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Last updated at Jan. 31, 2020 by Teachoo
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Ex 10.4, 3 If a unit vector ๐ โ makes angles ๐/3 with ๐ ฬ, ๐/4 , with ๐ ฬ and an acute angle ฮธ with ๐ ฬ , then find ฮธ and hence, the components of ๐ โ . Let us take a unit vector ๐ โ = ๐ฅ๐ ฬ + y๐ ฬ + z๐ ฬ So, magnitude of ๐ โ = |๐ โ | = 1 Angle of ๐ โ with ๐ ฬ = ๐ /๐ ๐ โ . ๐ ฬ = |๐ โ ||๐ ฬ | cos ๐/3 (x๐ ฬ + y๐ ฬ + z๐ ฬ). ๐ ฬ = 1 ร 1 ร 1/2 (x๐ ฬ + y๐ ฬ + z๐ ฬ). (1๐ ฬ + 0๐ ฬ + 0๐ ฬ) = 1/2 (x ร 1) + (y ร 0) + (z ร 0) = 1/2 x + 0 + 0 = 1/2 x = ๐/๐ Angle of ๐ โ with ๐ ฬ = ๐ /๐ ๐ โ . ๐ ฬ = |๐ โ ||๐ ฬ | cos ๐/4 (x๐ ฬ + y๐ ฬ + z๐ ฬ). ๐ ฬ = 1 ร 1 ร 1/โ2 (x๐ ฬ + y๐ ฬ + z๐ ฬ). (0๐ ฬ + 1๐ ฬ + 0๐ ฬ) = 1/โ2 (x ร 0) + (y ร 1) + (z ร 0) = 1/โ2 0 + y + 0 = 1/โ2 y = ๐/โ๐ Also, Angle of ๐ โ with ๐ ฬ = ฮธ ๐ โ. ๐ ฬ = |๐ โ ||๐ ฬ |รcosโก"ฮธ" (x๐ ฬ + y๐ ฬ + z๐ ฬ). (0๐ ฬ + .0๐ ฬ + 1๐ ฬ) = 1 ร 1 ร cos ฮธ (x ร 0) + (y ร 0) + (z ร 1) = cosฮธ 0 + 0 + z = cos ฮธ z = cos ฮธ Now, Magnitude of ๐ โ = โ(๐ฅ^2+๐ฆ2+๐ง2) 1 = โ((1/2)^2+(1/โ2)^2+๐๐๐ 2"ฮธ" ) 1 = โ(1/4+1/2+๐๐๐ 2"ฮธ" ) 1 = โ(3/4+๐๐๐ 2"ฮธ" ) โ(3/4+๐๐๐ 2"ฮธ" ) = 1 (โ(3/4+๐๐๐ 2"ฮธ" ))^2 = 12 3/4 + ๐๐๐ 2" ฮธ" = 1 ๐๐๐ 2 "ฮธ" = 1 โ 3/4 ๐๐๐ 2" ฮธ" = 1/4 cosโก"ฮธ" = ยฑ โ(1/4) cosโก"ฮธ" = ยฑ 1/2 Since ฮธ is given an acute angle So, ฮธ < 90ยฐ โด ฮธ is in 1st quadrant And, cos ฮธ is positive in 1st quadrant= So, cos ฮธ = 1/2 โด ฮธ = 60ยฐ = ๐ /๐ Also, z = cos ฮธ = cos 60ยฐ = ๐/๐ Hence x = 1/2 , y = 1/โ2 & z = 1/2 The required vector ๐ โ is 1/2 ๐ ฬ + 1/โ2 ๐ ฬ + 1/2 ๐ ฬ So, components of ๐ โ are ๐/๐ , ๐/โ๐ & ๐/๐ The required vector ๐ โ is 1/2 ๐ ฬ + 1/โ2 ๐ ฬ + 1/2 ๐ ฬ So, components of ๐ โ are ๐/๐ , ๐/โ๐ & ๐/๐
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