
Check magnitude: √((15/13)² + (36/13)²) = √(225+1296)/169 = √1521/169 = 39/13 = 3 ✓
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Fluency
Q2 — Vector Between Points and Midpoint
Given P = (−2, 3) and Q = (4, 7), find: (a) PQ (b) |PQ| (c) the midpoint M of PQ.
(a) PQ = OQ − OP = (4 − (−2))i + (7 − 3)j = 6i + 4j
(b) |PQ| = √(36 + 16) = √52 = 2√13
(c) OM = ½(OP + OQ) = ½((−2+4)i + (3+7)j) = ½(2i + 10j) = i + 5j
Midpoint M = (1, 5)
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Understanding
Q3 — Find Unknown Scalar
Find all values of k such that |ki + 3j| = 5.
|ki + 3j|² = k² + 9 = 25
k² = 16 ⇒ k = ±4
Both k=4 and k=−4 give vectors of magnitude 5: (4,3) and (−4,3), which point in different directions but have the same length.
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Fluency
Q4 — Vector Arithmetic
Given a = 3i + j and b = i − 2j, find: (a) 2a − b (b) |3a + 2b|
(a) 2a = 6i + 2j; b = i − 2j; 2a − b = (6−1)i + (2−(−2))j = 5i + 4j
(b) 3a = 9i + 3j; 2b = 2i − 4j; 3a + 2b = 11i − j
|3a + 2b| = √(121 + 1) = √122
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Understanding
Q5 — Triangle Midsegment Theorem
Triangle with A = (1, 0), B = (4, 4), C = (7, 3). M is the midpoint of AB and N is the midpoint of BC. Find MN and show it is parallel to AC with half its length.
OM = ½(OA + OB) = ½((1+4)i + (0+4)j) = (5/2)i + 2j
ON = ½(OB + OC) = ½((4+7)i + (4+3)j) = (11/2)i + (7/2)j
MN = ON − OM = (11/2 − 5/2)i + (7/2 − 2)j = 3i + (3/2)j
AC = OC − OA = 6i + 3j = 2(3i + (3/2)j) = 2MN
Since MN = ½AC, MN is parallel to AC and has half its length. MN : AC = 1 : 2. ✓
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Problem Solving
Q6 — Parallelogram Vertices
ABCD is a parallelogram with A = (2, 1), B = (5, 2), D = (3, 4). Find: (a) the coordinates of C (b) the intersection of the diagonals.
(a) In a parallelogram, AB = DC.
AB = OB − OA = 3i + j
OC = OD + DC = OD + AB = (3i + 4j) + (3i + j) = 6i + 5j
C = (6, 5)
(b) In a parallelogram the diagonals bisect each other. The intersection is the midpoint of both diagonals.
Midpoint of AC = ½((2+6)i + (1+5)j) = 4i + 3j
Midpoint of BD = ½((5+3)i + (2+4)j) = 4i + 3j ✓
Diagonals intersect at (4, 3).
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Fluency
Q7 — Dot Product and Perpendicularity
Given a = 2i + 3j and b = 6i − 4j, find a·b and determine whether a and b are perpendicular.
a·b = 2(6) + 3(−4) = 12 − 12 = 0
Since a·b = 0, a and b are perpendicular. ✓
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Fluency
Q8 — Angle Between Vectors
Find the angle between u = 3i + 4j and v = 5i, giving your answer to the nearest degree.
u·v = 3(5) + 4(0) = 15
|u| = √(9 + 16) = 5; |v| = 5
cosθ = (u·v)/(|u||v|) = 15/(5 × 5) = 15/25 = 3/5 = 0.6
θ = arccos(0.6) ≈ 53°
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Understanding
Q9 — Vector Projection
Find the vector projection of a = 4i + 3j onto b = 2i.
a·b = 4(2) + 3(0) = 8; |b|² = 4 + 0 = 4
projb(a) = (a·b/|b|²)b = (8/4)(2i) = 2 × 2i = 4i
Geometrically: the projection of (4,3) onto the x-axis is simply (4,0), which matches 4i. ✓
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Understanding
Q10 — Centroid of a Triangle
In triangle OAB, let OA = a and OB = b. M is the midpoint of AB. Show that the median OM passes through the point G = ⅓(a + b) and that the median from A also passes through G.
