空間中三平面共線

心裡有數
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在坐標空間中,已知三平面\({E_1}:{a_1}x + {b_1}y + {c_1}z = {d_1}\),\({E_2}:{a_2}x + {b_2}y + {c_2}z = {d_2}\),\({E_3}:{a_3}x + {b_3}y + {c_3}z = {d_3}\)共線,則\[\left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right| = 0\],反之不成立。

 

上述敘述並非什麼罕見的性質,國編本、88課綱、99課綱都有,但採用三階克拉瑪公式證明此性質;108課綱不再提三階克拉瑪公式,轉而強調三重積與三階行列式的關係,此處採用此方式證明上述性質,證明如下:

 

設平面\({E_1}\),\({E_2}\),\({E_3}\)的法向量分別為\(\mathop {{n_1}}\limits^ \rightharpoonup   = ({a_1},{b_1},{c_1})\),\(\mathop {{n_2}}\limits^ \rightharpoonup   = ({a_2},{b_2},{c_2})\),\(\mathop {{n_3}}\limits^ \rightharpoonup   = ({a_3},{b_3},{c_3})\),並設\({E_1}\),\({E_2}\),\({E_3}\)之交線為\(L\)且\(L\)之方向向量為\(\mathop v\limits^ \rightharpoonup  \)

因為\({E_1}\)與\({E_2}\)之交線為\(L\)且\(L\)之方向向量為\(\mathop v\limits^ \rightharpoonup  \),所以\(\mathop v\limits^ \rightharpoonup   \bot \mathop {{n_1}}\limits^ \rightharpoonup  \)且\(\mathop v\limits^ \rightharpoonup   \bot \mathop {{n_2}}\limits^ \rightharpoonup  \)

\( \Rightarrow \)\(\mathop v\limits^ \rightharpoonup  //\left( {\mathop {{n_1}}\limits^ \rightharpoonup   \times \mathop {{n_2}}\limits^ \rightharpoonup  } \right)\)

設\(\mathop v\limits^ \rightharpoonup   = k\left( {\mathop {{n_1}}\limits^ \rightharpoonup   \times \mathop {{n_2}}\limits^ \rightharpoonup  } \right)\),其中\(k \ne 0\)

又\(L\)在\({E_3}\)上,所以\(\mathop v\limits^ \rightharpoonup   \bot \mathop {{n_3}}\limits^ \rightharpoonup  \)

\( \Rightarrow \)\(\mathop v\limits^ \rightharpoonup   \cdot \mathop {{n_3}}\limits^ \rightharpoonup   = 0\)

\( \Rightarrow \)\(k\left( {\mathop {{n_1}}\limits^ \rightharpoonup   \times \mathop {{n_2}}\limits^ \rightharpoonup  } \right) \cdot \mathop {{n_3}}\limits^ \rightharpoonup   = 0\)

\( \Rightarrow \)\(\left( {\mathop {{n_1}}\limits^ \rightharpoonup   \times \mathop {{n_2}}\limits^ \rightharpoonup  } \right) \cdot \mathop {{n_3}}\limits^ \rightharpoonup   = 0\)

因此\[\left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right| = 0\]

 

反之不成立,例如:\({E_1}:x + y + z = 1\),\({E_2}:x + y + z = 2\),\({E_3}:x + y + z = 3\)。顯然,\({E_1}\),\({E_2}\),\({E_3}\)為平行的三個平面,所以不會共線,但\[\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&1&1\\1&1&1\end{array}\,} \right| = 0\]。

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