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If $(\bar{a} \times \bar{b}) \times \bar{c}=-5 \bar{a}+4 \bar{b}$ and $\bar{a} \cdot \bar{b}=3$, then the value of $\bar{a} \times(\bar{b} \times \bar{c})$ is
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Verified Answer
The correct answer is:
$4 \overline{\mathrm{b}}-3 \overline{\mathrm{c}}$
$\begin{aligned}
& (\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}} \\
& \text { But, }(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=-5 \overline{\mathrm{a}}+4 \overline{\mathrm{b}} \\
\therefore \quad & -5 \overline{\mathrm{a}}+4 \overline{\mathrm{b}}=(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}
\end{aligned}$
Comparing, we get
$\begin{aligned}
\therefore \quad \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=4 & \\
\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) & =(\overline{\mathrm{a} \cdot} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{c}} \\
& =4 \overline{\mathrm{b}}-3 \overline{\mathrm{c}} \quad \ldots[\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=3 \text { (given) }]
\end{aligned}$
& (\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}} \\
& \text { But, }(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=-5 \overline{\mathrm{a}}+4 \overline{\mathrm{b}} \\
\therefore \quad & -5 \overline{\mathrm{a}}+4 \overline{\mathrm{b}}=(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}
\end{aligned}$
Comparing, we get
$\begin{aligned}
\therefore \quad \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=4 & \\
\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) & =(\overline{\mathrm{a} \cdot} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{c}} \\
& =4 \overline{\mathrm{b}}-3 \overline{\mathrm{c}} \quad \ldots[\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=3 \text { (given) }]
\end{aligned}$
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