Find orthogonal vector that is in plane

Question

The equation of the plane is $\pi: -x + 2y - z = 2$. I need to find a support vector that is orthogonal to the plane and is also in the plane itself.

To me, that sounds contradictory, because if the vector is orthogonal, it would be parallel to the normal vector $(-1, 2, -1)$, but then it would not be in the plane and if the vector would be in the plane, then how can it be orthogonal to the plane?

Does anyone have a clue?

Answer 1

Your intuition is sort of right: you can't have a direction vector that is simultaneously orthogonal to a plane, but at the same time be a direction between vectors in the plane. It's not quite what they're asking though.

Note that the plane doesn't pass through the origin. When they say they want a vector in the plane, they're looking for a point in the plane, or if you like, a vector from the origin whose tip lies within the plane. Basically, we're looking for the unique point in the plane that is closest to the origin, so that the line between the origin and this point is perpendicular to the plane.

Normal vectors need to be parallel to $(-1, 2, -1)$, as you pointed out. So, the vector (or point) we're looking for takes the form $(-k, 2k, -k)$ for some $k \in \Bbb{R}$, but at the same time satisfies the equation $-x + 2y - z = 2$. Let's use this information to solve for $k$: $$-(-k) + 2(2k) - (-k) = 2 \iff 6k = 2 \iff k = \frac{1}{3}.$$ So, our vector comes to be $$\left(-\frac{1}{3}, \frac{2}{3}, -\frac{1}{3}\right).$$ This is the unique vector that is both orthogonal to the plane (i.e. the vector from $(0, 0, 0)$ to that point is orthogonal to the plane), and lies in the plane (i.e. the endpoint lies in the plane).

Answer 2

The problem is that in vector geometry vectors play both the role of directions and of points. Thinking of vectors as directed entities, points are represented by vectors rooted in 0.

What you are supposed to find is a vector, which is parallel to the normal vector (if rooted at a point on the plane), but if rooted in 0 has defines a point on the plane. In 2 dimensions you can depict it like this:

But how do we compute such a vector $\vec p$? Well we know two things:

1. $\vec p$ is a multiple of the normal vector $\vec n$ ie $\vec p = t \vec n$ for some $t \in \Bbb R$, because this is what parallel means.
2. $\vec p$ defines a point on the plane, ie satisfies the equation $\vec p \cdot \vec n = \alpha$ defining your plane.

In your case this means $$(p_1,p_2,p_3) = (-t, 2t, -t)$$ and $$-p_1 + 2p_2 -p_3 = 2$$ Thus the only thing to do is to solve for $t$.

Edit I am sorry for this image not being cropped, but somehow the StackExchange App only allows me to upload photos directly from the camera. I will clean this up.