If $\vec{a}=t \vec{b}$ where $t < 0$ is a scalar, then
If $\vec{a}=t \vec{b}$ where $t < 0$ is a scalar, then
$\vec{a}, \vec{b}$ are like vectors and $|\vec{a}|>|\vec{b}|$
$\vec{a}, \vec{b}$ are unlike vectors and $|\vec{a}|>|\vec{b}|$
$\vec{a}, \vec{b}$ are like vectors and $|\vec{a}| < |\vec{b}|$
$\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ are unlike vectors and either $|\overrightarrow{\mathrm{a}}| \geq|\overrightarrow{\mathrm{b}}|$ or $|\overrightarrow{\mathrm{a}}| < |\overrightarrow{\mathrm{b}}|$
Solution
Given $\vec{a}=t \vec{b}$ where $t < 0$ is a scalar
$\therefore \mathrm{t}$ is negative.
$\Rightarrow \mathrm{t}=-\mathrm{k}$ for some $\mathrm{k} \in \mathrm{Z}^{+}$
So, we have $\vec{a}=-k \vec{b}$ ......(i)
$\therefore \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ are unlike vectors.
Taking mod on both sides of eqn. (i), we get
$
|\vec{a}|=|-k \vec{b}| \Rightarrow|\vec{a}|=|k||\vec{b}|
$
If $0 < |k| < 1 \mid$ then $|\vec{a}| < |\vec{b}|$
And if $|\mathrm{k}| \geq 1$ then $|\vec{a}| \geq|\vec{b}|$