So this is a part of a larger question. The question starts as follows.
Let $v \in \mathbb{R}^n$ be a nonzero vector, and denote by $v^\perp \subset \mathbb{R}^n$ the set of vectors $w \in \mathbb{R}^n$ such that $v⋅w=0$
a) Show that $v^\perp$ is a subspace of $\mathbb{R}^n$
b) Given any vector $a \in \mathbb{R}^n$, show that $(a-\frac{a⋅v}{||v||^2}v) \in v^\perp$
c) Define the projection of $a$ onto $v^\perp$ by the formula $$P_{v^\perp}(a)=(a-\frac{a⋅v}{||v||^2}v). $$ Show that there is a unique number $t(a)$ such that $(a+t(a)v) \in v^\perp$. Show that $$a+t(a)v = P_{v^\perp}(a)$$
Parts a and b were relatively straight forward because you just had to apply the definition of subspace and $v^\perp$. However, I am confused on part c.
MY ATTEMPT AT PART C
Suppose that $a+t_1(a)v \in v^\perp$ and $a+t_2(a)v \in v^\perp$. Then applying the definition of $v^\perp$, we have
$$v⋅(a+t_1(a)v) = 0 $$
$$v⋅(a+t_2(a)v) = 0$$ Then $$v⋅(a+t_1(a)v) = v⋅(a+t_2(a)v)$$ $$v⋅a+t_1(a)v⋅v = v⋅a+t_2(a)v⋅v$$ Then, $$t_1(a)||v||^2=t_2(a)||v||^2$$ Thus, $$t_1(a)= t_2(a)$$ So here I have proved uniqueness. However, now I am unsure of how to prove that $$t(a) = -\frac{a⋅v}{||v||^2}$$ which is essentially what the question is asking me to do. Do I just assume that $t(a)$ is that number? Would I say that since $t(a)$ is unique, I can choose a single number to represent $t(a)$. I am unsure of how to finish this proof.
If $a+ t_1 v , a+t_2 v\in v^\bot$ then since $v^\bot$ is a subspace we have $(t_1-t_2) v \in v^\bot$. Since $\operatorname{sp}\{v\}, v^\bot$ are orthogonal we must have $(t_1-t_2) v =0$ and so $t_1=t_2$.
To compute $t(a)$ we look for a $t$ such that $a+tv \bot v$, or, equivalently $\langle a+tv, v \rangle = 0$. Expanding gives $\langle a, v \rangle + t \|v\|^2 = 0$ or $t = - { \langle a, v \rangle \over \|v\|^2 }$.