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In the cube abcd-a1b1c1d1 each vertex and each edge midpoint total 20 points, take any 2 points to form a straight line, take any one of these straight lines, the probability that it is perpendicular to the diagonal BD1 is () A. 21166B. 21190C. 18190D. 27166
From the meaning of the question, we can know that this question is a classical type. Take 2 out of 20 points, a total of C202 = 190, but any 2 of 3 points on each edge are repeated, and the denominator is 190-12c32 + 12 = 166. To be perpendicular to BD1, it should be parallel to or in the plane of a1dc1, and there are 9 parallel or coincident with a1c1, a total of 27, P = 27166
In the cube abcd-a1b1c1d1, (1) prove that the plane aa1c1c is perpendicular to the plane bdc1, (2) find the tangent of the angle between Aa1 and bdc1
（1）∵A1A⊥BD AC⊥BD
Ψ BD ⊥ plane aa1c
Plane aa1c1c ⊥ plane bdc1
(2) Connect AC to BD and O to C1O
The perpendicularity of C1O through point C is e
Because CE is in plane aa1c1c, BD ⊥ CE
The angle between plane bdc1 and plane bdc1 is ∠ cc1e
Aa1 is parallel to CC1
The angle between Aa1 and bdc1 is the angle between CC1 and bdc1
Let the side length of cube be 1
Then CE = root 3 / 3
C1E = radical 6 / 3
TG ∠ cc1e = root 2 / 2
In the cube abcd-a1b1c1d1, the angle between Ab1 and CD is______ ．
The angle between ∵ CD ∥ B1a1, ∵ Ab1 and CD is ∵ ab1a1, ∵ ab1a1 = 45 ° and the angle between ∵ Ab1 and CD is 45 °. So the answer is: 45 °
In the cube abcd-a1b1c1d1, m and N are the midpoint of CD and C1C respectively, then the angle between a1m and DN is different
Take D as the coordinate origin, establish the space rectangular coordinate system as shown in the figure,
Then d (0,0,0), n (0,2,1), m (0,1,0), A1 (2,0,2), DN = (0,2,1), a1m = (- 2,1, - 2)
DN &; a1m = 0, so DN ⊥ a1m, that is, a1m ⊥ DN, the angle between a1m and DN is 90 degrees,
So the answer is: 90 degrees
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