Examples

Chapter 10 Class 12 Vector Algebra
Serial order wise

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= π Μ(1 β (-4)) β j Μ (3 β 4) + π Μ (β3 β1) = π Μ(1 + 4) β j Μ (β1) + π Μ (β4) = 5π Μ + π Μ β 4π Μ Magnitude of π β Γ π β = β(52+1^2+(β4)2) |π β Γ π β | = β(25+1+16) = βππ Area of parallelogram ABCD = |π β Γ π β | = β42 Therefore, the required area is βππ . Let the unit vector be π β We know that π β = π₯π Μ + yπ Μ + zπ Μ Since the vector is in XY plane, there is no Z βcoordinate. Hence, π β = xπ Μ + yπ Μ + 0π Μ π β = ππ Μ + yπ Μ Taking a general vector π β, Making an angle π with the x β axis Unit vector in direction of x axis is π Μ & in y axis is π Μ Angle with X-axis Since π β makes an angle of ΞΈ with x-axis So, angle between π β & π Μ is ΞΈ We know that, π β . π β = |π β ||π β | cos ΞΈ, Putting π β = π β , π β = π Μ & ΞΈ = ΞΈ π β .π Μ = |π β ||π Μ | cos ΞΈ π β .π Μ = 1 Γ 1 Γ cos ΞΈ π β . π Μ = cos ΞΈ (π₯π Μ + yπ Μ + 0π Μ). π Μ = cos ΞΈ (π₯π Μ + yπ Μ + 0π Μ). (1π Μ + 0π Μ + 0π Μ) = cos ΞΈ π₯.1 + y.0 + 0.0 = cos ΞΈ (As π β is unit vector, |π β | = 1 & π Μ is a unit vector, |π Μ | = 1) x = cos ΞΈ Angle with Y-axis π β makes an angle of (90Β° β ΞΈ) with y-axis So, angle between π β & π Μ is (90Β° β ΞΈ) We know that, π β . π β = |π β ||π β | cos ΞΈ, Putting π β = π β , π β = π Μ & ΞΈ = (90Β° β ΞΈ) π β .π Μ = |π β ||π Μ | cos (90Β° β ΞΈ) π β .π Μ = 1 Γ 1 Γ cos (90Β° β ΞΈ) π β .π Μ = cos (90Β° β ΞΈ) π β .π Μ = sin ΞΈ (As π β is unit vector, |π β | = 1 & π Μ is a unit vector, |π Μ | = 1) (π₯π Μ + yπ Μ + 0π Μ). π Μ = sin ΞΈ (π₯π Μ + yπ Μ + 0π Μ). (0π Μ + 1π Μ + 0π Μ) = sin ΞΈ π₯.0 + y.1 + 0.0 = sin ΞΈ y = sin ΞΈ Thus, π β = xπ Μ + yπ Μ π β = cos ππ Μ + sin π π Μ This value will be true in all quadrants So, 0 β€ ΞΈ β€ 2Ο Therefore, π β = cos ππ Μ + sinππ Μ ; for 0 β€ ΞΈ β€ 2Ο