Cho tứ diện \(ABCD\) có \(AB = CD = a\), \(AC = BD = b\), \(AD = BC = c\). Tính góc giữa hai đường thẳng \(AC\) và \(BD\).
A.\(\angle \left( {AC;BD} \right) = \arccos \left| {\dfrac{{{a^2} - {c^2}}}{{{b^2}}}} \right|\)
B.\(\angle \left( {AC;BD} \right) = \arccos \left| {\dfrac{{2\left( {{a^2} + {c^2}} \right)}}{{{b^2}}}} \right|\)
C.\(\angle \left( {AC;BD} \right) = \arccos \left| {\dfrac{{2\left( {{a^2} - {c^2}} \right)}}{{3{b^2}}}} \right|\)
D.\(\angle \left( {AC;BD} \right) = \arccos \left| {\dfrac{{2\left( {{a^2} - {c^2}} \right)}}{{{b^2}}}} \right|\)
A.\(\angle \left( {AC;BD} \right) = \arccos \left| {\dfrac{{{a^2} - {c^2}}}{{{b^2}}}} \right|\)
B.\(\angle \left( {AC;BD} \right) = \arccos \left| {\dfrac{{2\left( {{a^2} + {c^2}} \right)}}{{{b^2}}}} \right|\)
C.\(\angle \left( {AC;BD} \right) = \arccos \left| {\dfrac{{2\left( {{a^2} - {c^2}} \right)}}{{3{b^2}}}} \right|\)
D.\(\angle \left( {AC;BD} \right) = \arccos \left| {\dfrac{{2\left( {{a^2} - {c^2}} \right)}}{{{b^2}}}} \right|\)
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Hóa Học lớp 12
