Scalar product - Defination

Chapter 10 Class 12 Vector Algebra
Concept wise

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Misc 12 Let π β = π Μ + 4π Μ + 2π Μ, π β = 3π Μ β 2π Μ + 7π Μ and π β = 2π Μ β π Μ + 4π Μ . Find a vector π β which is perpendicular to both π β and π β and π β β π β = 15 . Given π β = π Μ + 4π Μ + 2π Μ π β = 3π Μ - 2π Μ + 7π Μ π β = 2π Μ + π Μ + 4π Μ Let π β = xπ Μ + yπ Μ + zπ Μ Since π β is perpendicular to π β and π β π β . π β = 0 & π β . π β = 0 π β . π β = 0 (xπ Μ + yπ Μ + zπ Μ). (1π Μ + 4π Μ + 2π Μ) = 0 (x Γ 1) + (y Γ 4) + (z Γ 2) = 0 x + 4y + 2z = 0 π β . π β = 0 (xπ Μ + yπ Μ + zπ Μ). (3π Μ β 2π Μ + 7π Μ) = 0 (x Γ 3) + (y Γ -2) + (z Γ 7) = 0 3x β 2y + 7z = 0 Also, π β . π β = 15 (2π Μ β 1π Μ + 4π Μ). (xπ Μ + yπ Μ + zπ Μ) = 15 (2 Γ x) + (β1 Γ y) + (4 Γ 2) = 15 2x β y + 4z = 15 Now, we need to solve equations x + 4y + 2z = 0 3x β 2y + 7z = 0 & 2x β y + 4z = 15 Solving x + 4y + 2z = 0 3x β 2y + 7z = 0 π₯/(28 β (β4) ) = π¦/(6 β 7 ) = π§/(β2 β 12 ) π/(ππ ) = π/(βπ ) = π/(βππ ) Writing x & y in terms of z β΄ x = 32π§/(β14) = (βπππ)/π & y = (β1π§)/(β14) = π/ππ Putting values of x and y in (3) 2x β y + 4z = 15 2 ((β16π§)/7 ) β (π§/14 ) + 4z = 5 (β32)/7 z β 1/14 z + 4/1 z = 15 ((β64 β 1 + 56))/14 z = 15 (β9)/14 z = 15 z = 15 Γ 14/(β9) z = (βππ)/π Putting value of z in x & y, x = (β16π§)/7 = (β16)/7 Γ (β70)/3 = πππ/π y = π§/14 = 1/14 Γ (β70)/3 = (βπ)/π Therefore, the required vector π β = xπ Μ + yπ Μ + zπ Μ = 160/3 π Μ β 5/3 π Μ β 70/3 π Μ = π/π (160π Μ β 5π Μ β 70π Μ) Note: Answer given in the book is incorrect If we have made any mistake, please email at admin@teachoo.com