Find a normal vector using cross product

Question

Am I doing this right?

If a plane contains the points A = (2, 2, 3), B = (1, 0, 1) and C = (−1, 3, 4), find a normal vector by using cross product.

• 1) First I find a cross product for AB
• 2) Find a cross product for BC
• 3) Then find a cross product for AB and BC

Is this correct way to do this?

Answer 1

I don't know what you mean with your first two steps. All you have to do is to compute $AB\times AC$.

Answer 2

Consider the three points $A = (2, 2, 3), B = (1, 0, 1)$, and $C = (−1, 3, 4)$, which lie on a plane. Then the vectors $$ AB=\mathcal{O}B-\mathcal{O}A=\langle 1, 0, 1 \rangle - \langle 2, 2, 3\rangle = \langle 1-2, 0-2, 1-3 \rangle = \langle -1, -2, -2\rangle $$ and $$ BC=\mathcal{O}C-\mathcal{O}B= \langle −1, 3, 4 \rangle - \langle 1, 0, 1 \rangle = \langle −1-1, 3-0, 4-1 \rangle = \langle -2,3,3\rangle $$ are two vectors that are parallel to the plane, where $\mathcal{O}A$ is the position vector from the origin $(0,0,0)$ to the point $A$ (similar with $\mathcal{O}B$ and $\mathcal{O}C$). Note that $AB$ and $BC$ are non-parallel.

Then the cross product of $AB$ and $BC$ is: $$ \boxed{AB\times BC = \langle 0,7,-7 \rangle}, $$ which is perpendicular to the two vectors $AB$ and $BC$. So it is normal to the plane.