Angle Between Two Planes

The governing principle is that the angle between two planes equals the angle between their normal vectors, computed as \cos\theta = |n_1 \cdot n_2| / (|n_1||n_2|). This normal-vector method is a standard JEE Advanced approach because a plane's orientation is encoded entirely in its normal. The normals here are n_1 = (1, 1, 1) and n_2 = (1, -1, 2). Their dot product is 1(1) + 1(-1) + 1(2) = 1 - 1 + 2 = 2. The magnitudes are |n_1| = \sqrt{3} and |n_2| = \sqrt{6}, so the product is \sqrt{18}. Hence \cos\theta = 2/\sqrt{18}, giving \theta = \cos^{-1}(2/\sqrt{18}). Option \cos^{-1}(1/\sqrt{2}) corresponds to a 45-degree guess unsupported by the dot product. Option \cos^{-1}(3/\sqrt{18}) wrongly adds the components without sign. Option \cos^{-1}(0) would require perpendicular normals, but the dot product is nonzero. This uses the normal-vector angle theorem for planes, where the absolute value guarantees we report the acute rather than the obtuse angle between the two orientations. It is worth noting that the planes themselves intersect along a line, and the dihedral angle along that line of intersection is precisely the angle captured by their normals. Plausibility check: 2/\sqrt{18} \approx 0.471 lies strictly between 0 and 1, yielding a valid acute angle near 62 degrees, and since the dot product was positive the normals point into a consistent half-space, confirming the acute measure is the correct one.