Let us assume that $\vec{a}$ is a vector which is having a magnitude of $4.5$ unit due north. What can be the magnitude of the vector $-4\vec{a}$?

Hint: A vector quantity is a quantity which is having a magnitude as well as direction. Whereas the scalar quantities are the physical quantities which are having only the magnitude of the quantity. This quantity will not specify the direction of the quantity. Multiply the vector with a negative of four. The negative sign will give us the direction of the vector. This all will help you in solving this question.

Complete step by step answer:
It has been mentioned in the question that the vector $\vec{a}$ is having a magnitude as,
$\left| {\vec{a}} \right|=4.5$
Therefore the magnitude of the vector $-4\vec{a}$ can be found by multiplying the magnitude of the vector with the coefficient of the vector. This can be written as,
$-4\left| {\vec{a}} \right|=4.5\times -4=-18$
As the vector $\vec{a}$ is having a direction towards the north side, the negative of the vector will be in the direction opposite to this. This will be true if the negative has been given for the multiple of the same vector. Therefore the direction of the vector $-4\vec{a}$ will be in the opposite direction. That is in the southward direction.

Note: Distance is mentioned as a scalar quantity while displacement is found to be a vector quantity. The vectors can be multiplied with a scalar without changing its direction. The parallel vectors also will have the same direction. The dot product is a kind of taking the product of two vectors which will give the output as a scalar. The cross product will be giving an output as the vector itself. A scalar quantity cannot be used to perform a cross product.