How to show that there are infinitely many vectors in R3 with Euclidean norm 1

Question

There is a related question: "Finding vectors in Rn with Euclidean norm 1", although, I still cant seem to find a way to the answer.

The question asks: Show that there are infinitely many vectors in R3 with Euclidean norm 1 whose Euclidean inner product with < −1, 3,−5 > is zero.

I know that ||x, y, z|| = 1, which is the square root of x^2+y^2+x^2.

The dot product of the two is = 0 which leads to:(x, y, z).<-1, 3, -5> = 0 and I get -x+3y-5z = 0.

I am not sure where to go from here in order to get the solutions. I could make z = 0 which leads to 3y = x but this doesn't help me.

Could someone please give me some tips on where to go from here?

Answer 1

Alright let's take a general vector: $$ \vec v=(x,y,z) $$ now let's make it unitary, dividing it by it's norm: $$ \hat v=\left(\frac{x}{\sqrt{x^2+y^2+z^2}},\frac{y}{\sqrt{x^2+y^2+z^2}},\frac{z}{\sqrt{x^2+y^2+z^2}}\right) $$ now let's make it orthogonal to your other vector $\vec a=(-1,3,5)$ the ortogonality condition is ofcourse: $$ \vec a \cdot \hat v=0 \iff -x+3y-5z=0 $$ solve it for x: $$ x=3y-5z $$ now substitute this result in $\hat v$: $$ \hat v=\left(\frac{3y-5z}{\sqrt{y^2 + (3 y - 5 z)^2 + z^2}},\frac{y}{\sqrt{y^2 + (3 y - 5 z)^2 + z^2}},\frac{z}{\sqrt{y^2 + (3 y - 5 z)^2 + z^2}}\right) $$ this is a unitary vector orthogonal to $\vec a$. Since it still depends on 2 variables you can conclude that there are infinitely many vectors with those 2 properties.

You could have answered it without any calculation just by imagining that on the tip on vector $\vec a$ you can put another vector orthogonal to it of length 1 and you can rotate that vector using $\vec a$ as an axis of rotation still keeping it of length 1 and orthogonal in infinitely many ways.

moreover one could also demonstrate that it really only depends on one parameter. using the substitution: $$ \begin{cases} y=r \cos t\\ z=r \sin t \end{cases} $$

do the substitution and you'll see that the $r$ simplifies and the resulting vector will only depend on $t$.

Answer 2

Fix $x$ (say, let $x = 0.5$). Now, for every value $y \in [0, 1]$, the equation $||(x, y, z)||=1$ determines $z$ uniquely. And there are infinitely many such $y$'s.

Answer 3

You want $x,y,z$ that solve $$\begin{cases}-x+3y-5z=0\\ \sqrt{x^2+y^2+z^2}=1\end{cases}$$

So, use the first one to write a variable (say, $x$) in terms of the other two, substitute in the second equation and see if you get something that has clearly infinitely many solutions in $y$ and $z$ (perhaps because of a previous geometric knowledge)

Answer 4

The set of all vectors in $R^3$ of length 1 is equivalent to the set of all points on the unit sphere. Those can be written, in spherical coordinates, $\left(\rho, \theta, \phi\right)$ with $\rho= 1$, $\theta$ from 0 to $2\pi$, and $\phi$ from 0 to $\pi$.

Answer 5

Here’s a more conceptual, less computational, argument:

The condition that a vector $(x,y,z)$ be perpendicular to the given one is $-x+3y-5z=0$. The vectors $v_1=(3,1,0)$ and $v_2=(0,5,3)$ are two such linearly independent ones, so that the infinitely many vectors $v_1+\lambda v_2$ all point in different directions, but are all perpendicular to $(-1,3,-5)$. Now just divide each of these by its magnitude to get points on the unit sphere.