Scalar product - Defination

Chapter 10 Class 12 Vector Algebra
Concept wise

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Ex 10.3, 12 (Introduction) If π β .π β = 0 & π β . π β = 0, then what can be concluded about the vector π β? π β . π β = |π β | |π β | cos ΞΈ , ΞΈ in the angle b/w π β and π β Let π β = 0 β = 0π Μ + 0π Μ + 0π Μ π β = 0π Μ + 0π Μ + 0π Μ Let π β = 8π Μ β 5π Μ + 2π Μ π β. π β = 0.8 + 0.(β5) + 0.2 = 0 + 0 + 0 = 0 So, π β. π β = 0 π β = 0π Μ + 0π Μ + 0π Μ Let π β = 3π Μ β 4π Μ + 7π Μ π β. π β = 0.3 + 0(β4) + 0.7 = 0 + 0 + 0 = 0 So, π β. π β = 0 Hence, if π β = 0 β, then π β.π β = 0 for any vector π β Ex 10.3, 12 If π β .π β = 0 and π β . π β = 0, then what can be concluded about the vector π β ? Given, π β. π β = 0 |π β | |π β | cos 0 = 0 |π β |2 cos 0 = 0 |π β |2 Γ 1 = 0 |π β |2 = 0 |π β | = 0 So, π β = 0 β Now it is given, π β . π β = 0 0 β . π β = 0 is true, for any vector π β Therefore, if π β . π β = 0 and π β . π β = 0, then π β = 0 β and π β can be any vector