Consider the following line and plane equations in three-dimensional space:Line L: r = \ $$ r_0 \ + \ $$ t_d where $$ r_0 = (1,2,3) and d = (4,-1,2) Plane P: 3x-2y+z=5 Which of the following statements is true regarding the intersection of the line L and the plane P?(C:1)

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Question restatement: Given Line L: r = r0 + t d with r0 = (1,2,3) and d = (4,-1,2), and Plane P: 3x - 2y + z = 5, determine the true statement about their intersection.

Option a: The line L is parallel to the plane P and does not intersect it.
- To test parallelism, check if the line’s direction vector d is orthogonal to the plane normal n = (3, -2, 1). If d · n = 0, the line is parallel to the plane; otherwise not. Here, d · n = (4)(3) + (-1)(-2) + (2)(1) = 12 + 2 + 2 = 16, which is not zero. This means the line is not parallel to the plane, so a parallel, non-intersecting scenario is incorrect. Additionally, a nonzero dot product indicates the line may intersect the plane.

Option b: The line L intersects the plane P at a point, which can be found by substituting the parametric equations of the line into the plane equation.
- This approach is standard: substitute x = 1 + 4t, y = 2 - t, z = 3 + 2t into 3x - 2y + z = 5. Compute: 3(1 + 4t) - 2(2 - t) + (3 + 2t) = 5 → 3 + 12t - 4 + 2t + 3 + 2t = 5 → (3 - 4 + 3) + (12t + 2t + 2t) = 5 → 2 + 16t = 5 → 16t = 3 → t = 3/16. A real t value exists, so the line indeed intersects the plane at a single point. This option aligns with the calculation and standard intersection method.

Option c: The line L and the plane P do not intersect because their normal vectors are orthogonal.
- This statement is mixing concepts. Orthogonality of normals refers to the plane being parallel to the line when the line’s direction is orthogonal to the plane normal, i.e., d · n = 0. Here d · n = 16 ≠ 0, so the normals are not orthogonal in a way that would imply no intersection. In fact, a nonzero dot product suggests the line is not parallel to the plane and can intersect it, which contradicts the claim that they do not intersect due to orthogonal normals. Therefore this option is incorrect.

Option d: The line L lies entirely within the plane P.
- For a line to lie entirely in a plane, every point r = r0 + t d on the line must satisfy the plane equation. Check at t = 0: r0 = (1,2,3) gives 3(1) - 2(2) + 3 = 3 - 4 + 3 = 2 ≠ 5, so r0 is not in the plane. Since the line passes through a point not in the plane, it cannot lie entirely within the plane. Hence this option is false.

Summary of reasoning across options: The parallelism test using d · n shows the line is not parallel to the plane, eliminating option a and supporting the possibility of intersection. Substituting the line into the plane equation yields a solvable equation with t = 3/16, confirming a single intersection point, which validates option b. The geometric interpretations in options c and d contradict the computed result or the necessary conditions for line-plane relationships, so they are incorrect.

Consider the following line and plane equations in three-dimensional space:Line L: r = \ $$ r_0 \ + \ $$ t_d where $$ r_0 = (1,2,3) and d = (4,-1,2) Plane P: 3x-2y+z=5 Which of the following statements is true regarding the intersection of the line L and the plane P?(C:1)单项选择题

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