1. ## parallel vector problem

given $\displaystyle \vec{a}=\begin{pmatrix} h-2 \\ 5 \end{pmatrix}$ and $\displaystyle \vec{b}=\begin{pmatrix} 6 \\ 2h \end{pmatrix}$.Find the value of $\displaystyle h$ such that the vector $\displaystyle \vec{a}$ is parallel to the vector $\displaystyle \vec{b}$

2. If the vector $\displaystyle \mathbf{a}$ is parallel to vector $\displaystyle \mathbf{b}$, then they have the same direction, and one will be the scalar multiple of another.

Therefore

$\displaystyle \mathbf{a} = k\mathbf{b}$

$\displaystyle \left[\begin{matrix}h-2\\5\end{matrix}\right] = k\left[\begin{matrix}6\\2h\end{matrix}\right]$

$\displaystyle \left[\begin{matrix}h-2\\5\end{matrix}\right] = \left[\begin{matrix}6k\\2kh\end{matrix}\right]$.

This means you end up with the system of equations

$\displaystyle h - 2 = 6k$
$\displaystyle 5 = 2kh$

Rearranging the second gives

$\displaystyle k = \frac{5}{2h}$

and substituting into the first gives

$\displaystyle h - 2 = 6\left(\frac{5}{2h}\right)$

$\displaystyle h - 2 = \frac{15}{h}$

$\displaystyle h^2 - 2h = 15$

$\displaystyle h^2 - 2h - 15 = 0$

$\displaystyle (h - 5)(h + 3) = 0$

$\displaystyle h = 5$ or $\displaystyle h = -3$.