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Example 26 (Supplementary NCERT) Deleted for CBSE Board 2025 Exams

Example 27 (Supplementary NCERT) Deleted for CBSE Board 2025 Exams

Example 28 (Supplementary NCERT) Important Deleted for CBSE Board 2025 Exams

Example 29 (Supplementary NCERT) Important Deleted for CBSE Board 2025 Exams

Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2025 Exams

Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2025 Exams

Last updated at April 16, 2024 by Teachoo

= π Μ(1 β (-4)) β j Μ (3 β 4) + π Μ (β3 β1) = π Μ(1 + 4) β j Μ (β1) + π Μ (β4) = 5π Μ + π Μ β 4π Μ Magnitude of π β Γ π β = β(52+1^2+(β4)2) |π β Γ π β | = β(25+1+16) = βππ Area of parallelogram ABCD = |π β Γ π β | = β42 Therefore, the required area is βππ . 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 βcoordinate. Hence, π β = xπ Μ + yπ Μ + 0π Μ π β = ππ 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 ππ Μ + sin π π Μ This value will be true in all quadrants So, 0 β€ ΞΈ β€ 2Ο Therefore, π β = cos ππ Μ + sinππ Μ ; for 0 β€ ΞΈ β€ 2Ο