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If 3 vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are such that $\mathbf{a} \neq \mathbf{0}$ and $\mathbf{a} \times \mathbf{b}=2(\mathbf{a} \times \mathbf{c}),|\mathbf{a}|=1,|\mathbf{c}|=1,|\mathbf{b}|=4$ and angle between $\mathbf{b}$ and $\mathbf{c}$ is $\cos ^{-1}\left(\frac{1}{4}\right)$ and $\mathbf{b}-2 \mathbf{c}=\lambda \mathbf{a}$, then $\lambda=$
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The correct answer is:
$4$
$|\mathbf{b}-2 \mathbf{c}|^2=|\mathbf{b}|^2+4|\mathbf{c}|^2-4|\mathbf{b}||\mathbf{c}| \cos \left(\cos ^{-1} \frac{1}{4}\right)$
$\begin{aligned} & \Rightarrow \quad|\lambda \mathbf{a}|^2=4^2+4 \times 1-4 \times 4 \times 1 \times \frac{1}{4} \\ & \Rightarrow \quad \lambda^2|\mathbf{a}|^2=16 \Rightarrow \lambda^2 \times 1=16 \\ & \Rightarrow \quad \lambda= \pm 4\end{aligned}$
$\begin{aligned} & \Rightarrow \quad|\lambda \mathbf{a}|^2=4^2+4 \times 1-4 \times 4 \times 1 \times \frac{1}{4} \\ & \Rightarrow \quad \lambda^2|\mathbf{a}|^2=16 \Rightarrow \lambda^2 \times 1=16 \\ & \Rightarrow \quad \lambda= \pm 4\end{aligned}$
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