Minimum distance between Vector and its projection I'm struggling to get to an easy and simple algebraic solution for this question:

I actually thought about optimization with the Cauchy-Schwarz Inequality but it just got dirty and probably wrong.
But still, if you guys are willing to help out a little bit more, would there be even be a solution using calculus 1 or just linear algebra concepts? 
I'm a enthusiast freshman so I like to try places where I can apply both courses materials.
Thanks in advance, here's what I've got:

 A: A point $Q$ on the line can be written $q=q_0+t d$ where $q_0$ is one point of the line chosen arbitrarily. I omit all arrows as I'm lazy...
You have $$\Vert p- q \Vert^2=\Vert (p -q_0) -t d \Vert^2 =\Vert p- q_0 \Vert^2 -2t (p-q_0) \cdot d + t^2\Vert d \Vert^2$$ the right hand side is a polynomial of degree two which goes to $+\infty$ as $t \to \infty$. Hence it has a minimum when its derivative vanishes. That is for $t = T = \frac{(p-q_0) \cdot d}{\Vert d \Vert^2}$. You'll then verify that for that value of $t$, $q$ is the projection of $p$ on the line as $(p-q)\cdot d=0$.
A: decompose $p$ as the sum of a vector parallel to the line and one orthogonal to the line. that is $$p = t d + (p- td) $$  choose the number $t$ so that $(p-td)^\top d = 0.$ that is $t = \frac{p^\top d}{d^\top d.}$
now pick any point $q_1=t_1d$  on the line and consider $$p - q_1 = (p-td)+td-t_1d=\\(p-td)+(t-t_1)d $$ now we know that $(p-td)$ and $(t-t_1)d$are orthogonal, therefore $$|p-q_1|^2 = |p-td|^2 +|t-t_1||d|^2 \ge  |p-td|^2 \text{ for all } q_1 = t_1d$$ and the minimum is achieved at $t_1 = t.$
