Ex 10.4, 11 - Let |a|= 3, |b| = root2/3, Then a x b is unit vector

Ex 10.4, 11 - Chapter 10 Class 12 Vector Algebra - Part 2

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Ex 10.4, 11 Let the vectors π‘Ž βƒ— and 𝑏 βƒ— be such that |π‘Ž βƒ—| = 3 and |𝑏 βƒ—| = √2/3, Then π‘Ž βƒ— Γ— 𝑏 βƒ— is a unit vector, if the angle between π‘Ž βƒ— and 𝑏 βƒ— is (A) Ο€/6 (B) Ο€/4 (C) Ο€/3 (D) Ο€/2 |π‘Ž βƒ— | = 3 & |𝑏 βƒ— | = √2/3 π‘Ž βƒ— Γ— 𝑏 βƒ— = |π‘Ž βƒ— | |𝑏 βƒ— | sin ΞΈ 𝑛 Μ‚ Given, (π‘Ž βƒ— Γ— 𝑏 βƒ—) is a unit vector Magnitude of (π‘Ž βƒ— Γ— 𝑏 βƒ—) = |𝒂 βƒ— Γ— 𝒃 βƒ—| = 1 Now, |𝒂 βƒ—" Γ— " 𝒃 βƒ— | = |(|𝒂 βƒ— |" " |𝒃 βƒ— |" sin ΞΈ " 𝒏 Μ‚ )| , ΞΈ is the angle between π‘Ž βƒ— and 𝑏 βƒ—. |π‘Ž βƒ—" Γ— " 𝑏 βƒ— | = |π‘Ž βƒ— | |𝑏 βƒ— | sin ΞΈ |𝑛 Μ‚ | |π‘Ž βƒ—" Γ— " 𝑏 βƒ— | = |π‘Ž βƒ— | |𝑏 βƒ— | sin ΞΈ Γ— 1 |π‘Ž βƒ—" Γ— " 𝑏 βƒ— | = |π‘Ž βƒ— | |𝑏 βƒ— | sin ΞΈ 1 = 3 Γ— √2/3 sin ΞΈ 1 = √2 sinΞΈ sin ΞΈ = 1/√2 ΞΈ = sin-1 (𝟏/√𝟐) = 𝝅/πŸ’ Therefore, the angle between the vectors π‘Ž βƒ— and 𝑏 βƒ— is 𝝅/πŸ’ . Hence, (B) is the correct option 𝑛 Μ‚ 𝑖𝑠 π‘Ž 𝑒𝑛𝑖𝑑 π‘£π‘’π‘π‘‘π‘œπ‘Ÿ π‘π‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘–π‘π‘’π‘™π‘Žπ‘Ÿ π‘‘π‘œ π‘Ž βƒ— π‘Žπ‘›π‘‘ 𝑏 βƒ— π‘†π‘œ,"|" 𝑛 Μ‚"|"=1

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