Example 15

Transcript

Example 15 If 𝑎﷯ = 5 𝑖﷯ − 𝑗﷯ − 3 𝑘﷯ and 𝑏﷯ = 𝑖﷯ + 3 𝑗﷯ − 5 𝑘﷯ , then show that the vectors 𝑎﷯ + 𝑏﷯ and 𝑎﷯ − 𝑏﷯ are perpendicular. Two vectors 𝑝﷯ and 𝑞﷯ are perpendicular if their scalar product is zero, i.e. 𝒑﷯ . 𝒒﷯ = 0 𝑎﷯ = 5 𝑖﷯ − 𝑗﷯ − 3 𝑘﷯ 𝑏﷯ = 𝑖﷯ + 3 𝑗﷯ − 5 𝑘﷯ ( 𝑎﷯ + 𝑏﷯) = (5 + 1) 𝑖﷯ + (−1 + 3) 𝑗﷯ + (−3 + (−5)) 𝑘﷯ = 6 𝑖﷯ + 2 𝑗﷯ + 8 𝑘﷯ ( 𝑎﷯ − 𝑏﷯) = (5 − 1) 𝑖﷯ + (−1 − 3) 𝑗﷯ + (−3 − (−5)) 𝑘﷯ = 4 𝑖﷯ − 4 𝑗﷯ + 2 𝑘﷯ Now, we have to show that ( 𝑎﷯ + 𝑏﷯) and ( 𝑎﷯ − 𝑏﷯) are perpendicular to each other. So, we need to show ( 𝑎﷯ + 𝑏﷯) . ( 𝑎﷯ − 𝑏﷯) = 0 Taking LHS ( 𝒂﷯ + 𝒃﷯) . ( 𝒂﷯ − 𝒃﷯) = (6 𝑖﷯ + 2 𝑗﷯ − 8 𝑘﷯) . (4 𝑖﷯ − 4 𝑗﷯ + 2 𝑘﷯) = (6 × 4) + (2 × −4) + (−8 × 2) = 24 + (−8) + (-16) = 24 + (−24) = 0 Since ( 𝑎﷯ + 𝑏﷯) . ( 𝑎﷯ − 𝑏﷯) = 0 Hence, ( 𝑎﷯ + 𝑏﷯) is perpendicular to ( 𝑎﷯ − 𝑏﷯)