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
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