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Last updated at Jan. 31, 2020 by Teachoo
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Ex 10.3, 5 Show that each of the given three vectors is a unit vector: 1/7 (2๐ ฬ + 3๐ ฬ + 6๐ ฬ), 1/7 (3๐ ฬ โ 6๐ ฬ + 2๐ ฬ), 1/7 (6๐ ฬ + 2๐ ฬ โ 3๐ ฬ), Also, show that they are mutually perpendicular to each other. ๐ โ = 1/7 (2๐ ฬ + 3๐ ฬ + 6๐ ฬ) = 2/7 ๐ ฬ + 3/7 ๐ ฬ + 6/7 ๐ ฬ ๐ โ = 1/7 (3๐ ฬ โ 6๐ ฬ + 2๐ ฬ) = 3/7 ๐ ฬ โ 6/7 ๐ ฬ + 2๐/7 ๐ ฬ ๐ โ = 1/7 (6๐ ฬ + 2๐ ฬ - 3๐ ฬ) = 6/7 ๐ ฬ + 2/7 ๐ ฬ โ 3/7 ๐ ฬ Magnitude of ๐ โ = โ((2/7)^2+(3/7)^2+(6/7)^2 ) |๐ โ | = โ(4/49+9/49+36/49) = โ(49/49) = 1 Since |๐ โ | = 1 So, ๐ โ is a unit vector. Magnitude of ๐ โ = โ((3/7)^2+((โ6)/7)^2+(2/7)^2 ) |๐ โ | = โ(9/49+36/49+4/49)= โ(49/49) = 1 Since |๐ โ | = 1 So, ๐ โ is a unit vector. Magnitude of ๐ โ = โ((6/7)^2+(2/7)^2+((โ3)/7)^2 ) |๐ โ | = โ(36/49+4/49+9/49) = โ(49/49) = 1 Since |๐ โ | = 1, So, ๐ โ is a unit vector Now, we need to show that they are mutually perpendicular to each other. So, ๐ โ. ๐ โ = ๐ โ. ๐ โ = ๐ โ . ๐ โ = 0 Thus, they are mutually perpendiculars to each other. ๐ โ = 2/7 ๐ ฬ + 3/7 ๐ ฬ + 6/7 ๐ ฬ ๐ โ = 3/7 ๐ ฬ โ 6/7 ๐ ฬ + 2/7 ๐ ฬ ๐ โ. ๐ โ = 2/7 . 3/7 + 3/7 (โ6/7) + 6/7 . 2/7 = 6/49 โ 18/49 + 12/49 = 0 ๐ โ = 3/7 ๐ ฬ โ 6/7 ๐ ฬ + 2/7 ๐ ฬ ๐ โ = 6/7 ๐ ฬ + 2/7 ๐ ฬ โ 3/7 ๐ ฬ ๐ โ. ๐ โ = 3/7 . 6/7 + (โ6/7) 2/7 + 2/7 . ((โ3)/7) = 18/49 โ 12/49 โ 6/49 = 0 ๐ โ = 6/7 ๐ ฬ + 2/7 ๐ ฬ โ 3/7 ๐ ฬ ๐ โ = 2/7 ๐ ฬ + 3/7 ๐ ฬ + 6/7 ๐ ฬ ๐ โ. ๐ โ = 6/7 . 2/7 + 2/7. 3/7 + ((โ3)/7) 6/7 = 12/49 + 6/49 โ 18/49 = 0
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