Solutions — Introduction to 3D Vectors — Grade 12 Specialist Maths

Find the magnitude of each vector. Fluency

(a) u = (3, 4, 0): |u| = √(9 + 16 + 0) = √25 = 5

(b) v = (1, −2, 2): |v| = √(1 + 4 + 4) = √9 = 3

(c) w = (2, 3, −6): |w| = √(4 + 9 + 36) = √49 = 7

Find the unit vector in the direction of each vector. Fluency

(a) a = (2, −1, 2): |a| = √(4+1+4) = √9 = 3

â = (1/3)(2, −1, 2) = (2/3, −1/3, 2/3)

Check: (2/3)² + (1/3)² + (2/3)² = 4/9 + 1/9 + 4/9 = 9/9 = 1 ✓

(b) b = (0, 1, −1): |b| = √(0+1+1) = √2

b̂ = (1/√2)(0, 1, −1) = (0, 1/√2, −1/√2) = (0, √2/2, −√2/2)

Find the direction cosines of v = (2, −3, 6). Fluency

|v| = √(4 + 9 + 36) = √49 = 7

cos α = 2/7, cos β = −3/7, cos γ = 6/7

α = cos⁻¹(2/7) ≈ 73.4°

β = cos⁻¹(−3/7) ≈ 115.4°

γ = cos⁻¹(6/7) ≈ 31.0°

Verify identity: (2/7)² + (−3/7)² + (6/7)² = 4/49 + 9/49 + 36/49 = 49/49 = 1 ✓

Given A = (1, 2, −1) and B = (4, −1, 3). Fluency

(a) AB = B − A = (4−1, −1−2, 3−(−1)) = (3, −3, 4)

(b) |AB| = √(3² + (−3)² + 4²) = √(9 + 9 + 16) = √34

(c) AB̂ = (1/√34)(3, −3, 4) = (3/√34, −3/√34, 4/√34)

Let a = (2, −1, 3), b = (1, 4, −2), c = (−3, 0, 1). Understanding

(a) 2b = (2, 8, −4)

a + 2b = (2+2, −1+8, 3−4) = (4, 7, −1)

(b) 3a = (6, −3, 9)

3a − c = (6−(−3), −3−0, 9−1) = (9, −3, 8)

(c) a + b = (2+1, −1+4, 3−2) = (3, 3, 1)

|a + b| = √(9 + 9 + 1) = √19

A vector with magnitude 5 and direction cosines cos α = 3/5, cos β = −4/5, cos γ = 0. Understanding

The unit vector in the direction of v is (cos α, cos β, cos γ) = (3/5, −4/5, 0).

v = 5 × (3/5, −4/5, 0) = (3, −4, 0)

Verify direction cosine identity: (3/5)² + (−4/5)² + 0² = 9/25 + 16/25 + 0 = 25/25 = 1 ✓

Verify magnitude: |(3, −4, 0)| = √(9+16+0) = √25 = 5 ✓

Determine whether each pair of vectors is parallel. Understanding

(a) a = (2, −4, 6), b = (−1, 2, −3): Check a = λb: 2 = λ(−1) ⇒ λ = −2. Then −2(−1)=2 ✓, −4=−2(2)=−4 ✓, 6=−2(−3)=6 ✓. Parallel (λ=−2).

(b) c = (1, 2, 3), d = (2, 4, 5): 1/2 = 0.5, 2/4 = 0.5, 3/5 = 0.6. The ratios are not all equal. Not parallel.

(c) e = (0, 3, −6), f = (0, −1, 2): First components are both 0. Check remaining: 3/(−1) = −3 and −6/2 = −3. All ratios equal −3. Parallel (e = −3f).

Show that P = (1,0,2), Q = (3,1,−1), R = (7,3,−7) are collinear. Understanding

PQ = Q − P = (3−1, 1−0, −1−2) = (2, 1, −3)

PR = R − P = (7−1, 3−0, −7−2) = (6, 3, −9)

Note: (6, 3, −9) = 3(2, 1, −3), so PR = 3PQ.

Since PR is a scalar multiple of PQ and both vectors start at P, the three points are collinear.

Particle from A=(2,−1,4) to B=(5,3,−2), then BC = 2AB. Find C and total distance. Problem Solving

AB = B − A = (5−2, 3−(−1), −2−4) = (3, 4, −6)

|AB| = √(9 + 16 + 36) = √61

BC = 2AB = 2(3, 4, −6) = (6, 8, −12)

Position of C: OC = OB + BC = (5,3,−2) + (6,8,−12) = (11, 11, −14)

|BC| = 2|AB| = 2√61

Total distance = |AB| + |BC| = √61 + 2√61 = 3√61 ≈ 23.4 units

Direction cosines: cos α = 1/3, cos β = cos γ, z-component positive, |v| = 6. Find v. Problem Solving

Using the direction cosine identity with cos β = cos γ:

cos²α + 2cos²β = 1

(1/3)² + 2cos²β = 1

1/9 + 2cos²β = 1

2cos²β = 8/9

cos²β = 4/9 ⇒ cos β = ±2/3

Since the z-component is positive and cos γ = cos β, we need cos γ = 2/3 > 0, so cos β = 2/3 (and hence the y-component is also positive).

Unit vector: v̂ = (1/3, 2/3, 2/3)

v = 6 × (1/3, 2/3, 2/3) = (2, 4, 4)

Verify: |(2,4,4)| = √(4+16+16) = √36 = 6 ✓

Introduction to 3D Vectors — Full Worked Solutions