(20 13? Xiamen simulation) As shown in the figure, PA and QC are perpendicular to the plane of square ABCD, AB=PA=2QC=2, AC ∩ BD = O (Ⅰ) Verification: OP⊥ plane.

Answer: (1) Proof: connecting OQ, from the four points of pa∨QC, ∴P, A, Q and C * *, it is easy to know that BD⊥AC, BD⊥PA, PA ∩ AC = A.

∴BD⊥ Aircraft PACQ, BD ⊥ OP

According to the data in the question, PA=2, AO=OC=2, QC= 1, ∴ PAOC = AOQC, △PAO∽△OCQ, ∴ POA = ∠ OQC.

And ∵ ∠ POA+∠ OPA = 90, ∴ ∠ POA+∠ COQ = 90, ∴ OP ⊥ OQ.

(Or calculate OQ=3, OP=6, PQ=3, and get ∠ POQ = 90 from Pythagorean theorem, that is, OP⊥OQ).

∫BD∩OQ = o, ∴OP⊥ Plane QBD ..

(2) Solution: As shown in the right figure, take A as the origin and straight lines of AB, AD and AP as the X, Y and Z axes respectively, and establish a spatial rectangular coordinate system.

From the meaning of the question, we can get A (0 0,0,0), B (2 2,0,0), C (2 2,2,0), D (0 0,2,0), P (0 0,0,2), Q (2 2,2, 1), O (.

∴OP=(？ 1,? 1，2)，BP=(？ 2,0,2), bq = (0 0,2, 1), let the normal vector of the plane PBQ be n = (x, y, z),

∴n？ BP=0n？ Bq = 0, what? 2x+2z = 02y+z = 0。 Let's take y=- 1 and get n = (2,? 1,2),

According to (I), OP is the normal vector of plane BDQ, so cos < op, n > = op? N|OP has praised it and stepped on it. What's your comment on this answer? Put away comments//high quality or satisfaction or special types or recommended answers dottimewindow.iperformance & window.iperformance.mark ('c _ best',+newdate); Lawyer recommendation service: If your problem has not been solved, please describe your problem in detail and seek free professional advice through Baidu Law Professional Edition. Other similar problems 2022-08- 16 know that PA is perpendicular to the plane of square ABCD, and PA=AB. Find the dihedral angle formed by plane PCD and plane ABCD 2022-07-2 1 It is known that ABCD is square and PA is perpendicular to the plane ABCD. 20 10-09-04 We have known the plane PAB vertical plane ABC and the plane PAC vertical plane ABC, and verified that the PA vertical plane ABC4420 12-03-0 1 in the cube ABCD-a1b/kloc-0. S is the midpoint of A 1D 1, A 1B 1, AB, BB 1 respectively. PQS vertical plane B1RC122020-02-29 ABCD is square, PA vertical plane ABCD, PA = 3pc. 52012-10-15 as shown in the figure, PA⊥ plane ABC, PA = √ 2, AB = 1, BC = √ 3, AC = 2, verification: plane PBC⊥ 𕬥 plane PAB. Verify that PA is parallel to QBC plane; ; If it is known that PQ vertical plane 820 12-08-09 PA is perpendicular to the plane of square ABCD, and PA=AB, how to find dihedral angle 2 formed by plane PCD and plane ABCD? & gt recommended for you: F.context('cmsRight', [{'URL':'/d01373f082025aaf511aa256e9edab64034f1a07? x-BCE-process = image % 2f resize % 2Cm _ lfit % 2Cw _ 450% 2Ch _ 600% 2c limit _ 1% 2f quality % 2Cq _ 85% 2f format % 2Cf _ auto '， ' contractId':'A24KA00562 '，}，{ ' URL ':/builder/bjh-activity/articlesTask？ taskId = 1598082 & aside = 0 & footer = true & from = 0 '，' src ':'/203 FB 80 e 7 bec 54 e 773 ea 0680 ab 389 b 504 fc 26 a2d？ x-BCE-process = image % 2f resize % 2Cm _ lfit % 2Cw _ 450% 2Ch _ 600% 2c limit _ 1% 2f quality % 2Cq _ 85% 2f format % 2Cf _ auto '，' contractId ':' '，}，{ ' URL ':/s？ word = % E6 % AC % A7 % E6 % B4 % B2 % E6 % 9D % AF & sa = searchpromo _ ozb _ zhidao _ tuijian '，' src ':'/3 BF 33 a 87 e 950352 AAF 5 df 4954 143 fbf2b 2 1 18 b6b？ x-BCE-process = image % 2f resize % 2Cm _ lfit % 2Cw _ 450% 2Ch _ 600% 2c limit _ 1% 2f quality % 2Cq _ 85% 2f format % 2Cf _ auto '，' contractId ':' '，}])； Why is the cost of cancer treatment getting higher and higher? What impact will the "network toilet" have? The price of electric cars has dropped many times. Is the quality guaranteed? Is Huaqiang North's second-hand mobile phone reliable? I recommend f.context ('recbrand', [{"img": "\/86D6277F9E2F07083523F69DFB 24B 899A901F20d? x-BCE-process = image % 2f resize % 2Cm _ lfit % 2Cw _ 450% 2Ch _ 600% 2c limit _ 1% 2f quality % 2Cq _ 85% 2f format % 2Cf _ auto "，" url":"/hm.js？ 6859 ce 5a af 00 FB 00387 e 6434 E4 FCC 925 "； var s = document . getelementsbytagname(" script ")[0]； s.parentNode.insertBefore(hm，s)； })(); window . TT = 1720799737；