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$(\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}) \cdot(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}) \times(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}})$ is equal to
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Verified Answer
The correct answer is:
$3[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$
$(\vec{a}+2 \vec{b}-\vec{c}) \cdot(\vec{a}-\vec{b}) \times(\vec{a}-\vec{b}-\vec{c})$
$=(\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}) \cdot[\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{b}}$ $-\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}]$
$=(\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}) \cdot[0-(-\overrightarrow{\mathbf{c}})-(\overrightarrow{\mathbf{c}})+0-(-\overrightarrow{\mathbf{b}})+\overrightarrow{\mathbf{a}}]$
$[\because \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{b}}=0]$
$=(\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}) \cdot(\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{a}})\left\{\begin{array}{l}\because \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{c}} \\ \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{a}} \\ \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{b}}\end{array}\right.$
$=(\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}) \cdot(\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{a}})$
$=\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}+2 \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}$
$=0+2(1)-0+(1)+2(0)-(0) \quad\left\{\begin{array}{l}\because \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=0 \\ \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}=0 \\ \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}=0\end{array}\right.$
$=2+1=3 \cdot 1$
$=3 \cdot\{\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})\} \quad\left\{\begin{array}{l}\because \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}}=1 \\ \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}}=1 \\ \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{c}}=1\end{array}\right.$
$=3[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}] \quad \because[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]=\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})$
$=(\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}) \cdot[\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{b}}$ $-\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}]$
$=(\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}) \cdot[0-(-\overrightarrow{\mathbf{c}})-(\overrightarrow{\mathbf{c}})+0-(-\overrightarrow{\mathbf{b}})+\overrightarrow{\mathbf{a}}]$
$[\because \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{b}}=0]$
$=(\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}) \cdot(\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{a}})\left\{\begin{array}{l}\because \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{c}} \\ \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{a}} \\ \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{b}}\end{array}\right.$
$=(\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}) \cdot(\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{a}})$
$=\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}+2 \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}$
$=0+2(1)-0+(1)+2(0)-(0) \quad\left\{\begin{array}{l}\because \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=0 \\ \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}=0 \\ \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}=0\end{array}\right.$
$=2+1=3 \cdot 1$
$=3 \cdot\{\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})\} \quad\left\{\begin{array}{l}\because \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}}=1 \\ \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}}=1 \\ \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{c}}=1\end{array}\right.$
$=3[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}] \quad \because[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]=\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})$
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