Let $\overline{\mathrm{a}},\overline{\mathrm{b}},\overline{\mathrm{c}}$ be three vectors satisfying $\overline{\mathrm{a}}\times \overline{\mathrm{b}}=2\overline{\mathrm{a}}\times \overline{\mathrm{c},}\left|\overline{\mathrm{a}}\right|=\left|\overline{\mathrm{c}}\right|=1,\left|\overline{\mathrm{b}}\right|=4$ 

and $|\overline{\mathrm{b}}\times \overline{\mathrm{c}}|=\sqrt{15.}$If $\overline{\mathrm{b}}-\overline{2}\mathrm{c}=\mathrm{\lambda }\overline{\mathrm{a}}$ then $\mathrm{\lambda }$ is

$Given \left|\overline{a}\right|=\left|\overline{c}\right|=1,and \left|b\right|=4$

 $$\begin{gathered} If angle between \overline{ \mathrm{b}} and \overline{\mathrm{c}} \mathrm{is} \mathrm{\alpha } \mathrm{and} |\overline{\mathrm{b}}\times \overline{\mathrm{c}}|=\sqrt{15}. \\ \Rightarrow \left|\overline{\mathrm{b}}\right|\left|\mathrm{c}\right|\mathrm{sin\alpha }=\sqrt{15} \\ \Rightarrow 4\left(1\right)\mathrm{sin\alpha }=\sqrt{15} \\ \Rightarrow \mathrm{sin\alpha }\frac{\sqrt{15}}{4} \\ Now {\cos }^2\alpha =1-{\sin }^2\alpha \\ \Rightarrow {\cos }^2\alpha =\frac{1}{16} \\ \Rightarrow \mathrm{cos\alpha }=\frac{1}{4} \\ Given \overline{a}\times \overline{b}=2\overline{a}\times \overline{c} \\ \Rightarrow \overline{a}\times \overline{b}-\overline{a} \times 2\overline{c }=0 \\ \Rightarrow \overline{a}\times (\overline{b}-2\overline{c})=0 \\ \Rightarrow \overline{a} and (\overline{b}-2\overline{c}) are parrallel vectors \\ \Rightarrow \overline{b}-2\overline{c}=\lambda \overline{a} \\ \Rightarrow {\overline{(b}-2\overline{c})}^2={\left(\lambda \overline{a}\right)}^2 \Rightarrow {\left|\overline{b}\right|}^2+4{\left|\overline{c}\right|}^2-4\overline{b}.\overline{c}={\lambda }^2{\left|\overline{a}\right|}^2 \\ \Rightarrow 16+4-4\left(\left|\overline{b}\right|\right)\left(\left|\overline{c}\right|\right)\cos \alpha ={\lambda }^2\left(1\right) \\ \Rightarrow 20-4\left(4\right)\left(\frac{1}{4}\right)={\lambda }^2 \\ \Rightarrow 16={\lambda }^2 \\ \Rightarrow \lambda =\pm 4 \end{gathered}$$