Misc 6 - Find a vector of magnitude 5 units, parallel to - Miscellaneous

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Misc 6 Find a vector of magnitude 5 units, and parallel to the resultant of the vectors 𝑎﷯ = 2 𝑖﷯ + 3 𝑗﷯ − 𝑘﷯ and 𝑏﷯ = 𝑖﷯ − 2 𝑗﷯ + 𝑘﷯. 𝑎﷯ = 2 𝑖﷯ + 3 𝑗﷯ − 𝑘﷯ 𝑏﷯ = 𝑖﷯ − 2 𝑗﷯ + 𝑘﷯ ( 𝑎﷯ + 𝑏﷯) = (2 + 1) 𝑖﷯ + (3 − 2) 𝑗﷯ + (−1 + 1) 𝑘﷯ = 3 𝑖﷯ + 1 𝑗﷯ + 0 𝑘﷯ Let 𝑐﷯ = ( 𝑎﷯ + 𝑏﷯) ∴ 𝑐﷯ = 3 𝑖﷯ + 1 𝑗﷯ + 0 𝑘﷯ Magnitude of 𝑐﷯ = ﷮32+12+02﷯ 𝑐﷯﷯ = ﷮9+1﷯ = ﷮10﷯ Unit vector in direction of 𝑐﷯ = 1﷮ 𝑐﷯﷯﷯ × 𝑐﷯ 𝑐﷯ = 1﷮ ﷮10﷯﷯ × 3 𝑖﷯+1 𝑗﷯+0 𝑘﷯﷯ 𝑐﷯ = 𝟑﷮ ﷮𝟏𝟎﷯﷯ 𝒊﷯ + 𝟏﷮ ﷮𝟏𝟎﷯﷯ 𝒋﷯ + 0 𝒌﷯ Vector with magnitude 1 = 3﷮ ﷮10﷯﷯ 𝑖﷯ + 1﷮ ﷮10﷯﷯ 𝑗﷯ + 0 𝑘﷯ Vector with magnitude 5 = 5 × 3﷮ ﷮10﷯﷯ 𝑖﷯ + 1﷮ ﷮10﷯﷯ 𝑗﷯ + 0 𝑘﷯﷯ = 15﷮ ﷮10﷯﷯ 𝑖﷯ + 2﷮ ﷮10﷯﷯ 𝑗﷯ + 0 𝑘﷯. = 15﷮ ﷮10﷯﷯ 𝑖﷯ + 2﷮ ﷮10﷯﷯ 𝑗﷯. Hence the required vector is 𝟏𝟓﷮ ﷮𝟏𝟎﷯﷯ 𝒊﷯ + 𝟐﷮ ﷮𝟏𝟎﷯﷯ 𝒋﷯.

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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