Scalar product - Defination

Chapter 10 Class 12 Vector Algebra
Concept wise

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Ex 10.3, 5 Show that each of the given three vectors is a unit vector: 1/7 (2π Μ + 3π Μ + 6π Μ), 1/7 (3π Μ β 6π Μ + 2π Μ), 1/7 (6π Μ + 2π Μ β 3π Μ), Also, show that they are mutually perpendicular to each other. π β = 1/7 (2π Μ + 3π Μ + 6π Μ) = 2/7 π Μ + 3/7 π Μ + 6/7 π Μ π β = 1/7 (3π Μ β 6π Μ + 2π Μ) = 3/7 π Μ β 6/7 π Μ + 2π/7 π Μ π β = 1/7 (6π Μ + 2π Μ - 3π Μ) = 6/7 π Μ + 2/7 π Μ β 3/7 π Μ Magnitude of π β = β((2/7)^2+(3/7)^2+(6/7)^2 ) |π β | = β(4/49+9/49+36/49) = β(49/49) = 1 Since |π β | = 1 So, π β is a unit vector. Magnitude of π β = β((3/7)^2+((β6)/7)^2+(2/7)^2 ) |π β | = β(9/49+36/49+4/49)= β(49/49) = 1 Since |π β | = 1 So, π β is a unit vector. Magnitude of π β = β((6/7)^2+(2/7)^2+((β3)/7)^2 ) |π β | = β(36/49+4/49+9/49) = β(49/49) = 1 Since |π β | = 1, So, π β is a unit vector Now, we need to show that they are mutually perpendicular to each other. So, π β. π β = π β. π β = π β . π β = 0 Thus, they are mutually perpendiculars to each other. π β = 2/7 π Μ + 3/7 π Μ + 6/7 π Μ π β = 3/7 π Μ β 6/7 π Μ + 2/7 π Μ π β. π β = 2/7 . 3/7 + 3/7 (β6/7) + 6/7 . 2/7 = 6/49 β 18/49 + 12/49 = 0 π β = 3/7 π Μ β 6/7 π Μ + 2/7 π Μ π β = 6/7 π Μ + 2/7 π Μ β 3/7 π Μ π β. π β = 3/7 . 6/7 + (β6/7) 2/7 + 2/7 . ((β3)/7) = 18/49 β 12/49 β 6/49 = 0 π β = 6/7 π Μ + 2/7 π Μ β 3/7 π Μ π β = 2/7 π Μ + 3/7 π Μ + 6/7 π Μ π β. π β = 6/7 . 2/7 + 2/7. 3/7 + ((β3)/7) 6/7 = 12/49 + 6/49 β 18/49 = 0