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Last updated at Feb. 1, 2020 by Teachoo

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Misc 11 Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are 1/โ3, 1/โ3, 1/โ3 . Let the required vector be ๐ โ = ๐๐ ฬ + b๐ ฬ + c๐ ฬ Directions ratios are ๐, ๐, and ๐. Since the vector is equally inclined to axes OX, OY and OZ, thus the direction cosines are equal. ๐/(๐๐๐๐๐๐ก๐ข๐๐ ๐๐ ๐ โ ) = ๐/(๐๐๐๐๐๐ก๐ข๐๐ ๐๐ ๐ โ ) = ๐/(๐๐๐๐๐๐ก๐ข๐๐ ๐๐ ๐ โ ) ๐ = b = c โด The vector is ๐ โ = ๐๐ ฬ + ๐๐ ฬ + ๐๐ ฬ Magnitude of ๐ โ = โ(๐2+๐2+๐2) |๐ โ | = โ3๐2 |๐ โ | = โ3 ๐ Direction cosines are (๐/(โ3 ๐),๐/(โ3 ๐),๐/(โ3 ๐)) = (๐/(โ3 ๐),๐/(โ3 ๐),๐/(โ3 ๐)) = (๐/โ๐,๐/โ๐,๐/โ๐) Hence proved

Miscellaneous

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Misc 2

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Misc 4

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Misc 6

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Misc 8 Important

Misc 9

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About the Author

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.