The angle between a line and a plane
is the complement of the angle between the line and normal to the plane.

Thus, if θ is the angle between a lines
(x – α)/l = (y - β)/m = (z - γ)/n

And the plane ax + by + cz + d = 0.
Then, the angle between the normal to the plane and the line is (90°- θ).

cos (90°- θ) = (al + bm + cn)/ √ (a² +
b² + c²) √ (l² + m² +n²).

∴ sinθ = (al + bm + cn)/ √ (a² + b² + c²) √ (l²
+ m² +n²).

If the line is parallel to the plane,

∴ al + bm + cn = 0.

If the line is perpendicular to the
plane,

∴ l/a = m/b = n/c

Find the angle between the line (x - 1)/1
= (y + 2)/1 = (z - 4)/0 and the plane y +
z + 2 = 0.

**Solution:**

Let θ be the angle between the line (x - 1)/1
= (y + 2)/1 = (z - 4)/0 and the plane y +
z + 2 = 0.

∴ sinθ = (al + bm + cn)/ √ (a² + b² + c²) √ (l²
+ m² +n²).

⇒ sinθ = [1(0) + 1(1) + 1(0)]/ √ [(1)²
+ (1)² + (0)²] √ [(1)² + (1)² + (0)²]

⇒ sinθ = 1/ √2 √2

Then,

⇒ sinθ = ½