# Vectors on the same plane

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.

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

$$\vec a = \overrightarrow{AB}$$ $$\vec b = \overrightarrow{BC}$$ $$\vec c = \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$$
and you will get that $$\vec a \cdot (\vec b \times \vec c) = 0$$, regardless of $$\alpha$$ and $$\beta$$.