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1. Chapter 10 Class 12 Vector Algebra
2. Serial order wise
3. Miscellaneous

Transcript

Misc 3 A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl’s displacement from her initial point of departure.Let initial point of departure be O(0, 0) & OA = 4km, AB = 3km ∠ XAB = 30° Here, Displacement will be vector (𝑂𝐵) ⃗ So, we need to find coordinates of B Now, ∠ BAO = 90° – ∠ XAB = 90° – 30° = 60° Let us draw BD ⊥ OA In right triangle ADB sin 60° = 𝐵𝐷/𝐴𝐵 √3/2 = 𝐵𝐷/3 (3√3)/2 = BD BD = (𝟑√𝟑)/𝟐 In right triangle ADB cos 60° = 𝐴𝐷/𝐴𝐵 1/2 = 𝐴𝐷/3 3/2 = AD AD = 𝟑/𝟐 Now, OD = OA – AD OD = 4 – 𝟑/𝟐 = (𝟒(𝟐) − 𝟑)/𝟐 = 𝟓/𝟐 So, x-coordinate of point B = −OD = (−𝟓)/𝟐 Thus, coordinates of point B = ((−𝟓)/𝟐, (𝟑√𝟑)/𝟐) So, Displacement = (−𝟓)/𝟐 𝒊 ̂ + (𝟑√𝟑)/𝟐 𝒋 ̂ y-coordinate of point B = BD = (𝟑√𝟑)/𝟐 Given, 𝑎 ⃗ = 𝑏 ⃗ + 𝑐 ⃗ Let 𝑏 ⃗ = 1𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂ & 𝑐 ⃗ = 2𝑖 ̂ − 1𝑗 ̂ − 2𝑘 ̂ 𝑎 ⃗ = (𝑏 ⃗ + 𝑐 ⃗) = (1 + 2) 𝑖 ̂ + (2 − 1) 𝑗 ̂ + (3 − 2) 𝑘 ̂ = 3𝑖 ̂ + 1𝑗 ̂ + 1𝑘 ̂ So, 𝑎 ⃗ = 3𝑖 ̂ + 1𝑗 ̂ + 1𝑘 ̂ Magnitude of 𝑎 ⃗ = √(32+1^2+1^2 ) |𝑎 ⃗ | = √(9+1+1) = √11 Given, 𝑎 ⃗ = 𝑏 ⃗ + 𝑐 ⃗ Let 𝑏 ⃗ = 1𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂ & 𝑐 ⃗ = 2𝑖 ̂ − 1𝑗 ̂ − 2𝑘 ̂ 𝑎 ⃗ = (𝑏 ⃗ + 𝑐 ⃗) = (1 + 2) 𝑖 ̂ + (2 − 1) 𝑗 ̂ + (3 − 2) 𝑘 ̂ = 3𝑖 ̂ + 1𝑗 ̂ + 1𝑘 ̂ So, 𝑎 ⃗ = 3𝑖 ̂ + 1𝑗 ̂ + 1𝑘 ̂ Magnitude of 𝑎 ⃗ = √(32+1^2+1^2 ) |𝑎 ⃗ | = √(9+1+1) = √11 Magnitude of 𝑏 ⃗ = √(12+22+32) |𝑏 ⃗ | = √(1+4+9) = √14 Magnitude of 𝑐 ⃗ = √(22+(−1)2+(−2)2) |𝑐 ⃗ | = √(4+1+4) = √9 = 3 Now, |𝑏 ⃗ | + |𝑐 ⃗ | = √14 + 3 ≠ √11 ≠ |𝑎 ⃗ | So, |𝑎 ⃗ |≠ |𝑏 ⃗ | + |𝑐 ⃗ | Hence, false.

Miscellaneous 