The points A,B,C have position vectors; 2i+2j, -j+k and 2i+j-7k respectively, relative to the origin. Find the normal vector to the plane OAB.
I know that $$\vec {OA}$$ and $$\vec {OB}$$ are in the plane and the the cross product of them would give me the normal vector as i-j-k. However, I also know that $$\vec {AB}$$ and $$\vec {AC}$$ are also in the plane and when taking their cross product I get 22i-14j+2k.Are both answers correct? Shouldn’t both be scalar multiples of each other?
Best Answer
The normal vector of the plane defined by three arbitrary points is
$$\vec{n}_{ABC} = \vec{A} \times \vec{B} + \vec{B} \times \vec{C} + \vec{C} \times \vec{A} \tag{1}$$
A feature of this is that all points on the plane share the same dot product with the normal. So a point $\vec{P}$ belongs to the plane if $(\vec{n}_{ABC} \cdot \vec{P}) = (\vec{n}_{ABC} \cdot \vec{A})$ for example.
This means that if the plane ABC goes through the origin, then $(\vec{n} \cdot \vec{A}) = 0$ since point $\vec{P}$ is the origin.
But this is not the case here.
For this example $(\vec{n}_{ABC} \cdot \vec{A}) = 16 \neq 0$.
In summary, the normal vector of OAB is
$$ \vec{n}_{OAB} = \vec{O} \times \vec{A} + \vec{A} \times \vec{B} + \vec{B} \times \vec{O} = \vec{A} \times \vec{B} = 2(\hat{i} - \hat{j} - \hat{k})$$
and the normal vector of ABC is
$$ \vec{n}_{ABC} = \vec{A} \times \vec{B} + \vec{B} \times \vec{C} + \vec{C} \times \vec{A} = (22\hat{i} - 14\hat{j} - 2\hat{k})$$
which are clearly not in the same direction. If they were parallel, then their cross-product would be zero, but it is not.
If you check GeoGebra, clearly the ABC plane does not go through the origin