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If $\mathbf{a}=(1,2,3), \mathbf{b}=(2,-1,1), \mathbf{c}=(3,2,1)$ and $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}$, then
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Verified Answer
The correct answer is:
$\alpha=0, \beta=10, \gamma=-3$
Given,
$a=(1,2,3)=i+2 j+3 k$
$b=(2,-1,1)=2 i-j+k$
$\begin{array}{ll}\text { and } & c=(3,2,1)=3 i+2 j+k \\ \text { Now, } & a \times(b \times c)=\alpha a+\beta b+\gamma c \\ \Rightarrow & (a \cdot c) b-(a \cdot b) c=\alpha a+\beta b+\gamma c \\ \Rightarrow & \{(i+2 j+3 k) \cdot(3 i+2 j+k)\} b \\ & \quad-\{(i+2 j+3 k) \cdot(2 i-j+k)\} c\end{array}$
$\begin{aligned}
&=\alpha a+\beta b+\gamma c \\
\Rightarrow &(3+4+3) b-(2-2+3) c=\alpha a+\beta b+\gamma c \\
\Rightarrow & 0 a+10 b-3 c=\alpha a+\beta b+\gamma c
\end{aligned}$
On comparing, we get
$\alpha=0, \beta=10 \text { and } \gamma=-3$
$a=(1,2,3)=i+2 j+3 k$
$b=(2,-1,1)=2 i-j+k$
$\begin{array}{ll}\text { and } & c=(3,2,1)=3 i+2 j+k \\ \text { Now, } & a \times(b \times c)=\alpha a+\beta b+\gamma c \\ \Rightarrow & (a \cdot c) b-(a \cdot b) c=\alpha a+\beta b+\gamma c \\ \Rightarrow & \{(i+2 j+3 k) \cdot(3 i+2 j+k)\} b \\ & \quad-\{(i+2 j+3 k) \cdot(2 i-j+k)\} c\end{array}$
$\begin{aligned}
&=\alpha a+\beta b+\gamma c \\
\Rightarrow &(3+4+3) b-(2-2+3) c=\alpha a+\beta b+\gamma c \\
\Rightarrow & 0 a+10 b-3 c=\alpha a+\beta b+\gamma c
\end{aligned}$
On comparing, we get
$\alpha=0, \beta=10 \text { and } \gamma=-3$
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