[Math] Find nonzero vectors $\mathbf{v}$ and $\mathbf{w}$ that are perpendicular to (1, 0, 1) and to each other.

linear algebravector-spaces

I know that this question has been asked before but was closed due to a lack of context and I could not understand the answer provided. Currently I'm self-relearning linear algebra and this is Problem 13 of Chapter 1.2 of Gilber Strang's Introduction to Linear Algebra. I have been unable to solve this problem given the answer available at MIT's Linear Algebra site in the Math domain.

So far what I've done is create a system of equations which consist of:

$$\begin{cases}
\mathbf{v} \cdot (1,0,1)=0 \\
\mathbf{w} \cdot (1,0,1)=0 \\
\mathbf{v} \cdot \mathbf{w}=0
\end{cases}
$$

Which after solving resulted in the following conclusions: I've assumed that this is all occuring in $\mathbb{R³}$.

$v_1$ = $-v_3$ for the first equation in the system.

$w_1$ = $-w_3$ for the second equation in the system.

and finally after replacing the results of the first and second equations I get this third equation:

$\frac{(v_2w_2)}{-2}$ = $v_3w_3$

Now, I believe that something's off with what I've done. If I grab the solution in the book something looks extremely off. This is what the solution says:

The plane perpendicular to (1, 0, 1) contains all vectors (c, d, −c).
In that plane, v = (1, 0, −1) and w = (0, 1, 0) are perpendicular.

I think my solution is far from the one concluded in the book. I'd love to know what is wrong with my approach to see if I can grasp the concept better.

Best Answer

There is nothing wrong with your solution, but you have to take some decisions because the solution is not unique. For instance, set $w_3=0$ and $v_3=1$, and you get the result.

However, the point of linear algebra is dealing with abstract notions, such as subspaces, rather than concentrating on coordinates that depend on the choice of a particular basis.

Then, the rationale that (I guess) the book is expecting is something like that:

  1. Find the orthogonal subspace $V$ to the vector $x=(1,0,1)$, i.e. all vectors that are orthogonal to $x$. The main two forms to express a subspace are either writing the Cartesian equations or providing a basis.
  2. Choose two orthogonal vectors from $V$, which can belong to a basis, if you have one. Otherwise, in general, you should solve the system of Cartesian equations into parametric equations that include a system of generators.

Now you can do the particular problem in $\mathbb{R}^3$, where it turns out that computations are very easy:

  1. The orthogonal subspace $V$ has Cartesian equation $x_1+x_3=0$, i.e. it contains all vectors with $x_1=-x_3$.
  2. Choose an orthogonal basis from $V$, which is a plane, thus bidimensional. For instance, since $x_1=-x_3$ tells nothing about $x_2$, you can choose $v=(1,0,-1)$. In this particular case, just by guessing, it is easy to see that for $w=(0,1,0)$ also $x_1=-x_3$ holds and $v,w$ are orthogonal.

EDIT: For point 2, in a more systematic way, you have the single equation: $$x_1+x_3=0$$ Since you have three variables and one equation, you can choose two variables as free variables or parameters, e.g. $x_3=\lambda$, $x_2=\mu$. Now your solution is $x_1=-x_3=-\lambda$, sou you can express the plane parametrically: $$ \begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix} = \begin{pmatrix} -\lambda\\\mu\\\lambda \end{pmatrix} = \lambda \begin{pmatrix} -1\\0\\1 \end{pmatrix} +\mu \begin{pmatrix} 0\\1\\0 \end{pmatrix} $$ and your two desired vectors magically appear :)

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