mn ơi giảng giúp e câu này vs ạ

Ta có: `(A A^',(ABC))=(A A^',AH)=hat(A^'AH)=60^o`

Trong `DeltaA^'HA` vuông tại `H`, có:

`tan hat(A^'AH)=(A^'H)/(AH)=>A^'H=tan 60^o=sqrt3`

`cos hat(A^'AH)=(AH)/(A A^')=>A A^'=(AH)/(cos hat(A^'AH))=2`

Gắn hệ trục tọa độ `Oxyz` như hình, ta có:

`C(0;0;0);A(-2;0;0);B(0;b;0);H(-1;0;0);A^'(-1;0;sqrt3)`

Ta có: `vec(A^'C^')=vec(AC)=(2;0;0)=>C^'(1;0;sqrt3)`

Ta có: `vec(C^'B^')=vec(CB)=(0;b;0)=>B^'(1;b;sqrt3)`

`vec(A^'B^')=(2;b;0)`

`vec(A^' A)=(-1;0;-sqrt3)`

`=>` `[vec(A^'B^'),vec(A^' A)]=(-bsqrt3;2sqrt3;b)`

`=>` `vec(n_1)=(-bsqrt3;2sqrt3;b)`

`vec(A^'C^')=(2;0;0)`

`vec(A^' A)=(-1;0;-sqrt3)`

`=>` `[vec(A^'C^'),vec(A^' A)]=(0;-1;0)=-1*(0;1;0)`

`=>` `vec(n_2)=(0;1;0)`

Có: `[B,A A^',C^']=30^o=>`$((A A'B' B),(A C C'A'))$`=30^o`

`=>` `cos`$((A A'B' B),(A C C'A'))$`=sqrt3/2`

`=>` `(|vec(n_1)*vec(n_2)|)/(|vec(n_1)|*|vec(n_2)|)=sqrt3/2`

`=>` `(2sqrt3)/(sqrt(4b^2+12))=sqrt3/2`

`=>` `(1)/(sqrt(b^2+3))=1/2`

`=>` `sqrt(b^2+3)=2`

`=>` `b=1`

Ta có:

`vec(A^'H)=(0;0;-sqrt3)`

`vec(B^'C)=(-1;-1;-sqrt3)`

`=>` `[vec(A^'H),vec(B^'C)]=(-sqrt3;sqrt3;0)`

`vec(A^'C)=(1;0;-sqrt3)`

`=>` `d(A^'H,B^'C)=(|[vec(A^'H),vec(B^'C)]*vec(A^'C)|)/(|[vec(A^'H),vec(B^'C)]|)=sqrt2/2`

`=>` Đáp án: $\fbox{0,71}$