← Vectors in Three Dimensions › Dot Product and Cross Product in 3D › Solutions

Dot Product and Cross Product in 3D — Full Worked Solutions

1. Compute the dot product a · b for each pair. Fluency

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   (a) a · b = (2)(1) + (3)(−2) + (−1)(4) = 2 − 6 − 4 = −8

   (b) a · b = (0)(3) + (5)(0) + (2)(−1) = 0 + 0 − 2 = −2

   (c) a · b = (4)(2) + (−1)(4) + (2)(1) = 8 − 4 + 2 = 6

2. Find the angle θ between each pair of vectors. Fluency

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   (a) a = (1,0,1), b = (0,1,1)

   a · b = 0 + 0 + 1 = 1

   |a| = √(1+0+1) = √2;   |b| = √(0+1+1) = √2

   cosθ = 1/(√2 × √2) = 1/2

   θ = cos⁻¹(1/2) = 60.0°

   (b) a = (2,1,−2), b = (1,2,2)

   a · b = 2 + 2 − 4 = 0

   Since the dot product is zero, the vectors are perpendicular: θ = 90.0°

3. Determine whether each pair of vectors is perpendicular. Fluency

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   (a) u = (3,−2,1), v = (1,2,1):   u · v = 3 − 4 + 1 = 0. Perpendicular.

   (b) u = (1,4,−2), v = (2,−1,1):   u · v = 2 − 4 − 2 = −4 ≠ 0. Not perpendicular.

   (c) u = (2,0,−3), v = (3,1,2):   u · v = 6 + 0 − 6 = 0. Perpendicular.

4. Find the cross product a × b for each pair. Fluency

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   (a) a = (1,0,0), b = (0,1,0):

   i: (0)(0)−(0)(1) = 0;   j: −[(1)(0)−(0)(0)] = 0;   k: (1)(1)−(0)(0) = 1

   a × b = (0, 0, 1) = k (as expected from basis vector cross products)

   (b) a = (2,1,−1), b = (1,3,2):

   i: (1)(2)−(−1)(3) = 2+3 = 5

   j: −[(2)(2)−(−1)(1)] = −[4+1] = −5

   k: (2)(3)−(1)(1) = 6−1 = 5

   a × b = (5, −5, 5)

   Verify: a · (5,−5,5) = 10−5−5 = 0 ✓;   b · (5,−5,5) = 5−15+10 = 0 ✓

5. Area of parallelogram with a = (1,3,−2) and b = (2,1,4). Understanding

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   Compute a × b:

   i: (3)(4)−(−2)(1) = 12+2 = 14

   j: −[(1)(4)−(−2)(2)] = −[4+4] = −8

   k: (1)(1)−(3)(2) = 1−6 = −5

   a × b = (14, −8, −5)

   Area = |a × b| = √(14² + 8² + 5²) = √(196 + 64 + 25) = √285 ≈ 16.88 square units

6. Area of triangle with A = (1,0,0), B = (0,2,0), C = (0,0,3). Understanding

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   AB = B − A = (−1, 2, 0);   AC = C − A = (−1, 0, 3)

   Compute AB × AC:

   i: (2)(3)−(0)(0) = 6

   j: −[(−1)(3)−(0)(−1)] = −[−3−0] = 3

   k: (−1)(0)−(2)(−1) = 0+2 = 2

   AB × AC = (6, 3, 2)

   |AB × AC| = √(36+9+4) = √49 = 7

   Area of triangle = ½ × 7 = 3.5 square units

7. Unit vector perpendicular to a = (1,1,0) and b = (0,1,1). Understanding

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   Compute n = a × b:

   i: (1)(1)−(0)(1) = 1

   j: −[(1)(1)−(0)(0)] = −1

   k: (1)(1)−(1)(0) = 1

   n = (1, −1, 1);   |n| = √3

   n̂ = (1/√3, −1/√3, 1/√3)

   Check: a · n = 1−1+0 = 0 ✓;   b · n = 0−1+1 = 0 ✓

8. Find k so that a = (k,2,−1) and b = (3,k,4) are perpendicular. Understanding

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   Set a · b = 0:

   (k)(3) + (2)(k) + (−1)(4) = 0

   3k + 2k − 4 = 0

   5k = 4

   k = 4/5

   Check: a = (4/5, 2, −1), b = (3, 4/5, 4)

   a · b = 12/5 + 8/5 − 4 = 20/5 − 4 = 4 − 4 = 0 ✓

9. Cross product a = (3,−1,2), b = (1,4,1). Area of parallelogram. Problem Solving

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   Compute a × b:

   i: (−1)(1)−(2)(4) = −1−8 = −9

   j: −[(3)(1)−(2)(1)] = −[3−2] = −1

   k: (3)(4)−(−1)(1) = 12+1 = 13

   a × b = (−9, −1, 13)

   |a × b| = √(81+1+169) = √251 ≈ 15.84 square units

   Verify perpendicularity: a · (−9,−1,13) = −27+1+26 = 0 ✓;   b · (−9,−1,13) = −9−4+13 = 0 ✓

10. Forces F₁ = (2,−3,1) N and F₂ = (1,2,−4) N. Find: (a) angle between them, (b) vector perpendicular to both. Problem Solving

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    (a) F₁ · F₂ = (2)(1) + (−3)(2) + (1)(−4) = 2 − 6 − 4 = −8

    |F₁| = √(4+9+1) = √14;   |F₂| = √(1+4+16) = √21

    cosθ = −8 / (√14 × √21) = −8/√294

    √294 ≈ 17.146

    θ = cos⁻¹(−8/17.146) ≈ cos⁻¹(−0.4667) ≈ 117.8°

    (b) Compute F₁ × F₂:

    i: (−3)(−4)−(1)(2) = 12−2 = 10

    j: −[(2)(−4)−(1)(1)] = −[−8−1] = 9

    k: (2)(2)−(−3)(1) = 4+3 = 7

    A vector perpendicular to both forces: (10, 9, 7)

    Verify: (2,−3,1)·(10,9,7) = 20−27+7 = 0 ✓;   (1,2,−4)·(10,9,7) = 10+18−28 = 0 ✓