Vector of magnitude 5 units and in the direction opposite to

2i + 3j − 6k  is  ____________

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Transcript

Question 15 (OR 2nd Question) Vector of magnitude 5 units and in the direction opposite to 2๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ 6๐‘˜ ฬ‚ is ____________ Let ๐‘Ž โƒ— = 2๐‘– ฬ‚ + 3๐‘— ฬ‚ โ€“ 6๐‘˜ ฬ‚ Magnitude of ๐‘Ž โƒ— = โˆš(22+32+(โˆ’6)2) |๐‘Ž โƒ— | = โˆš(4+9+36) = โˆš49 = 7 Unit vector opposite to direction of ๐‘Ž โƒ— = โ€“1 ร— 1/|๐‘Ž โƒ— | . ๐‘Ž โƒ— = (โˆ’1)/7 (2๐‘– ฬ‚ + 3๐‘— ฬ‚ โ€“ 6๐‘˜ ฬ‚) Thus, Vector with magnitude 1 opposite to ๐‘Ž โƒ— = (โˆ’1)/7 (2๐‘– ฬ‚ + 3๐‘— ฬ‚ โ€“ 6๐‘˜ ฬ‚) Vector with magnitude 5 opposite to ๐‘Ž โƒ— = (โˆ’5)/7 (2๐‘– ฬ‚ + 3๐‘— ฬ‚ โ€“ 6๐‘˜ ฬ‚) = 5/7 (โ€“2๐‘– ฬ‚ โ€“ 3๐‘— ฬ‚ + 6๐‘˜ ฬ‚) Hence, the required vector is ๐Ÿ“/๐Ÿ• (โ€“2๐’Š ฬ‚ โ€“ 3๐’‹ ฬ‚ + 6๐’Œ ฬ‚)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.