So I keep self learning linear algebra but problems keep arising. I'm trying to solve this particular one:
Problem
Find all the vectors $v$ on $\mathbb{R}^3$ wich are orthogonal to the vector $s = [1,2,-1]$ and have norm (magnitude) $= 2$.
I try to solve the problem but I'm not sure how to express the solution, this is my solution:
Solution
I model $v$ as $[v_1,v_2,v_3]$ and given that $v$ is orthogonal to $s$ then the dot product of them will be $0$ and so:
$v_1+2v_2-v_3 = 0$; $v_1 = v_3-2v_2$; $v_2 = (1/2)(v_3-v_1)$ and $v_3 = v_1+2v_2$
But that only solves the first part of the problem. For the second I now the norm is 2 so:
$\|v\| = 2 = \sqrt{v_1^2 + v_2^2 + v_3^2} $
I'm stuck here, I don't know where to follow or if just defining that both conditions must be satisfied is enough. Thanks in advance for any guidance.
Best Answer
The orthogonal vectors $(v_1,v_2,v_3)$ have the form $$ v_1+2v_2-v_3=0. $$ This system has two free parameters, $a$ and $b$ say. Set $v_1=a$ and $v_2=b$. Then $v_3=a+2b$. The general orthogonal vector is then $$ {\bf x}=(a,b,a+2b). $$ where $a$ and $b$ are any scalars.
The norm of $\bf x$ is $$\tag{1}c=\sqrt{a^2+b^2+a^2+4ab+4b^2 } = \sqrt{2a^2+5b^2 +4ab }$$
Recalling that $\Vert \alpha{\bf x} \Vert=|\alpha|\Vert {\bf x} \Vert$, the required vectors are
$$ {2\over c}(a,b,a+2b) , $$ where $c$ is as in $(1)$ and not both $a$ and $b$ are zero.