Describe all vectors $v = \pmatrix{x\\y}$ that are orthogonal to $u = \pmatrix{a\\b}

Question

Describe all vectors $v = \pmatrix{x\\y}$ that are orthogonal to $u = \pmatrix{a\\b}$.

I know that vectors that are orthogonal will have a dot product of 0. So here's what I was thinking: \begin{align*} ax + by &= 0\\ yb &= -ax\\ y &= -ax/b \end{align*}

I then looked up the answer to check if I was right, and the solution says:

$v$ is of the form $k\pmatrix{b\\-a}$, where $k$ is a scalar.

Can anyone help me to understand how they came up with this answer?

Answer 1

The answers are equivalent. Your vector is $\vec v=\langle x, -ax/b\rangle$, which works for all $x\in\Bbb R$.

Factor out $x$: $\vec v=x\langle 1,-a/b\rangle$ and write $x=kb$.

Notice that your deductions work only if $b\neq 0$, you should treat that case separately.

Answer 2

Note that your method implicitly assume $b \neq 0$. Here's a more conceptual way to approach that problem, which avoids that kind of assumption, and which leads directly to the answer in the text:

Hint The map ${\bf u}^{\flat} : \Bbb R^2 \to \Bbb R$ defined by $${\bf u}^{\flat} : {\bf v} \mapsto {\bf u} \cdot {\bf v}$$ is nonzero so it has rank $1$. So, its kernel, which by construction is the set of vectors orthogonal to $\bf u$, $$\ker {\bf u}^{\flat} = \{{\bf v} : {\bf u} \cdot {\bf v} = 0\} ,$$ is a subspace of $\Bbb R^2$ of dimension $1$.

Thus, this set is spanned by any nonzero element it contains. Computing gives $${\bf u} \cdot \pmatrix{b\\-a} = \pmatrix{b\\-a} \cdot \pmatrix{a\\b} = 0,$$ and so the vectors orthogonal to $\bf u$ are precisely the vectors $$k \pmatrix{b\\-a}, \qquad k \in \Bbb R .$$