Find a vector $\overrightarrow {U}$ such that the line with equation $5x+2y=0$ lies along $\overrightarrow {U}$.

Question

Note that I will use dot product to find this but I couldn't do. Can you give a hint?

Answer 1

If you just want to find an arbitrary vector such that the line lies along it, observe the origin and $(2,-5)$ are both on the line, so you can choose the vector $(2,-5).$

If you are given a vector $\vec{v}$ and want to find its component parallel to the line, by the observation above you can find its projection $\vec{v}\cdot \frac{1}{\sqrt{29}}(2,-5)$ along the vector lies on the line.

Answer 2

We can write the equation of the line $5x+2y=0$ as scalar product:

$$(5,2)\cdot(x,y)=0$$

so any vector $(x,y)$ (that also is a point of the line) along the line is normal to the vector $(5,2)$; hence any point of the line defines a vector lying along the line, for instance the vector $(-2,5)$ is along the line.