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  1. Chapter 10 Class 12 Vector Algebra
  2. Serial order wise

Transcript

Misc 11 Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are 1/โˆš3, 1/โˆš3, 1/โˆš3 . Let the required vector be ๐‘Ÿ โƒ— = ๐‘Ž๐‘– ฬ‚ + b๐‘— ฬ‚ + c๐‘˜ ฬ‚ Directions ratios are ๐‘Ž, ๐‘, and ๐‘. Since the vector is equally inclined to axes OX, OY and OZ, thus the direction cosines are equal. ๐‘Ž/(๐‘š๐‘Ž๐‘”๐‘›๐‘–๐‘ก๐‘ข๐‘‘๐‘’ ๐‘œ๐‘“ ๐‘Ÿ โƒ— ) = ๐‘/(๐‘š๐‘Ž๐‘”๐‘›๐‘–๐‘ก๐‘ข๐‘‘๐‘’ ๐‘œ๐‘“ ๐‘Ÿ โƒ— ) = ๐‘/(๐‘š๐‘Ž๐‘”๐‘›๐‘–๐‘ก๐‘ข๐‘‘๐‘’ ๐‘œ๐‘“ ๐‘Ÿ โƒ— ) ๐‘Ž = b = c โˆด The vector is ๐‘Ÿ โƒ— = ๐‘Ž๐‘– ฬ‚ + ๐‘Ž๐‘— ฬ‚ + ๐‘Ž๐‘˜ ฬ‚ Magnitude of ๐‘Ÿ โƒ— = โˆš(๐‘Ž2+๐‘Ž2+๐‘Ž2) |๐‘Ÿ โƒ— | = โˆš3๐‘Ž2 |๐‘Ÿ โƒ— | = โˆš3 ๐‘Ž Direction cosines are (๐‘Ž/(โˆš3 ๐‘Ž),๐‘/(โˆš3 ๐‘Ž),๐‘/(โˆš3 ๐‘Ž)) = (๐‘Ž/(โˆš3 ๐‘Ž),๐‘Ž/(โˆš3 ๐‘Ž),๐‘Ž/(โˆš3 ๐‘Ž)) = (๐Ÿ/โˆš๐Ÿ‘,๐Ÿ/โˆš๐Ÿ‘,๐Ÿ/โˆš๐Ÿ‘) Hence proved

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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.