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So I keep self learning linear algebra but problems keep arising. I'm trying to solve this particular one:


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:


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.

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Multiply your non-zero vectors $v$ by $2/\Vert v\Vert$ – David Mitra Feb 16 '12 at 2:05
up vote 2 down vote accepted

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.

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Thank you very much David. Just one more thing. Why do you say the system has two free parameters? I mean, doesn't has three including also v3? – Randolf Rincón Fadul Feb 16 '12 at 2:43
@RandolfRincón-Fadul No, just two. If you knew the values of two of the coordinates, the third would be determined, no? – David Mitra Feb 16 '12 at 2:44
Ummm now I see, the free parameters are the parameters of the general solution. Thanks again. – Randolf Rincón Fadul Feb 16 '12 at 2:48
@RandolfRincón-Fadul Or, think of it this way: The set of vevtoors orthogonal to a given vector in $\Bbb R^3$ lie in a common plane through the origin. So they span a subspace of dimension two. – David Mitra Feb 16 '12 at 2:48

Hint: construct two vectors $u$ and $w$ such that $s$, $u$, $w$ are mutually orthogonal and $\|u\| = \|w\| = 1$. The Gram-Schmidt process may help with this. Then the general solution is $v = 2 \cos(\theta) u + 2 \sin(\theta) w$ for $0 \le \theta < 2 \pi$.

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