The acute angle between the xy -plane and the plane passing through the point (1, 2, 4) and parallel to the vectors with direction ratios 3, 2, -1 and 1, -2, -2 is...

Step-by-step solution

Let the vectors parallel to the plane be v ₁ = 3 i + 2 j - k and v ₂ = i - 2 j - 2 k . The normal vector n to the plane is given by the cross product of v ₁ and v ₂ : n = v ₁ v ₂ = vmatrix i & j & k 3 & 2 & -1 1 & -2 & -2 vmatrix n = i (-4 - 2) - j (-6 - (-1)) + k (-6 - 2) = -6 i + 5 j - 8 k The normal vector to the xy -plane is k . The acute angle between the two planes is the acute angle between their normal vectors, given by: = | n k | | n | | k | = |(-6 i + 5 j - 8 k ) k | (-6)^2 + 5^2 + (-8)^2 1^2 = |-8| 36 +