Ex 10.3

Chapter 10 Class 12 Vector Algebra
Serial order wise

### Transcript

Ex 10.3, 17 (Method 1) Show that the vectors 2π Μ β π Μ + π Μ, π Μ β 3π Μ β 5π Μ , 3π Μ β 4π Μ β 4π Μ form the vertices of a right angled triangle. Let A(2π Μ β π Μ + π Μ), B(π Μ β 3π Μ β 5π Μ) C(3π Μ β 4π Μ β 4π Μ) We know that two vectors are perpendicular to each other, i.e have an angle of 90Β° between them , if their scalar product is zero. (π΄π΅) β = (π Μ β 3π Μ β 5π Μ) β (2π Μ β π Μ + π Μ) = 1π Μ β 3π Μ β 5π Μ β 2π Μ + 1π Μ β 1π Μ = (1 β 2) π Μ + (β3 + 1) π Μ + (β5 β1) π Μ = β1π Μ β 2π Μ β 6π Μ (π΅πΆ) β = (3π Μ β 4π Μ β 4π Μ) β (π Μ β 3π Μ β 5π Μ) = 3π Μ β 4π Μ β 4π Μ β 1π Μ + 3π Μ + 5π Μ = (3 β 1) π Μ + (β4 + 3) π Μ + (β4 + 5) π Μ = 2π Μ β 1π Μ + 1π Μ (πΆπ΄) β = (2π Μ β π Μ + π Μ) β (3π Μ β 4π Μ β 4π Μ) = 2π Μ β 1π Μ + 1π Μ β 3π Μ + 4π Μ + 4π Μ = (2 β 3) π Μ + (β1 + 4) π Μ + (1 + 4) π Μ = β1π Μ + 3π Μ + 5π Μ Now, (π΅πΆ) β. (πΆπ΄) β = (2π Μ β 1π Μ + 1π Μ) . (-1π Μ + 3π Μ + 5π Μ) = (2 Γ β1) + (β1 Γ 3) + (1 Γ 5) = (β2) + (β3) + 5 = β5 + 5 = 0 Since, (π©πͺ) β. (πͺπ¨) β = 0 Therefore, (π΅πΆ) β is perpendicular to (πΆπ΄) β . Hence Ξ ABC is a right angled triangle Ex 10.3, 17 (Method 2) Show that the vectors 2π Μ β π Μ + π Μ, π Μ β 3π Μ β 5π Μ , 3π Μ β 4π Μ β 4π Μ form the vertices of a right angled triangle. Let A(2π Μ β π Μ + π Μ), B(π Μ β 3π Μ β 5π Μ) C(3π Μ β 4π Μ β 4π Μ) Considering βABC as a right angled triangle, By Pythagoras theorem, AB2 = BC2 + CA2 or |("AB" ) β |"2" = |("BC" ) β |"2" + |("CA" ) β |"2" (π΄π΅) β = (π Μ β 3π Μ β 5π Μ) β (2π Μ β π Μ + π Μ) = 1π Μ β 3π Μ β 5π Μ β 2π Μ + 1π Μ β 1π Μ = (1 β 2) π Μ + (β3 + 1) π Μ + (β5 β1) π Μ = β1π Μ β 2π Μ β 6π Μ (π΅πΆ) β = (3π Μ β 4π Μ β 4π Μ) β (π Μ β 3π Μ β 5π Μ) = 3π Μ β 4π Μ β 4π Μ β 1π Μ + 3π Μ + 5π Μ = (3 β 1) π Μ + (β4 + 3) π Μ + (β4 + 5) π Μ = 2π Μ β 1π Μ + 1π Μ (πΆπ΄) β = (2π Μ β π Μ + π Μ) β (3π Μ β 4π Μ β 4π Μ) = 2π Μ β 1π Μ + 1π Μ β 3π Μ + 4π Μ + 4π Μ = (2 β 3) π Μ + (β1 + 4) π Μ + (1 + 4) π Μ = β1π Μ + 3π Μ + 5π Μ Now, "Magnitude of " (π΄π΅) β" = " β((β1)2+(β2)2+(β6)2) " " |(π΄π΅) β |" = " β(1+4+36) " = " β41 Magnitude of (π΅πΆ) β = β(22+(β1)2+1) |(π΅πΆ) β | = β(4+1+1) = β6 Magnitude of (πΆπ΄) β = β((β1)2+32+52) |(πΆπ΄) β | = β(1+9+25) = β35 Now, |(π΅πΆ) β |2 + |(πΆπ΄) β |2 = (β6)2 + (β35)2 = 6 + 35 = 41 = (β41)2 = |(π΄π΅) β |2 Thus, |(π΄π΅) β |^2 = |(π΅πΆ) β |^2 + |(πΆπ΄) β |^2 So, ABC is a right angled triangle.