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Angle between line with direction ratios (2, 1, 2) and plane 3x − 2y + 6z + 1 = 0
Acos⁻¹(16/21)
Btan⁻¹(16/21)
Csin⁻¹(16/21)
Dsin⁻¹(1/3)
Answer & Solution
Correct answer: C. sin⁻¹(16/21)
1. sin θ = |a·n|/(|a||n|) where a is the line's direction vector and n is the plane's normal.
2. a·n = (2)(3) + (1)(-2) + (2)(6) = 6 − 2 + 12 = 16. |a| = √(4+1+4) = 3, |n| = √(9+4+36) = 7.
3. sin θ = 16/(3·7) = 16/21, so θ = sin⁻¹(16/21).
_Source: Maharashtra Balbharati Std XII Mathematics Part 1, Ch 6 "Line and Plane" §6.7_
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