Ex 10.4, 2

Transcript

Ex 10.4, 2 Find a unit vector perpendicular to each of the vector 𝑎﷯ + 𝑏﷯ and 𝑎﷯ − 𝑏﷯, where 𝑎﷯ = 3 𝑖﷯ + 2 𝑗﷯ + 2 𝑘﷯ and 𝑏﷯ = 𝑖﷯ + 2 𝑗﷯ − 2 𝑘﷯ . 𝑎﷯ = 3 𝑖﷯ + 2 𝑗﷯ + 2 𝑘﷯ 𝑏﷯ = 1 𝑖﷯ + 2 𝑗﷯ − 2 𝑘﷯ ( 𝑎﷯ + 𝑏﷯) = (3 + 1) 𝑖﷯ + (2 + 2) 𝑗﷯ + (2 − 2) 𝑘﷯ = 4 𝑖﷯ + 4 𝑗﷯ + 0 𝑘﷯ ( 𝑎﷯ − 𝑏﷯) = (3 − 1) 𝑖﷯ + (2 − 2) 𝑗﷯ + (2 − (−2)) 𝑘﷯ = 2 𝑖﷯ + 0 𝑗﷯ + 4 𝑘﷯ Now, we need to find a vector perpendicular to both 𝑎﷯ + 𝑏﷯ and 𝑎﷯ − 𝑏﷯, We know that ( 𝑎﷯ × 𝑏﷯) is perpendicular to 𝑎﷯ and 𝑏﷯ Replacing 𝑎﷯ by ( 𝑎﷯ + 𝑏﷯) & 𝑏﷯ by ( 𝑎﷯ − 𝑏﷯) ( 𝒂﷯ + 𝒃﷯) × ( 𝒂﷯ − 𝒃﷯) will be perpendicular to ( 𝒂﷯ + 𝒃﷯) and ( 𝒂﷯ − 𝒃﷯) Let 𝑐﷯ = ( 𝑎﷯ + 𝑏﷯) × ( 𝑎﷯ − 𝑏﷯) ∴ 𝑐﷯ = 𝑖﷯﷮ 𝑗﷯﷮ 𝑘﷯﷮ 4﷮2﷯﷮ 4﷮0﷯﷮ 0﷮4﷯﷯﷯ = 𝑖﷯ 4×4﷯−(0×0)﷯ − 𝑗﷯ 4×4﷯−(2×0)﷯ + 𝑘﷯ 4×0﷯−(2×4)﷯ = 𝑖﷯ (16 − 0) − 𝑗﷯ (16 − 0) + 𝑘﷯ (0 − 8) = 16 𝑖﷯ − 16 𝑗﷯ − 8 𝑘﷯ ∴ 𝑐﷯ = 16 𝑖﷯ − 16 𝑗﷯ − 8 𝑘﷯ Now, Unit vector of 𝑐﷯ = 1﷮𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑐﷯﷯ × 𝑐﷯ Magnitude of 𝑐﷯ = ﷮162+ −16﷯2+ −8﷯2﷯ 𝑐﷯﷯ = ﷮256+256+64﷯ = ﷮576﷯ = 24 Unit vector of 𝑐﷯ = 1﷮ 𝑐﷯﷯﷯ × 𝑐﷯ = 1﷮24﷯ × 16 𝑖﷯ − 16 𝑗﷯ − 8 𝑘﷯﷯ = 𝟐﷮𝟑﷯ 𝒊﷯ − 𝟐﷮𝟑﷯ 𝒋﷯ − 𝟏﷮𝟑﷯ 𝒌﷯ . Therefore the required unit vector is 2﷮3﷯ 𝑖﷯ − 2﷮3﷯ 𝑗﷯ − 1﷮3﷯ 𝑘﷯ . Note: There are always two perpendicular vectors So, another vector would be = − 𝟐﷮𝟑﷯ 𝒊﷯ − 𝟐﷮𝟑﷯ 𝒋﷯ − 𝟏﷮𝟑﷯ 𝒌﷯﷯ = −𝟐﷮𝟑﷯ 𝒊﷯ + 𝟐﷮𝟑﷯ 𝒋﷯ + 𝟏﷮𝟑﷯ 𝒌﷯ Hence, the perpendicular vectors are 2﷮3﷯ 𝑖﷯ − 2﷮3﷯ 𝑗﷯ − 1﷮3﷯ 𝑘﷯ & −2﷮3﷯ 𝑖﷯ + 2﷮3﷯ 𝑗﷯ + 1﷮3﷯ 𝑘﷯