Ex 10.4

Chapter 10 Class 12 Vector Algebra
Serial order wise

### Transcript

Ex 10.4, 12 Area of a rectangle having vertices A, B, C & D with position vectors βπ Μ + 1/2 π Μ + 4π Μ, π Μ + 1/2 π Μ + 4π Μ, π Μ β 1/2 π Μ + 4π Μ, βπ Μ β 1/2 π Μ + 4π Μ respectively is (A) 1/2 (B) 1 (C) 2 (D) 4 (ππ΄) β = βπ Μ + 1/2 π Μ + 4π Μ = β1π Μ + 1/2 π Μ + 4π Μ (ππ΅) β = π Μ + 1/2 π Μ + 4π Μ = 1π Μ + 1/2 π Μ + 4π Μ (ππΆ) β = π Μ β 1/2 π Μ + 4π Μ = 1π Μ - 1/2 π Μ + 4π Μ (ππ·) β = βπ Μ β 1/2 π Μ + 4π Μ = β1π Μ β 1/2 π Μ + 4π Μ Since rectangle is also a parallelogram Area of rectangle ABCD = |(π΄π΅) βΓ(π΅πΆ) β | (π¨π©) β = (πΆπ©) β β (πΆπ¨) β = ("1" π Μ" + " 1/2 " " π Μ" + 4" π Μ ) β ("β1" π Μ" + " 1/2 " " π Μ" + 4" π Μ ) = ("1β(β1)" π Μ" +" (1/2β1/2) π Μ" + (4 β 4)" ) π Μ = 2π Μ + 0π Μ + 0π Μ (π©πͺ) β = (πΆπͺ) β β (πΆπ©) β = ("1" π Μ" β " 1/2 " " π Μ" + 4" π Μ ) β ("1" π Μ" + " 1/2 " " π Μ" + 4" π Μ ) = ("(1β1)" π Μ" +" (β1/2β1/2) π Μ" + (4 β 4)" ) π Μ = 0π Μ β 1π Μ + 0π Μ |(π¨π©) βΓ(π©πͺ) β | = |β 8(π Μ&π Μ&π Μ@2&0&0@0&β1&0)| = π Μ (0 Γ 0 β (β1) Γ 0) β π Μ (2 Γ 0 β 0 Γ 0) + π Μ (2 Γ β1 β 0 Γ 0 ) = π Μ (0 β 0) β π Μ (0 β 0) + π Μ (β2 β 0) = 0π Μ β 0π Μ β 2π Μ Now, Magnitude of (π΄π΅) βΓ(π΅πΆ) β = β(02+02+(β2)2) |(π¨π©) βΓ(π©πͺ) β | = β4 = 2 Therefore, area of rectangle ABCD = |(π΄π΅) βΓ(π΅πΆ) β | = 2 Hence, (C) is the correct option