Vtyjkyuk

So this is a part of a larger question. The question starts as follows.

Let $v in mathbbR^n$ be a nonzero vector, and denote by $v^perp subset mathbbR^n$ the set of vectors $w in mathbbR^n$ such that $v⋅w=0$

a) Show that $v^perp$ is a subspace of $mathbbR^n$

b) Given any vector $a in mathbbR^n$, show that $(a-fraca⋅v^2v) in v^perp$

c) Define the projection of $a$ onto $v^perp$ by the formula $$P_v^perp(a)=(a-fraca⋅v^2v). $$ 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 $$t(a) = -fraca⋅v^2$$ 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 $a+ t_1 v , a+t_2 vin v^bot$ then since $v^bot$ is a subspace we have
$(t_1-t_2) v in v^bot$. Since $operatornamespv, v^bot$ 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 = - ^2 $.