Angle Between A Line And A Plane

The key relation is that the angle \phi between a line with direction vector b and a plane with normal n satisfies \sin\phi = |b \cdot n| / (|b||n|), because the line's inclination to the plane is the complement of its angle with the normal. This complement relationship is a frequently tested JEE Advanced subtlety. Here b = (2, -1, 2) and n = (1, 1, 1). The dot product is 2(1) + (-1)(1) + 2(1) = 2 - 1 + 2 = 3. The magnitudes are |b| = \sqrt{4+1+4} = 3 and |n| = \sqrt{3}, so |b||n| = 3\sqrt{3}. Therefore \sin\phi = 3/(3\sqrt{3}) = 1/\sqrt{3}. Option 1/(3\sqrt{3}) forgets to cancel the factor of 3 from the dot product. Option \sqrt{2}/3 misuses the components. Option 2/3 drops a term. This applies the line-plane complement theorem, which is easy to misremember because students often compute cosine and forget the ninety-degree shift between the line-normal angle and the line-plane angle. Recognizing that the normal points out of the plane while the line lies near it is the conceptual key to choosing sine over cosine here. Plausibility check: the value is positive and below 1, a legitimate sine yielding an acute inclination, and because the dot product of direction and normal is nonzero the line is genuinely oblique to the plane rather than parallel to it, which is consistent with a strictly positive sine.