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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 Μ‚ + 3𝑗 Μ‚ + 6π‘˜ Μ‚), 1/7 (3𝑖 Μ‚ – 6𝑗 Μ‚ + 2π‘˜ Μ‚), 1/7 (6𝑖 Μ‚ + 2𝑗 Μ‚ – 3π‘˜ Μ‚), Also, show that they are mutually perpendicular to each other. π‘Ž βƒ— = 1/7 (2𝑖 Μ‚ + 3𝑗 Μ‚ + 6π‘˜ Μ‚) = 2/7 𝑖 Μ‚ + 3/7 𝑗 Μ‚ + 6/7 π‘˜ Μ‚ 𝑏 βƒ— = 1/7 (3𝑖 Μ‚ βˆ’ 6𝑗 Μ‚ + 2π‘˜ Μ‚) = 3/7 𝑖 Μ‚ – 6/7 𝑗 Μ‚ + 2π‘˜/7 π‘˜ Μ‚ 𝑐 βƒ— = 1/7 (6𝑖 Μ‚ + 2𝑗 Μ‚ - 3π‘˜ Μ‚) = 6/7 𝑖 Μ‚ + 2/7 𝑗 Μ‚ – 3/7 π‘˜ Μ‚ Magnitude of π‘Ž βƒ— = √((2/7)^2+(3/7)^2+(6/7)^2 ) |π‘Ž βƒ— | = √(4/49+9/49+36/49) = √(49/49) = 1 Since |π‘Ž βƒ— | = 1 So, π‘Ž βƒ— is a unit vector. Magnitude of 𝑏 βƒ— = √((3/7)^2+((βˆ’6)/7)^2+(2/7)^2 ) |𝑏 βƒ— | = √(9/49+36/49+4/49)= √(49/49) = 1 Since |𝑏 βƒ— | = 1 So, 𝑏 βƒ— is a unit vector. Magnitude of 𝑐 βƒ— = √((6/7)^2+(2/7)^2+((βˆ’3)/7)^2 ) |𝑐 βƒ— | = √(36/49+4/49+9/49) = √(49/49) = 1 Since |𝑐 βƒ— | = 1, So, 𝑐 βƒ— is a unit vector Now, we need to show that they are mutually perpendicular to each other. So, 𝒂 βƒ—. 𝒃 βƒ— = 𝒃 βƒ—. 𝒄 βƒ— = 𝒄 βƒ— . 𝒂 βƒ— = 0 Thus, they are mutually perpendiculars to each other. π‘Ž βƒ— = 2/7 𝑖 Μ‚ + 3/7 𝑗 Μ‚ + 6/7 π‘˜ Μ‚ 𝑏 βƒ— = 3/7 𝑖 Μ‚ – 6/7 𝑗 Μ‚ + 2/7 π‘˜ Μ‚ 𝒂 βƒ—. 𝒃 βƒ— = 2/7 . 3/7 + 3/7 (βˆ’6/7) + 6/7 . 2/7 = 6/49 βˆ’ 18/49 + 12/49 = 0 𝑏 βƒ— = 3/7 𝑖 Μ‚ βˆ’ 6/7 𝑗 Μ‚ + 2/7 π‘˜ Μ‚ 𝑐 βƒ— = 6/7 𝑖 Μ‚ + 2/7 𝑗 Μ‚ – 3/7 π‘˜ Μ‚ 𝒃 βƒ—. 𝒄 βƒ— = 3/7 . 6/7 + (βˆ’6/7) 2/7 + 2/7 . ((βˆ’3)/7) = 18/49 βˆ’ 12/49 βˆ’ 6/49 = 0 𝑐 βƒ— = 6/7 𝑖 Μ‚ + 2/7 𝑗 Μ‚ βˆ’ 3/7 π‘˜ Μ‚ π‘Ž βƒ— = 2/7 𝑖 Μ‚ + 3/7 𝑗 Μ‚ + 6/7 π‘˜ Μ‚ 𝒄 βƒ—. 𝒂 βƒ— = 6/7 . 2/7 + 2/7. 3/7 + ((βˆ’3)/7) 6/7 = 12/49 + 6/49 βˆ’ 18/49 = 0

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