Question

The values of k for which $\left|k\overrightarrow{a}\right|<\left|\overrightarrow{a}\right|\mathrm{and}k\overrightarrow{a}+\frac{1}{2}\overrightarrow{a}$ is parallel to $\overrightarrow{a}$ holds true, are _______________.

Answer 1

Given:
$\left|k\overrightarrow{a}\right|<\left|\overrightarrow{a}\right|\mathrm{and}k\overrightarrow{a}+\frac{1}{2}\overrightarrow{a}$ is parallel to $\overrightarrow{a}$


$$\begin{gathered} \left|k\overrightarrow{a}\right|<\left|\overrightarrow{a}\right| \\ \Rightarrow \left|k\right|\left|\overrightarrow{a}\right|<\left|\overrightarrow{a}\right| \\ \Rightarrow \left|k\right|<1 \\ \Rightarrow -1<k<1...\left(1\right) \\ \mathrm{Also}, \\ k\overrightarrow{a}+\frac{1}{2}\overrightarrow{a}\mathrm{is}\mathrm{parallel}\mathrm{to}\overrightarrow{a} \\ \mathrm{But}\mathrm{at}k=-\frac{1}{2}, \\ k\overrightarrow{a}+\frac{1}{2}\overrightarrow{a}=-\frac{1}{2}\overrightarrow{a}+\frac{1}{2}\overrightarrow{a}=\overrightarrow{0} \\ \mathrm{Thus}, \\ k\ne -\frac{1}{2}...\left(2\right) \end{gathered}$$


Hence, the values of k for which $\left|k\overrightarrow{a}\right|<\left|\overrightarrow{a}\right|\mathrm{and}k\overrightarrow{a}+\frac{1}{2}\overrightarrow{a}$ is parallel to $\overrightarrow{a}$ holds true, are $\underline{\left(-1,1\right)~\left\{-\frac{1}{2}\right\}}$.