Unit vectors that are orthogonal to vectors

Question

I have to find all the unit vectors that are orthogonal to the vectors $\overrightarrow{a}=(2, -4, 3), \overrightarrow{b}=(-4, 8, -6)$ .

I calculated that the cross product $\overrightarrow{a} \times \overrightarrow{b}=0$.

Does this mean that the vector $(0, 0, 0)$ is a unit vector that is perpendicular to $\overrightarrow{a}$ and $\overrightarrow{b}$ ??

Answer 1

$\vec{a}=(2, -4, 3), \vec{b}=(-4, 8, -6)$ are elements of $\Bbb R^3$. Notice that $\vec b = -2\vec a$, thus the two vectors are collinear. So the space of vectors that are orthogonal the both of these vectors will just be the space of vectors orthogonal to the line that passes through both of them. Can you see that this space will be a plane?

So you just need to specify a plane with a vector equation. The vector equation of a plane is $\vec r(s,t) = \vec us + \vec vt + \vec c$.

We just need to find any two non-collinear vectors orthogonal to $\vec a$ or $\vec b$ (any we find orthogonal to one will automatically be orthogonal to the other).

So we need $(x,y,z) \cdot (2,-4,3) = 2x-4y+3z=0$. Being a linear equation in three variables, we should just be able to choose two of the variables and solve for the last (though don't choose them both zero or you'll just end up with the zero vector which is collinear with every other vector). Let's choose $x=3$ and $y=0$. Plugging in, we see that $z=-2$. So one vector orthogonal to $(2,-4,3)$ is $(3,0,-2)$. Now let's choose $x=0$ and $y=3$ (You could choose the the two numbers to be whatever you like, but notice I chose them so that I'd get integer solutions because no one likes unnecessary fractions). Then $z=4$. So another vector orthogonal to $(2,-4,3)$ is $(0,3,4)$. Notice that $(3,0,-2)$ and $(0,3,4)$ are not collinear (they are not scalar multiples of each other).

So let $\vec u=(3,0,-2)$ and $\vec v=(0,3,4)$. Then to find $\vec c$ we need any point on the line $\operatorname{span}(2,-4,3)$. $(0,0,0)$ is in that span. So let $\vec c = \vec 0$.

Then the equation representing our plane -- and thus every vector orthogonal to $\vec a$ and $\vec b$ -- is just $$\vec r(s,t) = (3,0,-2)s + (0,3,4)t$$

Answer 2

No, $(0, 0, 0)$ is not a unit vector (it has $0$ length).

Note that $\mathbf{a}$ and $\mathbf{b}$ are parallel, so the vectors perpendicular to $\mathbf{a}$ and $\mathbf{b}$ are in the plane perpendicular to $\mathbf{a}$.

Answer 3

Nope. In this case, it means that $a$ and $b$ are parallel vectors. :)

Answer 4

As everyone has pointed out, $\mathbf{a}$ and $\mathbf{b}$ are parallel, so we only need to use one of them. Using $\mathbf{a}$, the vecotrs orthogonal to it are all vectors $(x,y,z)$ such that $2x-4y+3z=0$. We can eliminate one of these variables, say $x$, so all the orthogonal vectors are those of the form $(2y-\frac{3}{2}z, y, z)$.

To make them unit vectors, you can divide by their length, which is $\sqrt{(2y-\frac{3}{2}z)^2 + y^2 +z^2}$.

So all the orthogonal unit vectors are the vectors of the form $$\frac{(2y-\frac{3}{2}z, y, z)}{\sqrt{(2y-\frac{3}{2}z)^2 + y^2 +z^2}}$$

Answer 5

suppose $(x, y, z)$ is orthogonal to $(2,-4,3).$ then the dot product $2x-4y+3z = 0$ two nonzero solutions are $(3, 0, -2)$ and $(2,1,0).$ any combination of the two vectors is also orthogonal to $(2,-4,3).$ to make a nonzero vector unit length, just divide by its length. for example the length of $(3,0, -2)$ is $\sqrt{3^2 + (-2)^2} = \sqrt {13}$ so that $(3/\sqrt{13}, 0, -2/\sqrt{13})$ is a unit vector.

$\bf edit:$

to find all the unit vectors, we need to find an orthogonal basis first. if you project $(3,0,-2)$ onto $(2,1,0)$ you get $(4,2,0).$ that is we have split $(3,0,-2) = (4,2,0) + (-1,2,-2).$ therefore an orthogonal basis is $\{(2,1,0), (-1,2,-2)\}.$ we make them orthonormal by dividing by their lengths so that $$\{u = (2/\sqrt 5, 1/\sqrt 5), v = (-1/3, 2/3, 2/3)\}$$ finally, an arbitrary unit orthogoanal vector is $$u\cos t + v \sin t, t \text{ any real number.}$$

Answer 6

As another approach, solve the system $<a,x>=0,$ using the standard Euclidean inner product i.e $2x_1 -4x_2+3x_3=0.$ This equation has two free variables, so it defines a plane, which is exactly what you would expect working in $\mathbb R^3.$ In case you don't know, a free variable means you can choose any value for it. So make two separate sets of choices, and the span of the two solutions is the plane you seek.

For more information, I'd consult an introductory linear algebra text.