It is known that in tetrahedral ABCD, ab ⊥ CD, AC ⊥ BD. verification: ad ⊥ BC But we haven't learned the three perpendicular theorem

It is known that in tetrahedral ABCD, ab ⊥ CD, AC ⊥ BD. verification: ad ⊥ BC But we haven't learned the three perpendicular theorem

In the tetrahedron, Ao ⊥ is made through the vertex a, and the bottom surface is at O, connecting Bo, CO and do and extending. Bo intersects CD at m, CO intersects BD at n, and do intersects BC at Q. because ab ⊥ CD and Bo are the projection of AB in the plane BCD, BM ⊥ CD is the same as cn ⊥ BD, so o point is the perpendicular center of triangular BCD, so DQ ⊥ BC is also because do is ad in the plane BCD

In tetrahedral ABCD, AB, AC and AD are perpendicular, and vector AB * vector CD = vector ac * vector BD = vector ad * vector BC In tetrahedral ABCD, AB, AC and AD are perpendicular, the following conclusions are incorrect: A. Vector AB * vector CD = vector ac * vector BD = vector ad * vector BC B. | vector AB + vector AC + vector ad| = | vector AB + vector AC - vector ad| C. (vector AB + vector AC + vector AD) * vector BC = 0 D. | vector AB + vector AC + vector ad| ^ 2 = | vector ab| ^ 2 + | vector ac| ^ 2 + | vector ad| ^ 2

A is correct, AB, AC and AD are perpendicular, so AB is perpendicular to plane ACD, so AB is perpendicular to CD, that is, 0. Similarly, the other two are equal to 0b. Correct, vector AB + vector AC is a vector in plane ABC, set as AE, and vector ad is perpendicular to plane ABC and also perpendicular to vector AE, that is, the direction of vector ad is turned 180 ° to

As shown in the figure, in triangle ABC, ad is perpendicular to AB, vector BC is equal to root 3, vector BD, and the modulus of vector ad = 1, then vector AC times vector ad is equal to

Ad is perpendicular to AB, ad should be perpendicular to BC,
Vector ad * AC = ad * (AD + DC) = ad * ad + ad * DC = 1 * 1 + 0 = 1

In triangle ABC, B = 60 degrees, C = 45 degrees, BC = 8, D is the upper point of BC, and vector BD = vector BC * (root 3-1) / 2, what is the length of ad?

Make AE ⊥ BC and set be = X
∵∠B=60º,∠C=45º
∴∠BAE=90º-∠B=30º,∠CAE=90º-∠C=45º=∠C
∴AB=2BE=2x,EC=AE=√﹙AB ²- BE ² ﹚=√3x
∴x+√3x=8,
‡ x = 4 (√ 3-1), i.e. be = 4 (√ 3-1),
BD = (√ 3-1) BC / 2 = 4 (√ 3-1),
‡ point d coincides with point E
∴AD=AE=√3x=12-4√3

As shown in Figure 1, in triangle ABC, angle a = 45 degrees, AC = root 2, ab = root 3 + 1, BC =?

BC^2=AB^2+AC^2-2AB*AC*cosA
=(√3+1)^2+(√2)^2-2*(√3+1)*√2*√2/2
=3+2√3+1+2-2(√3+1)
=6+2√3-2√3-2
=4
BC=2

As shown in the figure, in △ ABC, ad ⊥ AB, BC vector = BD vector of root 3, and the absolute value of Ad vector = 1, then AC vector × Ad vector equals______

Because BC = √ 3bd, ac-ab = √ 3 (ad-ab),
Therefore, AC = (1 - √ 3) * AB + √ 3 * ad,
Since ad, AB, ad * AB = 0,
So ac * ad = [(1 - √ 3) * AB + √ 3 * ad] * ad
=(1-√3)AB*AD+√3*AD^2
=0+√3*1
=√3 .