Last updated at Aug. 11, 2021 by Teachoo
Transcript
Ex 10.4, 11 Let the vectors π β and π β be such that |π β| = 3 and |π β| = β2/3, Then π β Γ π β is a unit vector, if the angle between π β and π β is (A) Ο/6 (B) Ο/4 (C) Ο/3 (D) Ο/2 |π β | = 3 & |π β | = β2/3 π β Γ π β = |π β | |π β | sin ΞΈ π Μ Given, (π β Γ π β) is a unit vector Magnitude of (π β Γ π β) = |π β Γ π β| = 1 Now, |π β" Γ " π β | = |(|π β |" " |π β |" sin ΞΈ " π Μ )| , ΞΈ is the angle between π β and π β. |π β" Γ " π β | = |π β | |π β | sin ΞΈ |π Μ | |π β" Γ " π β | = |π β | |π β | sin ΞΈ Γ 1 |π β" Γ " π β | = |π β | |π β | sin ΞΈ 1 = 3 Γ β2/3 sin ΞΈ 1 = β2 sinΞΈ sin ΞΈ = 1/β2 ΞΈ = sin-1 (π/βπ) = π /π Therefore, the angle between the vectors π β and π β is π /π . Hence, (B) is the correct option π Μ ππ π π’πππ‘ π£πππ‘ππ ππππππππππ’πππ π‘π π β πππ π β ππ,"|" π Μ"|"=1
Vector product - Defination
About the Author