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?

  • 3
    $\begingroup$ Why would you think that vector AC is in the plane? Why would point C have anything meaningful to do with plane OAB? $\endgroup$
    – Deepak
    Apr 14, 2022 at 10:47

1 Answer 1


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



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