Vector of magnitude 5 units and in the direction opposite to

2i + 3j āˆ’ 6kĀ  isĀ  ____________

Vector of magnitude 5 units and in direction opposite to  2i + 3j - 6k

Question 15 (OR 2nd Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 2

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Question 15 (OR 2nd Question) Vector of magnitude 5 units and in the direction opposite to 2š‘– Ģ‚ + 3š‘— Ģ‚ āˆ’ 6š‘˜ Ģ‚ is ____________ Let š‘Ž āƒ— = 2š‘– Ģ‚ + 3š‘— Ģ‚ – 6š‘˜ Ģ‚ Magnitude of š‘Ž āƒ— = √(22+32+(āˆ’6)2) |š‘Ž āƒ— | = √(4+9+36) = √49 = 7 Unit vector opposite to direction of š‘Ž āƒ— = –1 Ɨ 1/|š‘Ž āƒ— | . š‘Ž āƒ— = (āˆ’1)/7 (2š‘– Ģ‚ + 3š‘— Ģ‚ – 6š‘˜ Ģ‚) Thus, Vector with magnitude 1 opposite to š‘Ž āƒ— = (āˆ’1)/7 (2š‘– Ģ‚ + 3š‘— Ģ‚ – 6š‘˜ Ģ‚) Vector with magnitude 5 opposite to š‘Ž āƒ— = (āˆ’5)/7 (2š‘– Ģ‚ + 3š‘— Ģ‚ – 6š‘˜ Ģ‚) = 5/7 (–2š‘– Ģ‚ – 3š‘— Ģ‚ + 6š‘˜ Ģ‚) Hence, the required vector is šŸ“/šŸ• (–2š’Š Ģ‚ – 3š’‹ Ģ‚ + 6š’Œ Ģ‚)

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