Find the acute angle between the straight line x/1 = y/3 = z/0 and pla

1.	Find the acute angle between the straight line x/1 = y/3 = z/0 and plane 2x + y = 5.
Answer» Given equation x/1 = y/3 = z/0 and plane eq. 2x + y = 5 line (x - x₁)/l = (y - y₁)/m = (z - z₁)/n plane ax + by + cz + d = 0 here angle of them sin = (al + bm + cn)/√(a² + b² + c²) √(l² + m² + n²) Here, l = 1, m = 3, n = 0 a = 2, b = 1, c = 0 then, sin θ = (2 x 1 + 3 x 1 + 0 x 0)/√(2² + 1² + 0²) x √(1² + 3²) sin θ = (2 + 3 + 0)/√(4 + 1) √(1 + 9) sin θ = 5/(√5 x √10) sin θ = 5/√50 sin θ = 5/5√2 = 1√2 sin θ = sin π/4 θ = π/4

Find the acute angle between the straight line x/1 = y/3 = z/0 and plane 2x + y = 5.

Answer»

Given equation

x/1 = y/3 = z/0 and plane eq. 2x + y = 5

line (x - x₁)/l = (y - y₁)/m = (z - z₁)/n plane ax + by + cz + d = 0

here angle of them sin = (al + bm + cn)/√(a² + b² + c²) √(l² + m² + n²)

Here, l = 1, m = 3, n = 0

a = 2, b = 1, c = 0

then, sin θ = (2 x 1 + 3 x 1 + 0 x 0)/√(2² + 1² + 0²) x √(1² + 3²)

sin θ = (2 + 3 + 0)/√(4 + 1) √(1 + 9)

sin θ = 5/(√5 x √10)

sin θ = 5/√50

sin θ = 5/5√2 = 1√2

sin θ = sin π/4

θ = π/4