I'm stuck on this question that asks to find all values for a and b such that two vectors u = (-1, 2, a-2), and v = (b, 4, -2) are not perpendicular.
I know that if the dot product of u.v = 0 then the two vectors are perpendicular. I applied the dot product and got:
-b + +12 -2a.
And i rearranged the equation to find a and b, however when i put them back into the equation it gets messy. Any advice on how to tack this problem, thanks.
We have:
$$\begin{align} \vec{u} \text{ and } \vec{v} \text{ perpendicular} & \iff \vec{u} \cdot \vec{v} = 0, \\ & \iff -b + 12 - 2a = 0, \\ & \iff b + 2a = 12. \end{align}$$
Just as what you got. You can check for example, choosing $b = 0$, we get $2a = 12$ and hence, $a = 6$ and plugging this back into your original vector, we have that they are indeed perpendicular.
What you wanted was not perpendicular. So just negate our result above to have:
$$\vec{u} \text{ and } \vec{v} \text{ not perpendicular} \iff b + 2a \neq 12.$$
So if you fix your favorite $a$, then taking $b \neq 12 - 2a$ i.e. any $b$ not of this form renders $\vec{u}$ and $\vec{v}$ not perpendicular.
For example: Fix $a = 0$. We don't want $b = 12 - 2a = 12$. So take, say, $b = 1$, then
$$\vec{u} \cdot \vec{v} = (-1, 2, -2) \cdot (1, 4, -2) \neq 0.$$
But then since $a$ can be any real number, there are infinitely many solutions for which $\vec{u} \cdot \vec{v} \neq 0$ and they admit the form
$$b +2a \neq 12$$
or equivalently,
$$a \in \mathbb{R} \text{ and } b \neq 12 - 2a.$$