MathIsimple

Solid Geometry – Problem 5: Prove that plane

Question

In the quadrilateral pyramid BACC1A1B-ACC_1A_1, AA1AA_1\perp plane ABCABC, CC1CC_1\perp plane ABCABC, BAACBA\perp AC, and AB=AC=CC1=2AA1.AB=AC=CC_1=2AA_1.

(1) Prove that plane A1BC1BCC1A_1BC_1\perp BCC_1.

(2) Let PP be the reflection of point CC across plane A1BC1A_1BC_1. Find sinθ\sin\theta, where θ\theta is the angle between line APAP and plane A1BC1A_1BC_1.

Step-by-step solution

(1) Let M,NM,N be the midpoints of BC1BC_1, BCBC, respectively, and connect A1M,ANA_1M,AN. From AA1CC1AA_1\parallel CC_1, AA1=12CC1AA_1=\dfrac12CC_1, and MNCC1MN\parallel CC_1, MN=12CC1MN=\dfrac12CC_1, we get AA1MNAA_1\parallel MN and AA1=MNAA_1=MN. Hence quadrilateral A1MNAA_1MNA is a parallelogram, so A1MANA_1M\parallel AN.

Also, AB=ACANBCAB=AC\Rightarrow AN\perp BC, and ANCC1AN\perp CC_1. Since BCCC1=CBC\cap CC_1=C, we have ANAN\perp plane BCC1BCC_1, thus A1MA_1M\perp plane BCC1BCC_1. Because A1MA_1M\subset plane A1BC1A_1BC_1, plane A1BC1plane BCC1.\text{plane }A_1BC_1\perp\text{plane }BCC_1.

(2) Through CC, draw CQBC1CQ\perp BC_1, meeting BC1BC_1 at QQ. Then CQCQ\perp plane A1BC1A_1BC_1. Let AB=AC=CC1=2AA1=2AB=AC=CC_1=2AA_1=2. Set a 3D rectangular coordinate system with AA as origin, and lines containing AB,AC,AA1AB,AC,AA_1 as the x,y,zx,y,z-axes. Take B(2,0,0), C(0,2,0), C1(0,2,2), A1(0,0,1).B(2,0,0),\ C(0,2,0),\ C_1(0,2,2),\ A_1(0,0,1). From right-triangle projection relations, 2C1Q=QB2C_1Q=QB, so Q(23,43,43).Q\left(\frac23,\frac43,\frac43\right). Since PP is the reflection of CC across plane A1BC1A_1BC_1, point QQ is the midpoint of CPCP, thus P(43,23,83),AP=(43,23,83).P\left(\frac43,\frac23,\frac83\right),\quad \overrightarrow{AP}=\left(\frac43,\frac23,\frac83\right). A normal vector of plane A1BC1A_1BC_1 can be n=(1,1,2)\vec n=(1,-1,2). Therefore sinθ=APnAPn=31414.\sin\theta=\frac{|\overrightarrow{AP}\cdot\vec n|}{|\overrightarrow{AP}|\,|\vec n|}=\frac{3\sqrt{14}}{14}.

Final answer

(1) By midpoint construction and perpendicular-plane criteria, we find a line in plane A1BC1A_1BC_1 that is perpendicular to plane BCC1BCC_1, so plane A1BC1plane BCC1.\text{plane }A_1BC_1\perp\text{plane }BCC_1.

(2) After locating the reflection point PP through midpoint QQ, and using a normal vector of plane A1BC1A_1BC_1, we obtain sinθ=31414.\sin\theta=\dfrac{3\sqrt{14}}{14}. This is the sine of the angle between line APAP and the plane.

Marking scheme

1. Checkpoints (max 7 pts total)

Part (1): plane-perpendicular proof (3 pts)

  • Correct midpoint construction and parallel relation derivation. (1.5 pts)
  • Show a line in plane A1BC1A_1BC_1 is perpendicular to plane BCC1BCC_1. (1 pt)
  • Conclude two-plane perpendicularity. (0.5 pt)

Part (2): line-plane angle (4 pts)

  • Build coordinates and determine projection/symmetry point coordinates. (2 pts)
  • Find a valid normal vector of plane A1BC1A_1BC_1. (1 pt)
  • Apply vector formula and obtain 31414\frac{3\sqrt{14}}{14}. (1 pt)

Total (max 7)


2. Zero-credit items

  • Assuming midpoint/reflection coordinates without derivation.
  • No normal vector or equivalent angle computation in part (2).

3. Deductions

  • Reflection-point error (-1): confusing projection point and midpoint relation.
  • Angle formula misuse (-1): using cosine formula directly for line-plane angle.
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