Solutions — Vector Proofs in Geometry — Grade 11 Specialist Maths

Q1 — Triangle Midpoint Theorem

OA = a, OB = b, OC = c. M = midpoint of AB, N = midpoint of AC.

OM = ½(a + b), ON = ½(a + c)

MN = ON − OM = ½(a + c) − ½(a + b) = ½(c − b)

BC = c − b

Therefore MN = ½BC. Since MN is a scalar multiple of BC, MN ∥ BC, and |MN| = ½|BC|. ✓

Q2 — Diagonals of a Parallelogram

OA = a, OB = b, OD = d. ABCD is a parallelogram, so AB = DC.

Part (a): AB = b − a. DC = OC − d. Setting equal: OC = b − a + d.

Part (b):

Midpoint of AC: ½(a + OC) = ½(a + b − a + d) = ½(b + d)

Midpoint of BD: ½(b + d) = ½(b + d)

The midpoints of both diagonals are identical, so the diagonals bisect each other. ✓

Q3 — Centroid of a Triangle

OA = a, OB = b, OC = c.

Midpoints: M₁ (midpt of BC) = ½(b + c); M₂ (midpt of AC) = ½(a + c); M₃ (midpt of AB) = ½(a + b).

Median from A: r = (1 − t)a + ½tb + ½tc

Median from B: r = ½sa + (1 − s)b + ½sc

Equating coefficients — from c: ½t = ½s, so t = s. From b: ½t = 1 − t ⇒ t = 2/3.

OG = (1/3)a + (1/3)b + (1/3)c = ⅓(a + b + c)

By symmetry of the expression in a, b, c, the median from C also passes through this same point. All three medians are concurrent at G. ✓

Q4 — Diagonals of a Rhombus are Perpendicular

A at origin, AB = b, AD = d, with |b| = |d|. Then OC = b + d.

AC = b + d. BD = d − b.

AC · BD = (b + d) · (d − b) = |d|² − b · d + d · b − |b|² = |d|² − |b|² = 0

Since |b| = |d|, the dot product is zero, so AC ⊥ BD. ✓

Q5 — Angle in a Semicircle

O = origin, OA = a, OB = −a, OP = p with |p| = |a| = r.

PA = a − p. PB = −a − p.

PA · PB = (a − p) · (−a − p)

= −|a|² − a · p + p · a + |p|²

= −r² + r² = 0

Therefore angle APB = 90°. ✓ (Thales’ Theorem)

Q6 — Length of a Median

OA = a, OB = b. M = midpoint of AB: OM = ½(a + b).

|OM|² = ½(a + b) · ½(a + b) = ¼(|a|² + 2a · b + |b|²)

Therefore: |OM| = ½√(|a|² + 2a · b + |b|²)

Note: if a ⊥ b (i.e., OAB is right-angled at O), then a · b = 0 and the median from O equals ½√(|a|² + |b|²) = ½|AB| — in a right triangle, the median to the hypotenuse equals half the hypotenuse.

Q7 — Prove a Quadrilateral is a Parallelogram

A = (1, 2), B = (4, 3), C = (5, 6), D = (2, 5).

AB = B − A = (3, 1). DC = C − D = (3, 1). AB = DC ✓

AD = D − A = (1, 3). BC = C − B = (1, 3). AD = BC ✓

Both pairs of opposite sides are equal and parallel, so ABCD is a parallelogram. ✓

Q8 — Point on a Line Segment

A = (1, 2), B = (7, 8), P = (3, 4).

AB = (6, 6). AP = (2, 2) = ⅓(6, 6) = ⅓AB

Since AP = ⅓AB and 0 < ⅓ < 1, P lies between A and B on segment AB. ✓

AP : PB = ⅓ : ⅔ = 1 : 2

Q9 — Parallelogram: Diagonals and Intersection

A = (0, 0), B = (4, 0), C = (5, 3), D = (1, 3).

Part (a):

AB = (4, 0). DC = C − D = (4, 0). AB = DC ✓

AD = (1, 3). BC = C − B = (1, 3). AD = BC ✓

ABCD is a parallelogram. ✓

Part (b):

Midpoint of AC: ½(0 + 5, 0 + 3) = (2.5, 1.5)

Midpoint of BD: ½(4 + 1, 0 + 3) = (2.5, 1.5)

Both diagonals pass through (2.5, 1.5), confirming they bisect each other at this point. ✓

Q10 — Intersection of Medians

O = origin, OA = a, OB = b. M = ½b (midpoint of OB). N = ½a (midpoint of OA).

Median AM: r = (1 − t)a + ½tb

Median BN: r = ½sa + (1 − s)b

Equating coefficients: 1 − t = ½s ...(1) and ½t = 1 − s ...(2)

From (2): s = 1 − ½t. Substituting into (1): 1 − t = ½ − ¼t ⇒ ¾t = ½ ⇒ t = 2/3.

OG = (1 − 2/3)a + ½(2/3)b = ⅓(a + b) ✓

The centroid divides each median in the ratio 2 : 1 from vertex to midpoint. Here, along median AM, t = 2/3 means G is 2/3 of the way from A to M.

Vector Proofs in Geometry — Solutions