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If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three non-coplanar vectors and $\mathbf{d}$ is any unit vector, then $|(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \times \mathbf{c})+(\mathbf{b} \cdot \mathbf{d})(\mathbf{c} \times \mathbf{a})+(\mathbf{c} \cdot \mathbf{d})(\mathbf{a} \times \mathbf{b})|=$
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Verified Answer
The correct answer is:
$[\mathrm{abc}] \mid$
a, b, c are non-coplanar.
$\therefore \mathbf{b} \times \mathbf{c}, \mathbf{c} \times \mathbf{a}$ and $\mathbf{a} \times \mathbf{b}$ are also non-coplanar.
So, any vector can be expressed as a linear combination of these vectors.
Let, $\mathbf{d}=\lambda(\mathbf{b} \times \mathbf{c})+\mu(\mathbf{c} \times \mathbf{a})+m(\mathbf{a} \times \mathbf{b})$
$\mathbf{a} \cdot \mathbf{d}=\lambda[\mathbf{a} \mathbf{b} \mathbf{c}]$
$\begin{gathered}
\mathbf{b} \cdot \mathbf{d}=\mu(\mathbf{b} \mathbf{c} \mathbf{a}) \\
\mathbf{c} \cdot \mathbf{d}=m(\mathbf{c} \mathbf{a}) \\
\therefore \mathrm{d}=\frac{(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \times \mathbf{c})}{[\mathbf{a} \mathbf{b} \mathbf{c}]}+\frac{(\mathbf{b} \cdot \mathbf{d})(\mathbf{c} \times \mathbf{a})}{[\mathbf{a} \mathbf{b} \mathbf{c}]}+\frac{(\mathbf{c} \cdot \mathbf{d})(\mathbf{a} \times \mathbf{b})}{[\mathbf{a b} \mathbf{c}]} \\
\Rightarrow(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \times \mathbf{c})+(\mathbf{b} \cdot \mathbf{d})(\mathbf{c} \times \mathbf{a})+(\mathbf{c}-\mathbf{d})(\mathbf{a} \times \mathbf{b})= \\
\mathbf{d}[\mathbf{a} \mathbf{b} \mathbf{c}] \\
\Rightarrow(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \times \mathbf{c}+(\mathbf{b} \cdot \mathbf{d})(\mathbf{c} \times \mathbf{a})+(\mathbf{c} \cdot \mathbf{d})(\mathbf{a} \times \mathbf{b}) \mid \\
=[\mathbf{a} \mathbf{b} \mathbf{c}]
\end{gathered}$
$\therefore \mathbf{b} \times \mathbf{c}, \mathbf{c} \times \mathbf{a}$ and $\mathbf{a} \times \mathbf{b}$ are also non-coplanar.
So, any vector can be expressed as a linear combination of these vectors.
Let, $\mathbf{d}=\lambda(\mathbf{b} \times \mathbf{c})+\mu(\mathbf{c} \times \mathbf{a})+m(\mathbf{a} \times \mathbf{b})$
$\mathbf{a} \cdot \mathbf{d}=\lambda[\mathbf{a} \mathbf{b} \mathbf{c}]$
$\begin{gathered}
\mathbf{b} \cdot \mathbf{d}=\mu(\mathbf{b} \mathbf{c} \mathbf{a}) \\
\mathbf{c} \cdot \mathbf{d}=m(\mathbf{c} \mathbf{a}) \\
\therefore \mathrm{d}=\frac{(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \times \mathbf{c})}{[\mathbf{a} \mathbf{b} \mathbf{c}]}+\frac{(\mathbf{b} \cdot \mathbf{d})(\mathbf{c} \times \mathbf{a})}{[\mathbf{a} \mathbf{b} \mathbf{c}]}+\frac{(\mathbf{c} \cdot \mathbf{d})(\mathbf{a} \times \mathbf{b})}{[\mathbf{a b} \mathbf{c}]} \\
\Rightarrow(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \times \mathbf{c})+(\mathbf{b} \cdot \mathbf{d})(\mathbf{c} \times \mathbf{a})+(\mathbf{c}-\mathbf{d})(\mathbf{a} \times \mathbf{b})= \\
\mathbf{d}[\mathbf{a} \mathbf{b} \mathbf{c}] \\
\Rightarrow(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \times \mathbf{c}+(\mathbf{b} \cdot \mathbf{d})(\mathbf{c} \times \mathbf{a})+(\mathbf{c} \cdot \mathbf{d})(\mathbf{a} \times \mathbf{b}) \mid \\
=[\mathbf{a} \mathbf{b} \mathbf{c}]
\end{gathered}$
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