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?
Best Answer
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 $t$.