I bumbed into this question and I have been trying to solve it but I got stuck.
Determine $β$ and $α$ by using vectors such that $A$, $B$ and $C$ lie in the same plane, given that vector $\overrightarrow{AB} = -4\vec\imath - \vec\jmath - 2\vec k$ and vector $\overrightarrow{BC} = 4\vec\imath + (β + 3)\vec\jmath + (α - 6)\vec k$.
I know that to prove they are on the same plane: $$\vec a \cdot (\vec b \times \vec c) = 0$$
But how to break the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ into component vectors $\vec a$, $\vec b$ and $\vec c$ is what I don't even have a clue of. Please I need help. I'm just so dumb right now.
Any three points in 3D space will inevitably be on the same plane.
To show this, you can take:
$$\vec a = \overrightarrow{AB}$$ $$\vec b = \overrightarrow{BC}$$ $$\vec c = \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$$
and you will get that $\vec a \cdot (\vec b \times \vec c) = 0$, regardless of $\alpha$ and $\beta$.