The values of $x$ for which the angle between the vectors $x^2 \hat{i}+2 x \hat{j}+\hat{k}$ and $\hat{i}-2…
The values of $x$ for which the angle between the vectors $x^2 \hat{i}+2 x \hat{j}+\hat{k}$ and $\hat{i}-2 \hat{j}+x \hat{k}$ is obtuse, lie in the interval
$(-\infty, 0) \cup(3, \infty)$
$(0,3)$
$[0,3]$
$(-\infty, 0) \cup[3, \infty)$
Solution
Angle between $x^2 \hat{i}+2 x \hat{j}+\hat{k}$ and $\hat{i}-2 \hat{j}+x \hat{k}$ is
$\begin{aligned}
& \cos \theta=\frac{x^2-4 x+x}{\sqrt{\left(x^4+4 x+1\right)\left(1+4+x^2\right)}} \\
& \Rightarrow \cos \theta=\frac{x^2-3 x}{\sqrt{\left(x^4+4 x+1\right)\left(5+x^2\right)}}
\end{aligned}$ Given that angle between these is obtuse
$\therefore \cos \theta \text { is }-\mathrm{ve} \Rightarrow x^2-3 x \lt 0 \Rightarrow x \in(0,3)$
