Vector Algebra — Class 12 Maths Solution

We have, $|k\overrightarrow {\rm{a}} | < |\overrightarrow {\rm{a}} |$ and $k\overrightarrow {\rm{a}} + \frac{1}{2}\overrightarrow {\rm{a}}$

is parallel to $\overrightarrow {\rm{a}}$.

$\therefore$ $|k\vec a| < |\vec a| \Rightarrow |k||\vec a| < |\vec a|$
$\Rightarrow$ $|k| < 1 \Rightarrow - 1 < k < 1$

Also, since $k\overrightarrow {\rm{a}} + \frac{1}{2}\overrightarrow {\rm{a}}$ is parallel to $\overrightarrow {\rm{a}} ,$

then we see that at$k = \frac{{ - 1}}{2},k\overrightarrow {\rm{a}} + \frac{1}{2}\overrightarrow {\rm{a}}$

becomes a null vector and then it will not be parallel to $\overrightarrow {\rm{a}}$.

So, $k\overrightarrow {\rm{a}} + \frac{1}{2}\overrightarrow {\rm{a}}$ is parallel to $\overrightarrow {\rm{a}}$

holds true when $\left. {k \in } \right] - 1,1\left[ {k \ne \frac{{ - 1}}{2}} \right.$.