Ex 10.3, 5 - Show unit vector: 1/7 (2i + 3j + 6k), 1/7(3-6j+2k)

Ex 10.3, 5 - Chapter 10 Class 12 Vector Algebra - Part 2
Ex 10.3, 5 - Chapter 10 Class 12 Vector Algebra - Part 3 Ex 10.3, 5 - Chapter 10 Class 12 Vector Algebra - Part 4

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Ex 10.3, 5 Show that each of the given three vectors is a unit vector: 1/7 (2𝑖 ̂ + 3𝑗 ̂ + 6𝑘 ̂), 1/7 (3𝑖 ̂ – 6𝑗 ̂ + 2𝑘 ̂), 1/7 (6𝑖 ̂ + 2𝑗 ̂ – 3𝑘 ̂), Also, show that they are mutually perpendicular to each other. 𝑎 ⃗ = 1/7 (2𝑖 ̂ + 3𝑗 ̂ + 6𝑘 ̂) = 2/7 𝑖 ̂ + 3/7 𝑗 ̂ + 6/7 𝑘 ̂ 𝑏 ⃗ = 1/7 (3𝑖 ̂ − 6𝑗 ̂ + 2𝑘 ̂) = 3/7 𝑖 ̂ – 6/7 𝑗 ̂ + 2𝑘/7 𝑘 ̂ 𝑐 ⃗ = 1/7 (6𝑖 ̂ + 2𝑗 ̂ - 3𝑘 ̂) = 6/7 𝑖 ̂ + 2/7 𝑗 ̂ – 3/7 𝑘 ̂ Magnitude of 𝑎 ⃗ = √((2/7)^2+(3/7)^2+(6/7)^2 ) |𝑎 ⃗ | = √(4/49+9/49+36/49) = √(49/49) = 1 Since |𝑎 ⃗ | = 1 So, 𝑎 ⃗ is a unit vector. Magnitude of 𝑏 ⃗ = √((3/7)^2+((−6)/7)^2+(2/7)^2 ) |𝑏 ⃗ | = √(9/49+36/49+4/49)= √(49/49) = 1 Since |𝑏 ⃗ | = 1 So, 𝑏 ⃗ is a unit vector. Magnitude of 𝑐 ⃗ = √((6/7)^2+(2/7)^2+((−3)/7)^2 ) |𝑐 ⃗ | = √(36/49+4/49+9/49) = √(49/49) = 1 Since |𝑐 ⃗ | = 1, So, 𝑐 ⃗ is a unit vector Now, we need to show that they are mutually perpendicular to each other. So, 𝒂 ⃗. 𝒃 ⃗ = 𝒃 ⃗. 𝒄 ⃗ = 𝒄 ⃗ . 𝒂 ⃗ = 0 Thus, they are mutually perpendiculars to each other. 𝑎 ⃗ = 2/7 𝑖 ̂ + 3/7 𝑗 ̂ + 6/7 𝑘 ̂ 𝑏 ⃗ = 3/7 𝑖 ̂ – 6/7 𝑗 ̂ + 2/7 𝑘 ̂ 𝒂 ⃗. 𝒃 ⃗ = 2/7 . 3/7 + 3/7 (−6/7) + 6/7 . 2/7 = 6/49 − 18/49 + 12/49 = 0 𝑏 ⃗ = 3/7 𝑖 ̂ − 6/7 𝑗 ̂ + 2/7 𝑘 ̂ 𝑐 ⃗ = 6/7 𝑖 ̂ + 2/7 𝑗 ̂ – 3/7 𝑘 ̂ 𝒃 ⃗. 𝒄 ⃗ = 3/7 . 6/7 + (−6/7) 2/7 + 2/7 . ((−3)/7) = 18/49 − 12/49 − 6/49 = 0 𝑐 ⃗ = 6/7 𝑖 ̂ + 2/7 𝑗 ̂ − 3/7 𝑘 ̂ 𝑎 ⃗ = 2/7 𝑖 ̂ + 3/7 𝑗 ̂ + 6/7 𝑘 ̂ 𝒄 ⃗. 𝒂 ⃗ = 6/7 . 2/7 + 2/7. 3/7 + ((−3)/7) 6/7 = 12/49 + 6/49 − 18/49 = 0

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.