Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

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Example 26 (Supplementary NCERT) Deleted for CBSE Board 2024 Exams

Example 27 (Supplementary NCERT) Deleted for CBSE Board 2024 Exams

Example 28 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Example 29 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 by Teachoo

= 𝑖 ̂(1 − (-4)) − j ̂ (3 − 4) + 𝑘 ̂ (−3 −1) = 𝑖 ̂(1 + 4) − j ̂ (−1) + 𝑘 ̂ (−4) = 5𝒊 ̂ + 𝒋 ̂ − 4𝒌 ̂ Magnitude of 𝑎 ⃗ × 𝑏 ⃗ = √(52+1^2+(−4)2) |𝒂 ⃗ × 𝒃 ⃗ | = √(25+1+16) = √𝟒𝟐 Area of parallelogram ABCD = |𝑎 ⃗ × 𝑏 ⃗ | = √42 Therefore, the required area is √𝟒𝟐 . Let the unit vector be 𝒂 ⃗ We know that 𝑎 ⃗ = 𝑥𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂ Since the vector is in XY plane, there is no Z –coordinate. Hence, 𝑎 ⃗ = x𝑖 ̂ + y𝑗 ̂ + 0𝒌 ̂ 𝒂 ⃗ = 𝒙𝒊 ̂ + y𝒋 ̂ Taking a general vector 𝑎 ⃗, Making an angle 𝛉 with the x – axis Unit vector in direction of x axis is 𝑖 ̂ & in y axis is 𝑗 ̂ Angle with X-axis Since 𝑎 ⃗ makes an angle of θ with x-axis So, angle between 𝒂 ⃗ & 𝒊 ̂ is θ We know that, 𝑎 ⃗ . 𝑏 ⃗ = |𝑎 ⃗ ||𝑏 ⃗ | cos θ, Putting 𝑎 ⃗ = 𝑎 ⃗ , 𝑏 ⃗ = 𝑖 ̂ & θ = θ 𝒂 ⃗ .𝒊 ̂ = |𝒂 ⃗ ||𝒊 ̂ | cos θ 𝑎 ⃗ .𝑖 ̂ = 1 × 1 × cos θ 𝑎 ⃗ . 𝑖 ̂ = cos θ (𝑥𝑖 ̂ + y𝑗 ̂ + 0𝑘 ̂). 𝑖 ̂ = cos θ (𝑥𝑖 ̂ + y𝑗 ̂ + 0𝑘 ̂). (1𝑖 ̂ + 0𝑗 ̂ + 0𝑘 ̂) = cos θ 𝑥.1 + y.0 + 0.0 = cos θ (As 𝑎 ⃗ is unit vector, |𝑎 ⃗ | = 1 & 𝑖 ̂ is a unit vector, |𝑖 ̂ | = 1) x = cos θ Angle with Y-axis 𝑎 ⃗ makes an angle of (90° – θ) with y-axis So, angle between 𝒂 ⃗ & 𝒋 ̂ is (90° – θ) We know that, 𝑎 ⃗ . 𝑏 ⃗ = |𝑎 ⃗ ||𝑏 ⃗ | cos θ, Putting 𝑎 ⃗ = 𝑎 ⃗ , 𝑏 ⃗ = 𝑗 ̂ & θ = (90° – θ) 𝒂 ⃗ .𝒋 ̂ = |𝒂 ⃗ ||𝒋 ̂ | cos (90° – θ) 𝑎 ⃗ .𝑗 ̂ = 1 × 1 × cos (90° – θ) 𝑎 ⃗ .𝑗 ̂ = cos (90° – θ) 𝒂 ⃗ .𝒋 ̂ = sin θ (As 𝑎 ⃗ is unit vector, |𝑎 ⃗ | = 1 & 𝑗 ̂ is a unit vector, |𝑗 ̂ | = 1) (𝑥𝑖 ̂ + y𝑗 ̂ + 0𝑘 ̂). 𝑗 ̂ = sin θ (𝑥𝑖 ̂ + y𝑗 ̂ + 0𝑘 ̂). (0𝑖 ̂ + 1𝑗 ̂ + 0𝑘 ̂) = sin θ 𝑥.0 + y.1 + 0.0 = sin θ y = sin θ Thus, 𝑎 ⃗ = x𝑖 ̂ + y𝑗 ̂ 𝒂 ⃗ = cos 𝜃𝒊 ̂ + sin 𝜃 𝒋 ̂ This value will be true in all quadrants So, 0 ≤ θ ≤ 2π Therefore, 𝒂 ⃗ = cos 𝜃𝒊 ̂ + sin𝜃𝒋 ̂ ; for 0 ≤ θ ≤ 2π