If \(\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}\) are three vectors such that \(\overline{\mathrm{a}} \neq \overline{0}\) and \(\overline{\mathrm{a}} \times \overline{\mathrm{b}}=2 \overline{\mathrm{a}} \times \overline{\mathrm{c}},|\overline{\mathrm{a}}|=|\overline{\mathrm{c}}|=1,|\overline{\mathrm{~b}}|=4\) and \(|\overline{\mathrm{b}} \times \overline{\mathrm{c}}|=\sqrt{15}\). If \(\bar{b}-2 \bar{c}=\lambda \bar{a}\), then \(\lambda\) is

If angle between \(\overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\) is \(\alpha\) and
\(\begin{aligned}
& |\overline{\mathrm{b}} \times \overline{\mathrm{c}}|=\sqrt{15} \\
& \Rightarrow|\overline{\mathrm{~b}}||\overline{\mathrm{c}}| \sin \alpha=\sqrt{15} \\
& \Rightarrow(4)(1) \sin \alpha=\sqrt{15} \\
& \Rightarrow \sin \alpha=\frac{\sqrt{15}}{4} \\
& \Rightarrow \cos \alpha=\frac{1}{4}
\end{aligned}\)
\(\begin{aligned} & \text { Now, } \overline{\mathrm{b}}-2 \overline{\mathrm{c}}=\lambda \overline{\mathrm{a}} \\ & \Rightarrow|\overline{\mathrm{b}}-2 \overline{\mathrm{c}}|^2=\lambda^2|\overline{\mathrm{a}}|^2 \\ & \Rightarrow|\overline{\mathrm{~b}}|^2+4|\overline{\mathrm{c}}|^2-4 \overline{\mathrm{~b}} \cdot \overline{\mathrm{c}}=\lambda^2|\overline{\mathrm{a}}|^2 \\ & \Rightarrow 16+4-4(|\overline{\mathrm{~b}}| \overline{\mathrm{a}} \mid \cos \alpha)=\lambda^2 \\ & \Rightarrow 20-4\left(4 \times 1 \times \frac{1}{4}\right)=\lambda^2 \\ & \Rightarrow 16=\lambda^2 \\ & \Rightarrow \lambda= \pm 4\end{aligned}\)