Question

If vectors A^➙ = 4i^˄ + 3j^˄ – 2k^˄ and vectors B^➙ = 8i^˄ + 6j^˄ – 4k^˄ the angle between A^➙ and B^➙ is 

(A) 45° 

(B) 0° 

(C) 60° 

(D) 90°

Answer 1

Correct option: (B) 0°

Explanation:

cos θ = {(A^➙ ∙ B^➙) / (|A| |B|)}

A^➙ ∙ B^➙ = (4i˄ + 3j˄ – 2k˄) ∙ (8i˄ + 6j˄ – 4k˄)

= (4) (8) + (3) (6) + (– 2) (–4)

= 32 + 18 + 8

= 58

|A| = √(4² + 3² + 2²) = √29

|B| = √(8² + 6² (– 4)²) = √116

cos θ = [{58} / {√(29) ∙ √(116)}] = 1

hence θ = 0°