The least positive integral value of $\alpha$, for which the angle between the vectors $\alpha \hat{

Question

The least positive integral value of $\alpha$, for which the angle between the vectors $\alpha \hat{i}-2 \hat{j}+2 \hat{k}$ and $\alpha \hat{i}+2 \alpha \hat{j}-2 \hat{k}$ is acute, is $\_\_\_\_$ .\

Professor PC Thomas Classes · Answer

(5) : $\cos 	heta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$ As $	heta$ is acute, so $\cos 	heta>0$.\ $\frac{(\alpha \hat{i}-2 \hat{j}+2 \hat{k}) \cdot(\alpha \hat{i}+2 \alpha \hat{j}-2 \hat{k})}{\sqrt{\alpha^{2}+(-2)^{2}+(2)^{2}} 	imes \sqrt{\alpha^{2}+(2 \alpha)^{2}+(-2)^{2}}}>0$\ $\Rightarrow \frac{\alpha^{2}-4 \alpha-4}{\sqrt{\alpha^{2}+8} 	imes \sqrt{5 \alpha^{2}+4}}>0$\ $\Rightarrow \alpha^{2}-4 \alpha-4>0 \Rightarrow \alpha^{2}-4 \alpha+4-4-4>0$\ $\Rightarrow \quad(\alpha-2)^{2}-8>0 \Rightarrow(\alpha-2)^{2}>8$\ $\Rightarrow \alpha-2>2 \sqrt{2}$ or $\alpha-2<-2 \sqrt{2}$\ (Neglect)\ $\Rightarrow \quad \alpha>2+2 \sqrt{2}=4.83$\ ⇒ Least positive integral value of $\alpha=5$\