Mathematics - Vector Question with Solution | TestHub

Let $\theta$ be the angle between $\mathbf{b}$ and $\mathbf{c}$. Given, $|\mathrm{b}\times \mathrm{c}|=\sqrt{15}$ $$\begin{array}{l}\Rightarrow |\mathrm{b}\Vert \mathrm{c}|\sin \theta =\sqrt{15}\\ \Rightarrow \sin \theta =\frac{\sqrt{15}}{|\mathrm{b}|\mathrm{c}\mid }\\ \Rightarrow \sin \theta =\sqrt{\frac{15}{4\times 1}}=\frac{\sqrt{15}}{4}\\ \therefore \cos \theta =\sqrt{1-\frac{15}{16}}=\frac{1}{\sqrt{16}}=\frac{1}{4}\end{array}$$ Now given, $\mathrm{b}-2\mathrm{c}=\lambda \mathrm{a}\Rightarrow \left|\mathrm{b}-2{\mathrm{c}}^2\right|=\mid \lambda {\mathrm{a}}^2$ $$\Rightarrow |\mathrm{b}{|}^2+4|\mathrm{c}|-4(\mathrm{b}\cdot \mathrm{c})={\lambda }^2|\mathrm{a}{|}^2$$ $$\begin{array}{l}\Rightarrow 16+4-4|\mathrm{b}\Vert \mathrm{c}|\cos \theta ={\lambda }^2\\ \Rightarrow 20-16\cos \theta ={\lambda }^2\\ \Rightarrow 20-4={\lambda }^2\Rightarrow {\lambda }^2=16\left(\therefore \cos \theta =\frac{1}{4}\right)\\ \Rightarrow \lambda =\pm 4\end{array}$$