# Unit vectors that are orthogonal to vectors

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 $\overrightarrow{b}$ ??

• No, $(0,0,0)$ is never a unit vector. A zero cross product means that $\vec a$ and $\vec b$ are parallel (and indeed $\vec b=-2\vec a$), so anything that is perpendicular to one will be perpendicular to the other too. – Henning Makholm Feb 20 '15 at 0:39
• Ok... How can we find ALL the unit vectors that are orthogonal to the vectors?? @HenningMakholm – Mary Star Feb 20 '15 at 1:32

$\vec{a}=(2, -4, 3), \vec{b}=(-4, 8, -6)$ 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 $$\vec r(s,t) = (3,0,-2)s + (0,3,4)t$$

• I see!! Thank you!!! :-) – Mary Star Mar 1 '15 at 3:27
• I read again your answer and I got stuck... The space of vectors that are orthogonal to both of the vectors $\vec a$ and $\vec b$ is the space of vectors orthogonal to the line that passes through both of them, because this line is parallel to both vectors, right? Why is this space a plane? Why is the vector equation equal to $\vec r(s,t) = \vec us + \vec vt + \vec c$ ? – Mary Star Feb 17 '16 at 22:34
• $\vec b = -2\vec a$ so $\vec b$ and $\vec a$ are parallel to each other. Thus any vector perpendicular to one will be perpendicular to the other. This means that we really one need to consider the set of vectors orthogonal to one of those two vectors. That set of vectors has a special name -- the orthogonal complement of the line $\operatorname{span}(\vec a)$ (or $\vec b$ since they're parallel). Go back and look through your linear algebra text (or Google it) to see some of the properties of the orthogonal complement. Most importantly are the properties that the orthogonal... – user137731 Feb 18 '16 at 1:29
• complement of a subspace is a subspace and that if $W^{\perp}$ is the orthogonal complement of $W$ in the vector space $V$ then $W+W^{\perp} = V$ (assuming finite dimension). Therefore the space orthogonal to $\vec a$ (and $\vec b$) will be a vector space of dimension $2$ -- i.e. a plane. The equation $\vec r(s,t) = \vec us + \vec vt +\vec c$ is the standard (vector) equation of a plane. – user137731 Feb 18 '16 at 1:31
• Now, something I forgot to mention when I wrote this answer is that because you're just looking for the unit length vectors orthogonal to your pair of vectors, you'll have to restrict the plane to only the vectors of unit length -- i.e. you'll have to specify the unit circle of the plane $\vec r = \vec r(s,t)$. – user137731 Feb 18 '16 at 1:33

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 $\mathbf{a}$.

• How can we find ALL the unit vectors that are orthogonal to the vectors?? – Mary Star Feb 20 '15 at 1:35

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

• How can we find all the unit vectors that are orthogonal to the vectors?? – Mary Star Feb 20 '15 at 1:33
• As Abel notes, $\mathbf v = \frac{1}{\sqrt{13}}(3, 0, 2)$ is one such vector. Taking the cross product with $(2, -4, 3)$ and normalizing gives $\mathbf w = \frac{1}{\sqrt{233}} (8, -5, -12)$, and $\mathbf v$ and $\mathbf w$ are perpendicular, just like the unit vectors on the $x$- and $y$-axes of the plane. If you form all cosine/sine combinations of these (i.e., $\cos(t) \mathbf v + \sin(t) \mathbf w$), you get all possible unit vectors perpendicular to the ones given in your problem. – John Hughes Feb 20 '15 at 1:47
• Ok... Thanks!!! :-) – Mary Star Mar 1 '15 at 3:28

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 $$\frac{(2y-\frac{3}{2}z, y, z)}{\sqrt{(2y-\frac{3}{2}z)^2 + y^2 +z^2}}$$

• In that way do we find ALL the unit vectors that are orthogonal to the vectors?? – Mary Star Feb 20 '15 at 1:33
• In this case the cross product is zero, because the two vectors are parallel. – Henning Makholm Feb 20 '15 at 1:44
• The cross product is the zero vector, not the number $0$. – MarkG Feb 20 '15 at 2:09
• Ok... Thank you!!! :-) – Mary Star Mar 1 '15 at 3:28

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.

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 $$u\cos t + v \sin t, t \text{ any real number.}$$
• you do find all vectors orthogonal as the linear combinations of the two vectors in the post. they are linearly independent, so it is a basis for the space orthogonal to the line $(2t, -4t, 3t)$ – abel Feb 20 '15 at 1:41