First of all, the diagonal vector theorem applies not only to planes, but also to spaces. This can be seen from its proof that your understanding is only flat;

Secondly, the second picture is the angle between AC and BD, which can be deduced according to the diagonal vector theorem:

cos(AC，BD)=[(AD？ +BC？ )-(AB？ +CD？ )]/2|AC||BD|

To solve it!

Finally, solve your third problem!

According to the meaning of the question:

Connect CF, ef;

In tetrahedron: E-BCF, according to diagonal vector theorem:

cos & ltBE，CF & gt

=[(BF？ +CE？ )-(BC? +EF？ )]/2|BE||CF|

In diamond ABCD:

BF=EF= 1/2，BC= 1，CF=BE=√3/2

Then, replace:

cos & ltBE，CF & gt

=[(BF？ +CE？ )-(BC? +EF？ )]/2|BE||CF|

=(2/3)(CE- 1)

Inspection CE:

In the diamond ABCD, CE is the longest, then:

To find the length of CE, auxiliary lines need to be made to connect CH, where H is the midpoint of BA. Obviously:

CE=√[(√3/2 + √3/4)？ +? ( 1/4)]=√[(3√3/4)? +? ( 1/4)]

When e and the midpoint e' of CD coincide, CE is the shortest, at this time: CE=CE'= 1/2.

So:

- 1/2 & lt； cos & ltBE，CF & gt& lt 1/2

Namely:

cos & ltBE，CF & gt∈(π/3, 2π/3)

Option d