What is $\{\theta\in\mathbb{R}^d,\langle v,\theta-u\rangle =0\}$? A hyperplane?

Question

What is $\{\theta\in\mathbb{R}^d,\langle v,\theta-u\rangle =0\}$? Here $\theta\in\mathbb{R}^d$ and $v$ and $u$ fixed from $\mathbb{R}^d$, and $\langle a,b\rangle$ is dot product of two vectors $a$ and $b$.

My feeling is $\left\{\theta\in\mathbb{R}^d,\langle v,\theta-u\rangle =0\right\}$ is a hyperplane in $\mathbb{R}^d$ so that $\theta-u$ is orthogonal with $v\in\mathbb{R}^d$. The degree of freedom of $\theta\in\left\{\theta\in\mathbb{R}^d,\langle v,\theta-u\rangle =0\right\}$ is $d-1$. Is this correct?

Answer 1

It is an affine hyperplane, i.e. a translate (coset) of a $1$-codimensional vector subspace.

In particular, it is $u+v^\perp$ where $v^\perp$ denotes the orthogonal complement of $v$.

Answer 2

Yes, it is a hyperplane. The equation for the plane is $$v\cdot\theta +b = 0,$$ where $b=-v\cdot u$. It indeed has $d-1$ degrees of freedom.