Median from O: goes from O to M = ½(a + b).
Parametric form: r = t × ½(a + b). At t = 2/3: r = ⅓(a + b). So G lies on this median.
Median from A: goes from A (position a) to midpoint of OB = ½b.
Parametric form: r = a + s(½b − a) = (1−s)a + ½sb.
At G = ⅓(a + b): need (1−s) = 1/3 ⇒ s = 2/3. Check: ½×(2/3) = 1/3. ✓
Both medians pass through G = ⅓(a + b), the centroid.
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Understanding
Q11 — Prove ABCD is a Parallelogram
Points A = (1, 1), B = (5, 2), C = (4, 5), D = (0, 4). Use vectors to: (a) prove ABCD is a parallelogram (b) show the diagonals bisect each other.
(a) Compute sides:
AB = (5−1, 2−1) = (4, 1)
DC = (4−0, 5−4) = (4, 1)
Since AB = DC (equal magnitude and direction), AB is parallel to DC and equal in length. Therefore ABCD is a parallelogram. ✓
(b) Midpoint of diagonal AC = ½(1+4, 1+5) = (5/2, 3)
Midpoint of diagonal BD = ½(5+0, 2+4) = (5/2, 3)
Both diagonals share the same midpoint, so they bisect each other. ✓
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Problem Solving
Q12 — Angle in a Semicircle
Let O be the centre of a circle of radius r. A and B are the endpoints of a diameter, so OA = a and OB = −a (with |a| = r). P is any point on the circle, so OP = p with |p| = r. Prove that ∠APB = 90°.
We need to show PA ⊥ PB, i.e. PA·PB = 0.
PA = OA − OP = a − p
PB = OB − OP = −a − p
PA·PB = (a − p)·(−a − p)
= −a·a − a·p + p·a + p·p
= −|a|² + |p|² [since a·p cancels]
= −r² + r² = 0
Since PA·PB = 0, we have PA ⊥ PB, so ∠APB = 90°. ✓
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Fluency
Q13 — Write Parametric Equations
Write the parametric equations for the line through A = (3, −2) with direction d = i + 4j. State the coordinates of the point when t = 2.
Parametric equations: x = 3 + t, y = −2 + 4t
At t = 2: x = 3 + 2 = 5, y = −2 + 8 = 6. Point: (5, 6)
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Understanding
Q14 — Find Intersection of Two Lines
Find the intersection of ℓ1: r = (1, 3) + t(2, −1) and ℓ2: r = (5, 1) + s(1, 2).
Setting x and y equal:
x: 1 + 2t = 5 + s ⇒ 2t − s = 4 ...(1)
y: 3 − t = 1 + 2s ⇒ −t − 2s = −2 ⇒ t + 2s = 2 ...(2)
From (2): t = 2 − 2s. Substitute into (1): 2(2 − 2s) − s = 4 ⇒ 4 − 5s = 4 ⇒ s = 0, t = 2.
Intersection: (1 + 4, 3 − 2) = (5, 1)
Verify on ℓ2: at s=0: (5, 1) ✓
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Problem Solving
Q15 — Line Through Two Points — Full Analysis
A line passes through A = (1, 4) and B = (4, 2). (a) Write its parametric equations. (b) Find where it crosses the x-axis. (c) Find the foot of the perpendicular from the origin O = (0, 0) to the line.
(a) AB = (3, −2). Parametric: x = 1 + 3t, y = 4 − 2t
(b) Set y = 0: 4 − 2t = 0 ⇒ t = 2. Then x = 1 + 6 = 7.
The line crosses the x-axis at (7, 0).
(c) General point on line: Q = (1 + 3t, 4 − 2t).
OQ = (1 + 3t)i + (4 − 2t)j. Direction of line: d = (3, −2).
For perpendicularity: OQ·d = 0:
3(1 + 3t) + (−2)(4 − 2t) = 0
3 + 9t − 8 + 4t = 0 ⇒ 13t = 5 ⇒ t = 5/13
Foot of perpendicular: x = 1 + 15/13 = 28/13; y = 4 − 10/13 = 42/13.
Foot: (28/13, 42/13)