Please explain why there is a whole circle of unit vectors in this scenario.

Question

For a given vector

$w =(2,1,2)$

I am to find out the perpendicular unit vector $u$ to this vector $w$.

One such unit vector perpendicular to this vector is

$ U =(1,-2,0)/sqrt(5)$ which I found out as well.

However, the author mentions as a comment:

There is a whole plane of vectors perpendicular to $U$, and a whole circle of unit vectors in that plane.

This is the part I am confused about. In my mind, the given vector is an arrow in 3D space, then from any point in that arrow we can draw a perpendicular in any direction in 3D space, then how can all the perpendicular vectors lie in a plane?

Moreover, if this is true, then all the unit perpendicular vectors would form a cylinder with the vector $w$ as the axis.

What am I missing?

I am self studying his book and kind of new to Linear Algebra, so please explain. Thanks.

Answer 1

Vectors, in this context, are always taken to have their "tail" at the origin. If you could have vectors with any basepoint, then indeed, the set of "arrowheads" of the vector(s) in question would always fill up the whole space, even if you're only talking about one vector.

This convention is taken precisely so that we can draw an equivalence between vectors and points in the space: every point in space is determined by a unique vector based at the origin, and every vector based at the origin has its "tip" at exactly one point in space.

Answer 2

Note that

• $u= (1,-2,0)$ and $v=(2,1,-\frac{5}{2})$ are orthogonal to each other and orthogonal to $w = (2,1,2)$.
• So, $u$ and $v$ span a whole plane.
• If you normalize $u$ and $v$ to $e_u =\frac{1}{|u|}u$ and $e_v =\frac{1}{|v|}v$, then consider $c_t = \cos t \cdot e_u + \sin t \cdot e_v$: $$|c_t|^2 = \cos^2 t + \sin^2 t = 1 \mbox{ and } c_t \perp w$$ It is not difficult to see that $c_t$ forms a "circle of unit vectors" while $t$ runs through $[0,2\pi ]$.