Question

If $\overrightarrow{a}=x\hat{i}+(x-1)\hat{j}+\hat{k}$ and $\overrightarrow{b}=(x+1)\hat{i}+\hat{j}+a\hat{k}$ make an acute angle $\forall x\in R$, then find the values of a.

Answer 1

$\begin{array}{c}Given,\\ \overrightarrow{a}=x\hat{i}+(x-1)\hat{j}+\hat{k}\\ \overrightarrow{b}=(x+1)\hat{i}+\hat{j}+a\hat{k}\\ Now,\\ ifwetakedot.dotform:\\ \overrightarrow{a}.\overrightarrow{b}=(x\hat{i}+(x-1)\hat{j}+\hat{k}).\left(\right(x+1)\hat{i}+\hat{j}+a\hat{k})\\ \Rightarrow x(x+1)+(x-1)+a\\ \Rightarrow x^2+2x+a-1\\ conditiongiven,acuteangle:\overrightarrow{a}.\overrightarrow{b}>0\forall x\in R\\ x^2+2x+a-1>0|\begin{array}{c}a{x}^2+bx+c>0\\ D<0\\ d=b^2-4ac\end{array}\\ \\ (b=2,a=1,c=a-1)\\ \Rightarrow 4-4(a-1)<0\\ \therefore a>2\\ So,wecansaythata>2andthisisthecorrectanswer.\\ \\ \end{array}$