Solutions — Vector Operations and Scalar Multiples

Q1 — Basic Vector Arithmetic

Given a = 4i − j and b = −2i + 3j.

(a) a + b = (4 − 2)i + (−1 + 3)j = 2i + 2j

(b) a − b = (4 − (−2))i + (−1 − 3)j = 6i − 4j

(c) 2a = 2(4i − j) = 8i − 2j

3b = 3(−2i + 3j) = −6i + 9j

2a + 3b = (8 − 6)i + (−2 + 9)j = 2i + 7j

Q2 — Vector Between Two Points

A = (3, 7), B = (1, 2).

AB = OB − OA = (1i + 2j) − (3i + 7j)

AB = (1 − 3)i + (2 − 7)j = −2i − 5j

Note: BA = −AB = 2i + 5j. Always subtract the starting point’s position vector from the ending point’s.

Q3 — Scalar Multiplication

(a) 3(2i + 5j) = 6i + 15j

(b) −2(i − 4j) = −2i + 8j = −2i + 8j

The negative scalar reverses direction: multiply each component by −2.

(c) 0.5(−6i + 10j) = −3i + 5j

Q4 — Midpoint

P = (1, 4), Q = (7, 2).

OP = i + 4j, OQ = 7i + 2j

OM = ½(OP + OQ) = ½((1 + 7)i + (4 + 2)j) = ½(8i + 6j) = 4i + 3j

Midpoint M = (4, 3).

Q5 — Finding an Unknown Vector

Given p + r = q, rearranging: r = q − p

r = (3i − j) − (i + 2j) = (3 − 1)i + (−1 − 2)j = 2i − 3j

Check: p + r = (i + 2j) + (2i − 3j) = 3i − j = q ✓

Q6 — Parallel Vectors

a = 6i − 4j, b = −9i + 6j.

Test b = ka: from the i-component, k = −9/6 = −3/2.

Check j-component: (−3/2) × (−4) = 6 ✓

Therefore b = −(3/2)a. The vectors are parallel; since k < 0, they point in opposite directions.

|a| = √(36 + 16) = √52 = 2√13. |b| = √(81 + 36) = √117 = 3√13 = (3/2)|a| ✓

Q7 — Collinear Points

A = (1, 2), B = (4, 5), C = (10, 11).

AB = (4 − 1)i + (5 − 2)j = 3i + 3j

AC = (10 − 1)i + (11 − 2)j = 9i + 9j

AC = 3(3i + 3j) = 3AB

Since AC = 3AB, the vectors are parallel and share point A. Therefore A, B, C are collinear. ✓

C lies beyond B from A (k = 3 > 1); the order along the line is A, B, C.

Q8 — Linear Combination

We seek scalars m and n such that r = ma + nb:

7i + j = m(2i + j) + n(i − j) = (2m + n)i + (m − n)j

Equating components: 2m + n = 7 ...(1) and m − n = 1 ...(2)

Adding (1) and (2): 3m = 8 ⇒ m = 8/3

From (2): n = m − 1 = 8/3 − 3/3 = 5/3

Check (1): 2(8/3) + 5/3 = 16/3 + 5/3 = 21/3 = 7 ✓

r = (8/3)a + (5/3)b

Q9 — Parallelogram Vertex

ABCD is a parallelogram with A = (1, 1), B = (4, 3), C = (6, 7).

In parallelogram ABCD (vertices in order), AB = DC.

AB = OB − OA = (4 − 1)i + (3 − 1)j = 3i + 2j

Since DC = AB: OC − OD = 3i + 2j

OD = OC − AB = (6i + 7j) − (3i + 2j) = 3i + 5j

Therefore D = (3, 5).

Check (diagonals bisect each other):

Midpoint of AC = ½(1 + 6, 1 + 7) = (7/2, 4)

Midpoint of BD = ½(4 + 3, 3 + 5) = (7/2, 4) ✓

Q10 — Midpoints and Parallel Vectors

A = (0, 4), B = (6, 2), C = (8, 8). M is the midpoint of AB; N is the midpoint of BC.

Step 1: Find M and N.

OM = ½(OA + OB) = ½(6i + 6j) = 3i + 3j ⇒ M = (3, 3)

ON = ½(OB + OC) = ½(14i + 10j) = 7i + 5j ⇒ N = (7, 5)

Step 2: Find MN.

MN = ON − OM = (7 − 3)i + (5 − 3)j = 4i + 2j

Step 3: Find AC and compare.

AC = OC − OA = (8 − 0)i + (8 − 4)j = 8i + 4j = 2(4i + 2j) = 2MN

Therefore MN = ½AC: MN is parallel to AC and has half its length. ✓

This is the Triangle Midsegment Theorem: the segment joining midpoints of two sides is parallel to the third side and half its length.

Vector Operations and Scalar Multiples — Solutions