Scalar product - Defination

Chapter 10 Class 12 Vector Algebra
Concept wise

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Ex 10.3, 15 (Introduction) If the vertices A, B, C of a triangle ABC are (1,2,3), (β1, 0, 0), (0, 1, 2) respectively, then find β ABC. [β ABC is the angle between the vectors (π΅π΄) β and (π΅πΆ) β]. Consider a triangle ABC as shown β  ABC is not the angle between vectors (π΄π΅) β and (π΅πΆ) β But the angle between vectors (π΅π΄) β and (π΅πΆ) β β΄ β  ABC = Angle between vectors (π΅π΄) β and (π΅πΆ) β Ex 10.3, 15 If the vertices A, B, C of a triangle ABC are (1,2,3), (β1, 0, 0), (0, 1, 2) respectively, then find β ABC. [β ABC is the angle between the vectors (π΅π΄) Μ and (π΅πΆ) Μ]. A (1, 2, 3) B (β1, 0, 0) C (0, 1, 2) β ABC = Angle b/w (π΅π΄) β and (π΅πΆ) β We use formula π β. π β = |π β | |π β | cos ΞΈ , ΞΈ is the angle b/w π β & π β We find (π΅π΄) β and (π΅πΆ) β (π΅π΄) β = (1 β (-1)) π Μ + (2 β 0) π Μ + (3 β 0) π Μ = 2π Μ + 2π Μ + 3π Μ (π΅πΆ) β = (0 β (β1)) π Μ + (1 β 0) π Μ + (2 β 0) π Μ = 1π Μ + 1π Μ + 2π Μ Now, (π©π¨) β . (π©πͺ) β = ("2" π Μ" + " 2π Μ" + " 3π Μ) . ("1" π Μ" + " 1π Μ" + " 2π Μ) = 2.1+2.1+3.2 = 2+2+6 = 10 Magnitude of (π΅πΆ) β = β(12+12+22) |(π©πͺ) β | = β(1+1+4) = βπ Also, (π΅π΄) β . (π΅πΆ) β = |(π΅π΄) β | . |(π΅πΆ) β | cos ΞΈ 10 = β17 Γ β6 Γ cos ΞΈ β17 Γ β6 Γ cos ΞΈ = 10 cos ΞΈ = 10/(β17 Γβ6) cos ΞΈ = 10/β102 ΞΈ = cosβ1 (10/β102) Thus β ABC = cosβ1 (ππ/βπππ).