Introduction to Vectors — Solutions
-
Q1 — Magnitudes
(a) |5i + 12j| = √(52 + 122) = √(25 + 144) = √169 = 13
This is a 5–12–13 Pythagorean triple — worth memorising.
(b) |−3i + 4j| = √((−3)2 + 42) = √(9 + 16) = √25 = 5
Signs don’t matter inside the square root since we square the components (3–4–5 triple).
(c) |8i − 6j| = √(82 + (−6)2) = √(64 + 36) = √100 = 10
A scaled 3–4–5 triple: 8 = 2×4, 6 = 2×3, so magnitude = 2×5 = 10.
-
Q2 — Position Vectors
The position vector of point (a, b) is ai + bj.
(a) A(3, 7): OA = 3i + 7j
(b) B(−2, 5): OB = −2i + 5j
(c) C(0, −4): OC = 0i + (−4)j = −4j
C lies on the negative y-axis, so its position vector has no i-component.
-
Q3 — Unit Vectors
(a) a = 4i + 3j
|a| = √(42 + 32) = √(16 + 9) = √25 = 5
รข = (4i + 3j) / 5 = (4/5)i + (3/5)j
Verification: √((4/5)2 + (3/5)2) = √(16/25 + 9/25) = √(25/25) = 1 ✓
(b) b = −5i + 12j
|b| = √((−5)2 + 122) = √(25 + 144) = √169 = 13
b̂ = (−5i + 12j) / 13 = (−5/13)i + (12/13)j
Verification: √(25/169 + 144/169) = √(169/169) = 1 ✓
-
Q4 — Negative Vectors
a = 2i − j
−a = −(2i − j) = −2i + j
Verification:
|a| = √(22 + (−1)2) = √(4 + 1) = √5
|−a| = √((−2)2 + 12) = √(4 + 1) = √5
|−a| = |a| = √5 ✓
The negative vector reverses direction (it points the other way) but has identical magnitude. This is a fundamental property: |−v| = |v| for any vector v.
-
Q5 — Finding a Point from a Vector
We use the relationship: PQ = OQ − OP, rearranged as OQ = OP + PQ.
OP = 1i + 3j (position vector of P(1, 3))
PQ = 5i + 2j (given)
OQ = (1 + 5)i + (3 + 2)j = 6i + 5j
Therefore Q = (6, 5).
Check: PQ = OQ − OP = (6 − 1)i + (5 − 3)j = 5i + 2j ✓
-
Q6 — Vector Between Two Points
A(2, 5) and B(6, 2).
AB = OB − OA = (6i + 2j) − (2i + 5j) = (6 − 2)i + (2 − 5)j = 4i − 3j
|AB| = √(42 + (−3)2) = √(16 + 9) = √25 = 5
The rule is: AB = (xB − xA)i + (yB − yA)j. Always subtract the tail coordinates from the head coordinates.
-
Q7 — Parallel Unit Vectors
v = 3i − 4j
|v| = √(9 + 16) = √25 = 5
Unit vector in the direction of v: v̂ = (3/5)i − (4/5)j
Unit vector in the opposite direction: −v̂ = −(3/5)i + (4/5)j
Both have magnitude 1. These are the only two unit vectors parallel to v because a unit vector parallel to v must be ±v̂.
-
Q8 — Vector with Given Magnitude
Direction vector: d = 5i + 12j
|d| = √(25 + 144) = √169 = 13
Unit vector: d̂ = (5/13)i + (12/13)j
Force vector (magnitude 13 N, in this direction):
F = 13 × d̂ = 13 × [(5/13)i + (12/13)j] = 5i + 12j N
In this case, the direction vector already had magnitude 13 (equal to the desired force), so the answer is simply the direction vector itself. This is a happy coincidence of the numbers chosen.
General method: F = |F| × d̂.
-
Q9 — Angle of a Unit Vector
Verification that u is a unit vector:
|u| = √((1/√2)2 + (1/√2)2) = √(1/2 + 1/2) = √1 = 1 ✓
Finding the angle:
Any unit vector in the plane can be written as (cos θ)i + (sin θ)j, where θ is the angle with the positive x-axis.
Comparing: cos θ = 1/√2 and sin θ = 1/√2
θ = 45° (or π/4 radians)
This vector bisects the positive x- and y-axes. It makes equal angles with both axes, which makes sense since both components are equal.
-
Q10 — Equal Vectors and Unknown Components
a = (2t − 1)i + 3j and b = 5i + (t + 5)j
For equal vectors, ALL corresponding components must be equal:
i-component equation: 2t − 1 = 5 ⇒ 2t = 6 ⇒ t = 3
j-component equation: 3 = t + 5 ⇒ t = −2
Since t = 3 and t = −2 cannot both be true, there is no value of t that makes these vectors equal.
This is a valid and instructive result: equal vectors require simultaneous equality of ALL components. If the two component equations give different values of t, the vectors can never be equal.
If the problem had read a = (2t − 1)i + (t + 2)j and b = 5i + 5j, then both equations give t = 3, yielding a = b = 5i + 5j.