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.

  • 2
    $\begingroup$ If A, B and C are points in 3D space then they are always on the same plane. $\endgroup$ Commented Jun 23, 2019 at 9:34
  • $\begingroup$ Can I say vector AB is vector a, vector BC is vector b and vector AC is vector c. Is it correct? $\endgroup$
    – Atoms
    Commented Jun 23, 2019 at 9:41
  • $\begingroup$ Look at my answer. $\endgroup$ Commented Jun 23, 2019 at 9:46
  • $\begingroup$ Isn't the question rather about to find $\alpha, \beta$ such that $A,B,C$ are collinear, i.e. $AB\parallel BC$? $\endgroup$
    – Berci
    Commented Jun 23, 2019 at 9:56
  • 1
    $\begingroup$ I have solved it. Thanks $\endgroup$
    – Atoms
    Commented Jun 23, 2019 at 10:55

1 Answer 1


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 $\beta$.

  • 1
    $\begingroup$ This will go a long way. Let me get back to work. Thanks $\endgroup$
    – Atoms
    Commented Jun 23, 2019 at 9:48
  • $\begingroup$ @Atoms If I answered your question, you can mark it as accepted. $\endgroup$ Commented Jun 23, 2019 at 15:31

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